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1
Differential Equations
By: Patrick Bourque
Designed for students of MATH 2420 at The University of Texas at Dallas.
2
Contents
1 First Order Equations.
5
1.1
Separable Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5
1.2
First Order Linear Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
14
1.3
Bernoulli Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
23
1.4
Homogenous Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
27
1.5
Shift to Homogenous . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
31
α
1.6
The z Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
34
1.7
Equations of the form: y’=G(ax+by+c) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
36
1.8
Exact Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
40
1.9
Integrating Factors for non-exact Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
43
1.10 Orthogonal Trajectories . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
48
2 Second Order Equations.
3
4
55
2.1
Wronskian, Fundamental Sets and Able’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
55
2.2
Reduction of Order . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
62
2.3
Equations of the form y”=f(x,y’) and y”=f(y,y’) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
65
2.4
Homogenous Linear Equations with Constant Coefficients. . . . . . . . . . . . . . . . . . . . . . . . . . . .
70
2.5
The Method of Undetermined Coefficients
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
74
2.6
Variation of Parameters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
79
2.7
Cauchy Euler Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
85
2.8
Everyone Loves a Slinky: Springs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
91
2.9
Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
99
Series Solution
101
3.1
Series Solutions Around Ordinary Points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101
3.2
Method of Frobenius: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106
3.3
The Gamma Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113
3.4
Bessel’s Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115
Laplace Transform
119
4.1
Calculating Laplace and Inverse Laplace Transforms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119
4.2
Solving Initial Value Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125
4.3
Unit Step Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131
4.4
Convolution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134
3
4
CONTENTS
4.5
Delta Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140
5 First Order Systems of Differential Equations
143
5.1
Linear Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143
5.2
Locally Linear Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149
5.3
Linear Systems and the Laplace Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153
Chapter 1
First Order Equations.
1.1
Separable Equations
A Differential Equation is Separable if it can be written as:
f (x)dx = g(y)dy
The Solution is found by integrating both sides.
An Example: Solve:
ex ydx = (e2x + 1)dy
y(0) = 1
Solution:
ex
dx =
2x
e +1
Z
Z
dy
y
Using the substitution u = ex , du = ex dx on the integral on the left
Z
du
=
2
u +1
Z
dy
y
arctan(ex ) + C = ln |y|
arctan(u) + C = ln |y|
Applying our initial conditions
π
+C =0
4
arctan(1) + C = ln(1)
arctan(ex ) −
π
= ln |y|
4
Solving for y gives the solution to the differential equation:
5
C=
−π
4
6
CHAPTER 1. FIRST ORDER EQUATIONS.
x
y = earctan(e
)− π
4
♠
An Application:
A parachutist falling toward Earth is subject to two forces: the parachutist weight (w = 32m) and the drag of the
parachute. The drag of the parachute the drag is proportional to the velocity of the parachute and in this case is equal
to 8|v|. The parachutist weight is 128lb and initial velocity is zero. Find formulas for the parachutists velocity v(t) and
distance x(t).
Since x(t) increases as the parachutist falls, the downward direction is the positive direction. The force from the
parachutist’s weight acts in the positive direction while the drag from the parachute acts in the negative direction. Since
the parachutist falls down (the positive direction) velocity is always positive so |v| = v. The resultant force will be the
force of the weight of the parachutist minus the force of the drag of the parachute:
F = 128 − 8v
And the mass of the parachutist is:
128 = 32m
m=4
By Newton’s second law:
F = ma = 4
dv
dt
since
a=
dv
dt
This gives a differential equation:
4
dv
= 128 − 8v
dt
This equation is Separable:
dv
= dt
32 − 2v
Integrating
−1
ln(32 − 2v) = t + C
2
or
ln(32 − 2v) = −2t + C
Applying the initial condition v(0) = 0
C = ln(32)
Solving for v(t) gives:
v(t) = 16 − 16e−2t
The parachutist terminal velocity is given by:
1.1. SEPARABLE EQUATIONS
7
lim v(t) = lim 16 − 16e−2t = 16ft/sec
t→∞
t→∞
We can now find an equation: x(t) for how far the parachutist has fallen:
Z
x(t) =
Z
v(t)dt =
(16 − 16e−2t )dt = 16t + 8e−2t + K
Since x(0) = 0 we see that K = −8. Thus,
x(t) = 16t + 8e−2t − 8
1.
Show
y = cosh(x) =
ex + e−x
2
is a solution to
1
y2
K=
where K is the curvature
|y 00 |
K=
1+
(y 0 )2
23
2.
Find the values of r so that y = erx is a solution to
y 000 − 6y 00 + 11y 0 − 6y = 0
3.
Find the values of k so that y = sin(kx) is a solution to
y 00 + 100y = 0
4.
Find the values of n so that y = xn is a solution to
x2 y 00 + 7xy 0 + 8y = 0
5.
The Clairaut Equation is
y = xy 0 + f (y 0 )
Show y = kx + f (k) is a solution for some constant k.
Use the above result to find a solution to
8
CHAPTER 1. FIRST ORDER EQUATIONS.
y = xy 0 + (y 0 )3
6.
Show
(
y=
ex − 1
1−e
x≥0
−x
x<0
is a solution to
y 0 = |y| + 1
Remember, you must use the definition of the derivative to calculate y 0 (0).
7.
Solve the differential equation
(x2 + 1)dy = (4x + xy 2 )dx
y(0) = 2
8.
Solve the differential equation
2
(2xy + 2x)dx = e−x dy
p
y( ln(5)) = 0
9.
Solve the differential equation
p
dy
= 16x2 y − 4x2 y 2
dx
y(2) = 1
10.
Solve the differential equation
y
p
dy
2
= ln x + x − 1
dx
y(1) = e
11.
Solve the differential equation
xdy = (y 2 + 4y + 5)
p
x3 − 1dx
y(0) = −2
12.
Solve the differential equation
(x3 − x2 + x − 1)dy = (3x2 y − 2xy + y)dx
y(0) = e
13.
Solve the differential equation
√
y + xydy = arcsin(x)dx
y(0) = 1
1.1. SEPARABLE EQUATIONS
14.
9
A
Solve the differential equation
cos(2x)dy = (1 + sin(2x))(cos(x) − sin(x))(1 + y 2 )dx
y
15.
Solve the initial value problem
p
dy
1 − x2 = 2xy
dx
y(1) = 2
16.
Solve the initial value problem
4x ln(x) + 4xy ln(x) = y
dy
dx
y(e) = 0
17.
Solve the initial value problem
15e−y sin3 (x) = cos6 (x)
dy
dx
y(0) = ln(2)
18.
Solve the initial value problem
1+
√
y = (1 + sin(x))
dy
dx
y(0) = 0
19.
Solve
(ex − e−x )(y 2 + 1) = (2yex + 2ye−x )
20.
Solve
3
5
(y 2 + 1)dx = (x 4 + x 4 )dy
21.
Solve:
dy
dx
=
√
sin(x)
cos2 (x) + cos2 (x) y
22.
Solve:
x2 (y +
23.
Solve:
√
y)dx = (x4 + 2x2 + 1)dy
dy
dx
π
=0
4
10
CHAPTER 1. FIRST ORDER EQUATIONS.
dy
xex (y 2 + 4y + 5)
=
dx
x2 + 2x + 1
y(0) = −1
24.
Solve
xe3x
dy
= 2(3x + 1)2
y
dx
25.
Solve the differential equation
√
√
y+1
x−2
dx
dy = √
√
y−1
x+2
26.
Solve the differential equation
(xy + 2x + y + 2)dx = (x2 y 2 + 2xy 2 )dy
27.
Solve the differential equation
4
dy
=1
xy + xy 2 + x2 y + x2 y 2 dx
28.
Use the Second Fundamental Theorem of Calculus to verify
y = Ce−
Rx
a
g(u)du
is a solution to
y 0 + g(x)y = 0
29.
Use the substitution u = yex to transform the equation into a separable equation and then solve it
ydx + (1 + y 2 e2x )dy = 0
30.
Use the substitution y = zex to transform the equation into a separable equation and then solve it
p
dy
= y + e2x − y 2
dx
31.
Use the substitution z = y 2 + x − 1 to transform the equation into a separable equation and then solve it
2y
dy
= y2 + x − 1
dx
1.1. SEPARABLE EQUATIONS
11
Some times it is useful to convert a differential equation to polar coordinates before solving it. The conversions to
polar coordinates is:
x = r cos θ
y = r sin θ
Calculating the total differential of both x and y we get:
dx = cos θdr − r sin θdθ
dy = sin θdr + r cos θdθ
Making
dy
sin θdr + r cos θdθ
=
dx
cos θdr − r sin θdθ
Use this conversion to polar coordinates to solve the next two problems:
32.
Solve by converting to polar coordinates
x+y =x
dy
dx
33.
Solve by converting to polar coordinates
(2xy + 3y 2 )dx = (2xy + x2 )dy
34.
Solve by converting to polar coordinates
dy
y 3 + x2 y − x − y
= 3
dx
x + xy 2 − x + y
35.
Salt water containing .25 pounds of salt per gallon is being pumped into a tank initially containing 100 gallons of
water and 10 pounds of salt at a rate of 4 gallons per minute. The mixture in the tank is kept well stirred and fluid flows
out of the tank at a rate of 4 gallons per minute. Find a formula that represents the amount of salt in the tank at any
time.
36.
The logistic differential equation that models the size of a population of species in an environment of fixed size is given
by the following differential equation:
dP
= kP (M − P )
dt
where M is the carrying capacity of the environment: that is, M is the maximum population of the species that can fit
in the environment and k > 0 is a constant depending on the reproduction rate of the species. For example if P represent
the population of a bacteria in a petri dish then M would be the maximum population of the bacteria in the dish. We
also see from the differential equation if a population P is less than M then
approach the carrying capacity and if P is greater than M then
the carrying capacity.
dP
dt
dP
dt
> 0 and the population will increase and
< 0 and the population will decrease and approach
12
CHAPTER 1. FIRST ORDER EQUATIONS.
Show the population is increasing fastest when the population is half the carrying capacity then solve this differential
equation for the population P (t) as a function of time and show:
lim P (t) = M
t→∞
37.
It has been calculated that the world population cannot exceed 20 billion people. In 1970 the population was 3.7 billion
and in 2014 the population grew to 6.8 billion. Write a logistic differential equation representing the world population
and solve it for the world population as a function of time. When will the population exceed 10 billion?
38.
Another type of population model is the Gopertz growth model. It is similar to the logistic equation in that the model
assumes the population will increase at a rate proportional to the size of the population.That means the population will
increase a a rate of kP (t). The like the logistic model the Gopertz growth model also takes into account the maximum
population a species can have in an environment of fixed size
andresources. Instead of using (M − P (t)) as a factor like
the logistic model does the Gopertz growth model uses ln PM(t) as a factor, with M being the maximum population
(carrying capacity). The Gopertz growth model is
dP
= kP ln
dt
M
P
k>0
We also see from the differential equation if a population P is less than M then
increase and approach the carrying capacity and if P is greater than M then
and approach the carrying capacity.
Show the population is increasing fastest when the population is
M
e
dP
dt
dP
dt
> 0 and the population will
< 0 and the population will decrease
and then solve this differential equation for the
population P (t) as a function of time and show:
lim P (t) = M
t→∞
39.
200 fish of a particular species are introduced to a lake which can sustain no more than 3000 fish. After 2 years the fish
population had increased to 800 fish. If the population follows the Gopertz growth model how long after the introduction
of the fish to the lake will the population reach 2000 fish. Repeat the calculation using the logistic model and compare
the results.
40.
Differential equations can also be used to model the genetic change or evolution of a species. A commonly used hybrid
selection model is
dy
= ky(1 − y)(a − by)
dt
With y represents the portion of a population that has a certain characteristic and a, b, k constants and t is time
measured in generations.
At the beginning of a study of a population of a particular species it is found that half population had the advantageous
trait T and three generations later 60 percent of the population had trait T. Use the hybrid selection model with a = 2
and b = 1 to determine the number of generations it will take until more than 80 percent of the population has trait T.
1.1. SEPARABLE EQUATIONS
13
41.
Newton’s law of cooling states that an object with temperature T in a medium of constant temperature M will
experience a change in temperature proportional to the difference in the temperature of the object and the medium
(M − T ). This gives the differential equation:
dT
= k(M − T )
dt
A cup of 170◦ coffee is place in a 75◦ room. After 10 minutes the coffee is measured to have a temperature of 150◦ .
How long will it take for the coffee to cool to 120◦ ?
42.
In the study of learning it has been shown that a person’s ability to learn a task is governed by the differential equation
3
3
dy
2py 2 (1 − y) 2
√
=
dt
n
Where y represents the level that a student has learned a skill as a function of time and n and p are constants
depending on the person learning the skill and the diffuculty of learning the skill. Solve this differential equation with
the initial condition y(0) = 0 and p = 1, n = 4.
43.
Torricelli’s Law states that water draining from a tank of volume V (t) through a hole of area a in the bottom will
have an exiting velocity of
v(t) =
p
2gy(t)
where y(t) measures the height of the water level above the hole in the tank. The change in volume in the tank is
given by
p
dV
= −av(t) = −a 2gy(t)
dt
If A(y) denotes the area of the cross section of the tank at height y then for any slice of water at a height of y and
thickness dy will have volume
Z
V (y) =
y
A(y)dy
0
Using the second fundamental theorem of calculus to differentiate this integral gives
dy
dV
= A(y)
dt
dt
equating this result to the previous formula for
dV
dt
A(y)
gives Torricelli’s Law:
p
dy
= −a 2gy(t)
dt
Use the above results to find how long it takes a spherical tank with radius of 60 inches to be drained through a 1
inch hole in the bottom.
44.
Show that if y1 and y2 are solutions to
14
CHAPTER 1. FIRST ORDER EQUATIONS.
y 0 + P (x)y = Q1 (x)
and
y 0 + P (x)y = Q2 (x)
respectively then y = y1 + y2 is a solution to
y 0 + P (x)y = Q1 (x) + Q2 (x)
45.
Show that if y1 and y2 are solutions to
y 0 + P (x)y 2 = Q1 (x)
and
y 0 + P (x)y 2 = Q2 (x)
respectively then y = y1 + y2 is not a solution to
y 0 + P (x)y 2 = Q1 (x) + Q2 (x)
46.
There are about 3300 families of human languages spoken in the world. Assuming that all languages have evolved
from a single language and that one family of languages evolves into 1.5 families of language every 6000 years how long
ago was the original language spoken?
47.
For every point P(x, y) on a curve in the first quadrant, the rectangle containing the points O(0, 0) and P(x, y) as
vertices is divided by the curve into two regions: upper region A and lower region B. If the curve contains the point Q(1,
3), and region A always has twice the area of region B, find the equation of the curve.
48.
Find a function f (x) with the following properties: The average value of f on [1, x] is equal to twice the functions
value at x and f (2) = 1.
1.2
First Order Linear Equations
A Differential Equation is First Order Linear if it has the form:
dy
+ P (x)y = Q(x)
dx
To solve this equation we recognize the left hand side:
dy
dx
+ P (x)y looks close to the derivative of the product of
some function times y. Idea: multiply both sides of the equation by some function I(x) to make the left hand side the
derivative of the product of I(x) times y. Multiplying both sides by I(x) gives:
I(x)
dy
+ I(x)P (x)y = I(x)Q(x)
dx
1.2. FIRST ORDER LINEAR EQUATIONS
15
If the left hand side is the derivative of the product I(x) · y:
d
dy
I(x) · y = I(x) ·
+ I 0 (x) · y
dx
dx
Then:
I(x)
dy
dy
+ I(x)P (x) · y = I(x) ·
+ I 0 (x) · y
dx
dx
So
I(x)P (x) · y = I 0 (x) · y
I 0 (x)
= P (x)
I(x)
Integrating gives:
Z
ln |I(x)| =
P (x)dx
Solving for the Integrating Factor I(x) gives:
R
I(x) = e
P (x)dx
After multiplying both sides of the original differential equation by I(x) the left hand side is the derivative of the
product I(x) · y so the equation:
I(x)
dy
+ I(x)P (x)y = I(x)Q(x)
dx
Becomes:
d
I(x) · y = I(x)Q(x)
dx
Integrating gives:
I(x) · y
Z
=
I(x)Q(x)dx
And the solution is given by:
y=
1
I(x)
Z
I(x)Q(x)dx + C
An Example: Solve:
dy
+ sin(x)y = sec2 (x)
dx
Writing the differential equation in standard form:
cos(x)
dy
+ tan(x)y = sec3 (x)
dx
Creating the integrating factor
16
CHAPTER 1. FIRST ORDER EQUATIONS.
I=e
R
tan(x)dx
= eln(sec(x)) = sec(x)
Our Solution is:
y=
1
sec(x)
Z
sec(x) sec3 (x)dx + C
Z
Z
sec2 (x)(1 + tan2 (x))dx + C
sec4 (x)dx + C = cos(x)
y = cos(x)
Z
Z
2
2
2
y = cos(x)
sec (x)dx + tan (x) sec (x)dx + C
Using the substitution u = tan(x), du = sec2 (x)dx on the second integral
Z
Z
sec2 (x)dx + u2 du + C
y = cos(x)
u3
y = cos(x) tan(x) +
+C
3
1
3
y = cos(x) tan(x) + tan (x) + C
3
♠
Example
In this next example we will transform a nonlinear differential equation into a linear equation by converting it to polar
coordinates.
(x2 + y 2 + x)
dy
=y
dx
Let x = r cos θ, y = r sin θ
Therefore dx = cos θdr − r sin θdθ and dy = sin θdr + r cos θdθ
Under this substitution our differential equation becomes:
(r2 + r cos θ)(sin θdr + r cos θdθ) = r sin θ(cos θdr − r sin θdθ)
Multiplying things out gives
r2 sin θdr + r3 cos θdθ + r cos θ sin θdr + r2 cos2 θdθ = r sin θ cos θdr − r2 sin2 θdθ
Which reduces to
r cos θdθ + sin θdr + dθ = 0
1.2. FIRST ORDER LINEAR EQUATIONS
17
dr
+ cot θr = − csc θ
dθ
This is first order linear with integrating factor
R
I = e cot θdθ = sin θ
And the solution is...
r=
1
sin θ
Z
−
sin θ csc θdθ + C
r=
C −θ
sin θ
which reduces to
Or
r sin θ = C − θ
Converting back to rectangular coordinate system gives the solution:
y = C − arctan
y
x
49.
Solve:
xy 0 − 3y = x4
y(1) = 1
y 0 + ex y = ex
y(0) = 2e
50.
Solve:
51.
Solve:
0
y + tan(x)y = tan(x)
π
y
=1
4
52.
Solve:
p
1 − x2
dy
+y =1
dx
y(0) = 4
53.
Solve:
3
dy
6x2 − 4x + 8
ex +12x
+ 3
y
=
dx x − x2 + 4x − 4
(x − 1)2 (x2 + 4)
18
CHAPTER 1. FIRST ORDER EQUATIONS.
54.
Solve:
cos(x) − sin(x)
dy
+
y = sec3 (x)
dx cos(x) + sin(x)
y(0) = 4
55.
Solve:
(1 − x2 )
dy
− xy = 1
dx
y
√
3
1
=
2
2
56.
Solve:
dy
4x + 1
+
y = ex
dx
x
y(1) = 0
57.
Solve:
(1 + x2 )
dy
+ (4x2 − 4x + 2)y = 9 ln(x)
dx
y(1) = 0
58.
Solve:
dy
+ sin(x)y = sin(2x)
dx
59.
Solve:
dy
y
1
+
= 2x
dx 1 + e−x
e + 2xex + x2
y(0) = 0
60.
Solve:
dy
− 2xy = (2 + x−2 )
dx
y(1) = 0
61.
Solve:
(1 + x2 )
dy
− 2xy = (1)
dx
y(0) = 1
62.
Solve:
cos2 (x)y 0 + y = 1
63.
Solve:
y(0) = −3
1.2. FIRST ORDER LINEAR EQUATIONS
19
sin(x)
dy
+ cos(x)y = ln(x)
dx
64.
Solve:
(ex + e−x )
dy
+
dx
(ex + e−x )2
y=1
(ex − e−x )
65.
Solve:
(1 + x4 )
dy
− 4x3 y = (x5 + x) arctan(x2 )
dx
y(1) = π
66.
Solve:
(1 + x2 )
5
dy
+ xy = (x2 + 1) 2
dx
y(0) = 1
67.
Solve:
(1 + x2 )
dy
− 4xy = x2
dx
68.
Solve:
6x
dy
+
y=x
dx x4 + 5x2 + 4
69.
Solve:
y−x
dy
dy
= y 2 ey
dx
dx
70.
Solve:
dy
2 + tan2 (x)
+
y = cos(x)
dx
x + tan(x)
71.
Solve:
dy
+ (3x2 + 2x)y = 3x5 + 5x4 + 2x3
dx
72.
dy
1 + cos3 (x)
+
y = cos2 (x)
dx sin(x) cos(x)(1 + cos(x))
73.
Find all values of k so that the solution y approaches 0 as x approaches ∞
20
CHAPTER 1. FIRST ORDER EQUATIONS.
y0 +
k
y = x2
x
74.
Express the solution to
dy
= 1 + 2xy
dx
in terms of the error function:
2
erf(x) = √
π
Z
x
2
e−t dt
0
75.
The solution to the differential equation
dy
+ P (x)y = Q(x)
dx
is
y(x) = Ce−2x + t + 1
Find functions P (x) and Q(x).
76.
Salt water containing .25 pounds of salt per gallon is being pumped into a tank initially containing 100 gallons of
water and 10 pounds of salt at a rate of 4 gallons per minute. The mixture in the tank is kept well stirred and fluid flows
out of the tank at a rate of 2 gallons per minute. Find a formula that represents the amount of salt in the tank at any
time.
77.
At t = 0 one unit of a drug is administered to a patient who is hooked up to an IV drip supplying him with more of
the drug so that one unit of the drug is always present in his system. If the patient’s liver removes 15 percent of the drug
each hour how much of the drug must be administered by the IV each hour to keep 1 unit of the drug present?
78.
The following equation is not separable or linear. Use the substitution u = e2y to transform it into a linear equation
and solve it.
2xe2y
dy
= 3x4 + e2y
dx
79.
The following equation is not separable or linear. Use the substitution u = ey to transform it into a linear equation
and solve it.
dy
2
e−y
+ =
dx x
1 + x3
80.
The following equation is not separable or linear. Use the substitution u = tan(y) to transform it into a linear equation
and solve it.
1.2. FIRST ORDER LINEAR EQUATIONS
21
sec2 (y)
3
dy
− tan(y) = x4
dx x
81.
The following equation is not separable or linear. Use the substitution u =
1
4+y
to transform it into a linear equation
and solve it.
1
dy
2
−
= ln(x)
(4 + y)2 dx x(4 + y)
82.
The following equation is not separable or linear. Use the substitution y = eu to transform it into a linear equation
and solve it.
x
dy
− 4x2 y + 2y ln(y) = 0
dx
83.
The following equation is not separable or linear. Use the substitution z = ln(y) to transform it into a linear equation
and solve it.
x
dy
+ 2y ln(y) = 4x2 y
dx
84.
The following equation is not separable or linear. Use the substitution z = ln(y) to transform it into a linear equation.
dy
+ f (x)y = g(x)y ln(y)
dx
85.
The following equation is not separable or linear. Use the substitution u = y 2 to transform it into a linear equation
and solve it.
2xy
dy
+ 2y 2 = 3x − 6
dx
86.
The differential equation governing the velocity v of a falling object subject to air resistance is
dv
= mg − kv
k>0
v(0) = v0
dt
Solve this differential equation and determine the limiting velocity of the object.
m
87.
If A(t) represents the amount of money in an account then the change in the amount in the account is given by:
dA
= Deposits − Withdraws + Interest
dt
With a constant interest rate r the interest on the account is rA (remember A is the amount of money you will be
getting interest on). This makes the differentia equation:
dA
= Deposits − Withdraws + rA
dt
22
CHAPTER 1. FIRST ORDER EQUATIONS.
In first order linear form:
dA
− rA = Deposits − Withdraws
dt
Use this differential equation to solve the following:
A person opens an account yielding 3 percent interest is opened with an initial investment of 1000 dollars. On the
first year they deposit 100 dollars, the year month 110 dollars, the third year 120 dollars.... So each year they deposit 10
dollars more than the year before. How much will they have in the account in ten years?
88.
An equation of the form:
y 0 + P (x)y = 0
is called a first order linear homogenous (Q(x)=0) differential equation. It can be solve by separation of variables
while
y 0 + P (x)y = Q(x)
cannot. Show that if yh (x) is the solution to the homogenous equation
y 0 + P (x)y = 0
then
Z
y(x) = yh (x)
Q(x)
dx
yh (x)
is the solution to the nonhomogenous equation
y 0 + P (x)y = Q(x)
Use this technique to solve:
xy 0 + y = e4x
The Riccati Differential Equation is an equation of the form:
dy
= P (x)y 2 + Q(x)y + R(x)
dx
If u(x) is a solution to the equation then the substitution y = u +
1
v
will transform the equation into a first order
linear equation.
89.
Solve the Riccati Equation:
dy
= −8xy 2 + 4x(4x + 1)y − (8x3 + 4x2 − 1)
dx
90.
u(x) = x
is one solution
1.3. BERNOULLI EQUATION
23
Solve the Riccati Equation:
dy
y
= x3 (y − x)2 +
dx
x
u(x) = x
is one solution
91.
Show the nonlinear differential equation
(y 0 )2 + y · y 0 = x2 + xy
can be factored into
(y 0 + y + x)(y 0 − x) = 0
Set each factor to zero and solve each of the differential equations. Then show each solution you obtain is also a
solution to the original differential equation.
1.3
Bernoulli Equation
The Bernoulli Differential Equation is a nonlinear equation of the form:
After the substitution z = y 1−n
dy
+ P (x)y = Q(x)y n
dx
the Bernoulli Equation will be First Order Linear. Differentiating gives:
dy
dz
= (1 − n)y −n
dx
dx
Multiplying both sides of the Bernoulli equation by (1 − n)y −n gives:
dy
+ (1 − n)P (x)y 1−n = (1 − n)Q(x)
dx
Under the substitution this equation becomes:
(1 − n)y −n
dz
+ (1 − n)P (x)z = (1 − n)Q(x)
dx
Which is a First Order Linear differential equation.
An Example: Solve:
2
dy
+ 2xy = e3x +2x y 4
dx
Let
z = y 1−4 = y −3
So
Multiplying both sides of the differential equation by −3y −4
dz
dy
= −3y −4
dx
dx
gives:
2
dy
+ (−6x)y −3 = −3e3x +2x
dx
Under our substitution our differential equation becomes:
−3y −4
2
dz
+ (−6x)z = −3e3x +2x
dx
24
CHAPTER 1. FIRST ORDER EQUATIONS.
This is First Order Linear. Our Integrating Factor is:
I=e
R
(−6x)dx
2
= e−3x
The solution is:
z=
Z
1
2
e−3x (−3e3x
2
+2x
)dx + C
e−3x2
Z
2
2
−3 2x
z = e3x − 3 e2x dx + C = e3x
e +C
2
Back substituting gives:
2
−3 3x2 +2x
Ce3x − 3e3x
1
3x2
=
e
+
Ce
=
y3
2
2
y3 =
Ce3x2
2
− 3e3x2 +2x
So
r
y=
3
Ce3x2
2
− 3e3x2 +2x
♠
92.
Solve:
y
dy
xy 2
+ 2
=x
dx x + 1
93.
Solve:
dy
√
+ cot(x)y = cos3 (x) y
dx
94.
Solve:
x
dy
y
− = x arcsin(x)y 5
dx 2
95.
Solve:
dy
= y 4 cos(x) + y tan(x)
dx
96.
Solve:
dy
9x + 2
+ 2
y = ln(x)y −1
dx 9x + 4x
2
+2x
1.3. BERNOULLI EQUATION
25
97.
Solve:
1
dy
x
y=
+
dx 2 + 2x2
xy
98.
Solve:
dy
sec(x) tan(x)
sin3 (x) 2
−
y=
y
dx
1 + sec(x)
cos4 (x)
99.
Solve:
4 cos2 (x)
dy
+ y = 4 sin(x) cos(x)y −3
dx
100.
Solve:
1
dy
+
y = (1 − cos(2x))y 3
dx sin(2x)
101.
Solve:
dy
cos(x)
+
y = cos3 (x)y −3
dx 4 + 4 sin(x)
102.
Solve:
x
2
dy
− y = ex y 5
dx
103.
Solve:
5x
dy
+ 2
y = (x2 + 1)y 2
dx x + 1
104.
Solve:
x
dy
−y 2
+y =
dx
x
105.
Solve:
dy
+ tan(x)y = sec4 (x)y 3
dx
106.
Solve:
26
CHAPTER 1. FIRST ORDER EQUATIONS.
2
dy
cos(x)
+
y = cos3 (x)y −1
dx 1 + sin(x)
107.
Solve:
dy
cos(x) − sin(x)
+
y = (cos(x) − sin(x))y −2
dx 3 cos(x) + 3 sin(x)
108.
Solve:
dy
1
=
dx
xy + x2 y 3
109.
Use the substitution u = ey to transform the equation into a Bernoulli equation and then solve it
dy
1
+ = 6xe3y
dx x
Another differential equation that can be transformed into a first order linear differential equation is Lagrange’s Equation.
This is an equation of the form:
y = xF (y 0 ) + G(y 0 )
To simplify notation let y 0 = p and our equation becomes
y = xF (p) + G(p)
Differentiating with respect to x gives
y 0 = xF 0 (p)
dp
dp
+ F (p) + G0 (p)
dx
dx
Or
p = xF 0 (p)
Solving for
dp
dp
+ F (p) + G0 (p)
dx
dx
dx
dp
p − F (p) = (xF 0 (p) + G0 (p))
dp
dx
dp
p − F (p)
=
dx
xF 0 (p) + G0 (p)
F 0 (p)
G0 (p)
dx
=
x+
dp
p − F (p)
p − F (p)
In standard form
dx
+
dp
This is now a first order linear equation
F 0 (p)
G0 (p)
x=
F (p) − p
p − F (p)
1.4. HOMOGENOUS EQUATION
27
110.
Solve the Lagrange Equation
y = 2xy 0 + 4(y 0 )3
You may leave your solution in parametric form with p the parameter.
111.
Solve the Lagrange Equation
y = 2xy 0 − 9(y 0 )2
You may leave your solution in parametric form with p the parameter.
1.4
Homogenous Equation
A differential equation is Homogenous if it has the form:
dy
y
=f
dx
x
The substitution z =
y
x
or y = xz will transform the Homogenous equation into a Separable equation.
dy
dz
=z+x
dx
dx
Under this substitution the Homogenous equation becomes:
z+x
dz
= f (z)
dx
This reduces to the Separable equation
dz
dx
=
f (z) − z
x
An Example: Solve
y 3 + 2x2 y
dy
=
dx
xy 2 + x3
Solution:
Multiplying the numerator and denominator on the right hand side by
dy
=
dx
Let z =
y
x
gives:
y3
y
x3 + 2 x
y2
x2 + 1
so y = xz
dy
dz
=z+x
dx
dx
Under this substitution our differential equation becomes:
z+x
1
x3
dz
z 3 + 2z
= 2
dx
z +1
28
CHAPTER 1. FIRST ORDER EQUATIONS.
x
z
dz
= 2
dx
z +1
z+1
dx
dz =
z
x
Z 1
z+
z
Z
=
dx
x
z2
+ ln(z) = ln(x) + C
2
1 y2
+ ln
2 x2
y
= ln(x) + C
x
Since we cannot solve for y this implicit solution is our final answer.
♠
Notice the sum of the exponents in both terms in the numerator and both terms in the denominator is 3. Whenever
the sum of the exponents in each term in the problem is the same constant you should consider using the homogenous
substitution to solve the differential equation. The first practice problem should help reinforce this idea.
112.
Solve:
y 4 + x2 y 2 + x4
dy
=
dx
x3 y
113.
Solve:
2y 5 + x2 y 3 + 3x4 y
dy
=
dx
2xy 4 + 3x5
114.
Solve:
dy
13y 6 + 18x2 y 4 + 3x5 y
=
dx
12xy 5 + 16x3 y 3 + 2x6
115.
Solve:
y
dy
= +
dx
x
r
y
+1
x
116.
Solve:
dy
6y 3 − 5xy 2 + 2x2 y − x3
=
dx
5xy 2 − 4x2 y + x3
117.
Solve:
1.4. HOMOGENOUS EQUATION
29
y 4 + xy 3 + 2x2 y 2 + x4
dy
=
dx
xy 3
118.
Solve:
dy
=
dx
r
32
y
y
y
+ +
x
x
x
119.
Solve:
dy
=
dx
y
x
y 2 ln
+ x2
y
x
xy ln
120.
Solve:
x sin
y
y dy
= y sin
+x
x dx
x
121.
Solve:
dy
y
= + tan4
dx
x
y
x
122.
Solve:
dy
=
dx
√
√ x+ y 2
√
x
123.
Solve:
y
y
dx + x arctan
dy = 0
x − y arctan
x
x
124.
Solve:
2
dy
=
dx
2
y tan
y
x
xy tan2
+ x2
y
x
125.
Solve:
dy
2y 2 + 2xy + 2x2
=
dx
xy
30
CHAPTER 1. FIRST ORDER EQUATIONS.
126.
Solve:
y 3 + 2xy 2 + x2 y + x3
dy
=
dx
x(x + y)2
127.
Solve:
y(ln(y) − ln(x) + 1)
dy
=
dx
x
128.
Solve:
x
p
dy
= y + x2 + y 2
dx
y(1) = 0
129.
Solve:
dy
= csc2
dx
y
y
+
x
x
130.
Solve:
x 2
2xye( y )
dy
=
x 2
x 2
dx
y 2 + y 2 e( y ) + 2x2 e( y )
131.
Solve:
y
dy
=
√
dx
x + xy
132.
Solve:
√
√
√
√
( x + y + x − y)dx + ( x − y − x + y)dx = 0
y(1) = 1
133.
Solve:
(2(y − 1)(x + y − 1) + (y − 1)2 )dx + (4(y − 1)(x + y − 1) + (x + y − 1)2 )dy = 0
134.
Solve:
x
e y (y − x)
x
dy
+ y(1 + e y ) = 0
dx
135.
Use the substitution z =
y
x
to solve the following differential equation
1.5. SHIFT TO HOMOGENOUS
31
xdy = (x2 + y 2 + y)dx
136.
Use the substitution z =
y
x2
to solve the following differential equation
dy
2y
=
+ cos
dx
x
y
x2
137.
Use the substitution z =
y
x2
to solve the following differential equation
y2
dy
2y
=x 1+ 2 + 4
dx
x
x
138.
Use the substitution u = x2 + y 2 v = xy to solve the following differential equation
(x2 + y 2 )(xdy + ydx) − xy(xdx + ydy) = 0
139.
Use the substitution u = x3 v = y 2 to solve the following differential equation
3x5 − y(y 2 − x3 )
dy
=0
dx
140.
Use the substitution u = xy to solve the following differential equation
xdy + ydx = x2 y 2 dx
1.5
Shift to Homogenous
A differential equation can shifted to produce a Homogenous differential equation if it is of the form:
dy
a1 x + b1 y + c1
=
dx
a2 x + b2 y + c2
If the constant terms c1 and c2 are both zero then the above equation is already Homogenous. This provides the
motivation of shifting x and y by a constant so that the resulting equation has constant terms c1 and c2 both equal to
zero. To shift this into a Homogenous differential equation use the substitutions:
x = x + d1
With d1 and d2 chosen so that the two equations:
y = y + d2
32
CHAPTER 1. FIRST ORDER EQUATIONS.
a1 x + b1 y + c1 = 0
a2 x + b2 y + c2 = 0
reduce to:
a1 x + b1 y = 0
a 2 x + b2 y = 0
The idea is motivated by the fact that if the constants c1 and c2 are both zero then:
dy
a1 x + b1 y + c1
=
dx
a2 x + b2 y + c2
is Homogenous.
So choose d1 and d2 to be the solutions to:
a1 d1 + b1 d2 + c1 = 0
a2 d1 + b2 d2 + c2 = 0
An Example: Solve:
x − 2y − 2
dy
=
dx
2x + y + 6
x, y > −2
Solving system of equations:
x − 2y − 2 = 0
x = 2 + 2y
2(2 + 2y) + y = −6
2x + y + 6 = 0
5y = −10
We will use the substitution x = x − 2 and y = y − 2 making
dy
dx
=
y = −2
dy
dx
Our differential equation becomes:
dy
x − 2y
=
dx
2x + y
Multiplying the numerator and denominator on the right hand side by
1
x
gives:
1 − 2 xy
dy
=
dx
2 + xy
y
y = zx
x
Under this substitution our differential equation becomes:
Letz =
z+x
x
dy
dz
=z+x
dx
dx
dz
1 − 2z
=
dx
2+z
dz
−z 2 − 4z + 1
=
dx
z+2
z+2
dx
dz =
−z 2 − 4z + 1
x
Integrating gives:
x = −2
1.5. SHIFT TO HOMOGENOUS
33
−1
ln | − z 2 − 4z + 1| = ln |x| + C
2
Multiplying by −2 and negating the contents inside the absolute value bars on the left gives
ln |z 2 + 4z − 1| = ln |x|−2 + C
z 2 + 4z − 1 =
C
x2
Completing the square on the left gives
C
+5
x2
(z + 2)2 =
(z + 2)2 =
C + 5x2
x2
Extracting a square root gives
p
|z + 2| =
C + 5x2
|x|
Since both x and y are greater than -2, x > 0 and z + 2 > 0. So we get
p
z+2=
C + 5x2
x
p
C + 5x2
2x
z=
−
x
x
y
=
x
p
C + 5x2
2x
−
x
x
y=
p
C + 5x2 − 2x
y+2=
p
C + 5(x + 2)2 − 2(x + 2)
and our final solution is
y=
p
C + 5(x + 2)2 − 2x − 6
The same differential equation can be solved using the substitution u = x − 2y − 2 and v = 2x + y + 6 but the solution
is a bit longer (try it and see which solution you prefer).
141.
Solve:
dy
−8x + 3y + 2
=
dx
−9x + 5y − 1
142.
Solve:
34
CHAPTER 1. FIRST ORDER EQUATIONS.
dy
x−y−3
=
dx
x+y−1
143.
Solve:
dy
x + 3y + 4
=
dx
4y − 3x + 1
144.
Solve:
dy
x−y+1
=
dx
x+y
145.
Solve:
x+y+3
dy
=
dx
2y − x + 3
146.
Solve:
dy
2x + y
=
dx
−y − x + 1
147.
Solve:
dy
x+y+1
=
dx
x−y+3
148.
Solve:
dy
3 − 2x − y
=
dx
x+y−1
149.
Solve:
dy
5x + 4y − 3
=
dx
6x − y + 8
150.
Solve:
dy
1 (x + y − 1)2
=
dx
2 (x + 2)2
1.6. THE Z α SUBSTITUTION
1.6
35
The z α Substitution
In the previous section we learned how to shift an equation into a homogenous equation but our shifting method only
worked for equations of the form:
dy
a1 x + b1 y + c1
=
dx
a2 x + b2 y + c2
where the exponents of the x and y term were 1. Now we will study a substitution that works when the exponents
are not 1.
An Example: Solve:
(x2 y 2 − 2)dy + (xy 3 )dx = 0
This differential equation is almost homogenous: the sum of the exponents in x2 y 2 and xy 3 are both 4. In fact, if the
−2 was not involved in the equation it would be homogenous. This is a good candidate for a z α substitution.
y = zα
dy = αz α−1 dz
Under this substitution the differential equation becomes:
(x2 z 2α − 2)αz α−1 dz + xz 3α dx = 0
α(x2 z 3α−1 − 2z α−1 )dz + xz 3α dx = 0
We will now choose α so that the exponents on each term sum to the same value. So we choose α so that:
3α + 1 = α − 1
α = −1
If α = −1 our equation becomes:
−(x2 z −4 − 2z −2 )dz + xz −3 dx = 0
−(x2 z −4 − 2z −2 )
dz
+ xz −3 = 0
dx
dz
xz −3
= 2 −4
dx
x z − 2z −2
xz
dz
= 2
dx
x − 2z 2
Dividing both numerator and denominator by x2 gives
z
dz
x
=
2
dx
1 − 2 xz 2
This is now a homogenous equation so we make the following substitution:
u=
z
x
z = ux
dz
du
=u+x
dx
dx
36
CHAPTER 1. FIRST ORDER EQUATIONS.
Under this substitution our equation becomes:
u+x
du
u
=
dx
1 − 2u2
This is now a separable differential equation
x
du
2u3
=
dx
1 − 2u2
1 − 2u2
dx
du =
3
2u
x
Integrating gives
−1
− ln|u| = ln |x| + C
4u2
Since u =
z
x
and z = y −1 u =
1
xy
the solution becomes
1
−x2 y 2
− ln | | = ln |x| + C
4
xy
Or
−x2 y 2
+ ln |xy| = ln |x| + C
4
151.
Solve:
(2x2 y − 1)dy + 2xy 2 dx = 0
152.
Solve:
(x2 y + 1)dy + xy 2 dx = 0
153.
Solve:
(xy 3 + 1)dy + y 4 dx = 0
154.
Solve:
(x2 y 4 + 4)dy + xy 5 dx = 0
155.
Solve:
(xy + 2)dy − y 2 dx = 0
1.7. EQUATIONS OF THE FORM: Y’=G(AX+BY+C)
1.7
37
Equations of the form: y’=G(ax+by+c)
If a differential equation is of the form:
dy
= G(ax + by + c)
dx
Then use the following substitution to transform the equation into a Separable Differential Equation:
1 dz
dy
=
−a
dx
b dx
z = ax + by + c
Under this substitution our differential equation becomes
1 dz
− a = G(z)
b dx
which reduces to the Separable Differential Equation:
dz
= dx
a + bG(z)
An Example: Solve:
dy
=
dx
Let
x+y
1 + ln(x + y) − 1
dy
dz
=
−1
dx
dx
z =x+y
Under this substitution our differential equation becomes:
dz
− 1 = z(1 + ln(z)) − 1
dx
Z
dz
=
z(1 + ln(z))
Z
dx
ln(1 + ln(z)) = x + C
1 + ln(x + y) = Cex
ln(x + y) = Cex − 1
x + y = eCe
x
−1
And our final solution is:
y = eCe
156.
Solve:
x
−1
−x
38
CHAPTER 1. FIRST ORDER EQUATIONS.
dy
= (x + y − 4)2
dx
157.
Solve:
3
dy
= (2x + 3y − 1) + 4(2x + 3y − 1)−3 − 2
dx
158.
Solve:
2
dy
1
−1
=
dx
(x + 2y + 1)e(x+2y+1)2
159.
Solve:
dy
= tan2 (x + y)
dx
160.
Solve:
dy
= sin2 (y − x)
dx
161.
Solve:
dy
= csc2 (4x + y + 1) − 4
dx
162.
Solve:
dy
= sin(x + y)
dx
163.
Solve:
dy
1
=√
dx
x+y
164.
Solve:
p
dy
= e2x+2y − 1 − 1
dx
165.
Solve:
2
166.
dy
= sec(2y − 4x + 1) + tan(2y − 4x + 1) + 4
dx
1.7. EQUATIONS OF THE FORM: Y’=G(AX+BY+C)
39
Solve:
p
x − y + 1 + (x − y)2
dy
p
=
dx
1 + (x − y)2
167.
Solve:
dy
4(x − y) ln(x − y) − 1
=
dx
4(x − y) ln(x − y)
168.
Solve:
p
dy
= 1 + e2y−2x − 1
dx
169.
Solve:
dy
2e−x−y
= x+y
dx
e
− e−x−y
170.
Solve:
3
dy
= (4 + (4x + y)2 ) 2 − 4
dx
171.
Solve:
e
−y
dy
+1
dx
= xex
172.
Solve:
dy
=
dx
3
cos (x + y) − 1
cos (x + y) + 1
3
173.
Solve:
dy
= sin(2x + 2y) − sin2 (x + y)
dx
174.
Use the substitution z = y + x to solve
dy
2y
+
+ 3 = x2 (x + y)3
dx
x
175.
Use the substitution z = y + x to solve
dy
2x + y
4x
+
=
dx
x
x+y
40
CHAPTER 1. FIRST ORDER EQUATIONS.
176.
Use the substitution z = y − x to solve
dy
+ x(y − x) + x3 (y − x)2 = 1
dx
1.8
Exact Equations
In Multivariable we analyzed the functions of the form z = f (x, y) by studying their level sets. The level sets of this three
dimensional function can be graphed in two dimensional space by replacing z with different constants and analyzing the
resulting two dimensional graphs. The total differential of:
f (x, y) = C
is
fx dx + fy dy = 0
We also know that mixed partials are equal meaning:
fxy = fyx
for all
f (x, y) = C
We will now study the Exact Differential Equation. An Exact Differential Equation is an equation that was created by
calculating the total differential of some function of two or more variable set equal to a constant. In calculus we learned
to go from
f (x, y) = C
to
fx dx + fy dy = 0
In differential equations we will learn to go from
fx dx + fy dy = 0
to
f (x, y) = C
The equation
M dx + N dy = 0
is Exact if
M y = Nx
If an equation is exact then M is the partial derivative of some function f (x, y) = C with respect to x and N is the
partial derivative of the same function with respect to y. So to find f up to a constant we need to integrate M with
respect to x (obtaining a constant which will be a function g(y)) or we need to integrate N with respect to y (obtaining
a constant which will be a function g(x)). That is:
Z
f (x, y) =
M dx + g(y)
We then solve for g(y) using the fact that fy = N . That is:
∂
fy =
∂y
Z
M dx + g 0 (y) = N
And the solve for g(y) and insert it into our formula for f (x, y) and write the final answer as a level set f (x, y) = C
An Example: Solve:
1.8. EXACT EQUATIONS
41
(8xy 3 + 2xy + 3x2 )dx + (12x2 y 2 + x2 + 4y 3 )dy = 0
Solution:
M = 8xy 3 + 2xy + 3x2
N = 12x2 y 2 + x2 + 4y 3
Test for exactness:
My = 24xy 2 + 2x
Nx = 24xy 2 + 2x
Since My = Nx our differential equation is exact, making M = 8xy 3 + 2xy + 3x2 the partial derivative of the function
F we are solving for with respect to x. That is:
Fx = 8xy 3 + 2xy + 3x2
So
Z
F =
(8xy 3 + 2xy + 3x2 )dx = 4x2 y 3 + x2 y + x3 + g(y)
Since this equation is exact Fy = N . This creates the equation:
Fy = 12x2 y 2 + x2 + g 0 (y) = 12x2 y 2 + x2 + 4y 3
So
g 0 (y) = 4y 3
g(y) = y 4
Our final solution is:
4x2 y 3 + x2 y + x3 + y 4 = C
177.
Solve:
6
dy = 0
6xy + 6y 4 − 24x2 dx + 3x2 + 24xy 3 +
y
178.
Solve:
−y
x
dx
+
dy = 0
x2 + y 2
x2 + y 2
y(1) = 1
179.
Solve:
180.
Solve:
y cos(y) + y sin(x) + xy cos(x) dx + x sin(x) + x cos(y) − xy sin(y) dy = 0
42
CHAPTER 1. FIRST ORDER EQUATIONS.
sin(y) + y sin(x) + 2e
dx + x cos(y) − cos(x) + sin(y) dy = 0
2x
181.
Solve:
(2xexy + x2 yexy )dx + (x3 exy )dy = 0
y(1) = ln(4)
182.
Solve:
3x2 sin(y) + y cos(y) dx + x3 cos(y) + x cos(y) − xy sin(y) dy = 0
183.
Solve:
2
2
xy cos(xy) + 1 dx + x y cos(xy) + x sin(xy) dy = 0
184.
Solve:
xy
2 xy
xy(e + 1) + y dx + x e + x dy = 0
185.
Solve:
xex − ex + y dx + 1 + ln(y) + x dy = 0
186.
Solve:
x(x2 − y 2 )
y(y 2 − x2 )
dx +
dy = 0
(x2 + y 2 )2
(x2 + y 2 )2
187.
Find the function M (x, y) that makes the following differential equation exact.
1
xy
M (x, y)dx + xe + 2xy +
dy = 0
x
188.
The following differential equation arises from the total differential of a function F with variables: x, y and z set equal
to a constant. Find this function F (x, y, z)
(2yz 2 + 6xz)dx + (2xz 2 + 30y 2 z)dy + (4xyz + 3x2 + 10y 3 )dz = 0
189.
The following differential equation arises from the total differential of a function F with variables: x, y and z set equal
to a constant. Find this function F (x, y, z)
(yz + 1)dx + (xz + 2y)dy + (xy + 3z 2 )dz = 0
1.9. INTEGRATING FACTORS FOR NON-EXACT EQUATIONS
1.9
43
Integrating Factors for non-exact Equations
If a differential equation is not exact: My 6= Nx sometimes we can multiply both sides of the equation by an integrating
factor I to make it exact. We need a formula for this integrating factor. Starting with:
M dx + N dy = 0
For this to be exact
∂
∂y (IM )
=
and multiplying both sides by I gives
∂
∂x (IN ).
IM dx + IN dy = 0
Calculating these partials gives:
Iy M + My I = Ix N + Nx I
This is a partial differential equation that we cannot solve. So we will solve a special case: the case where I is a
function of one variable. Case 1: I is a function of x making Iy = 0. Now our partial differential equation is a bit easier
to solve:
M y I = I x N + Nx I
Or
I(My − Nx ) = Ix N
Which becomes:
M y − Nx
Ix
=
I
N
Integrating both sides with respect to x gives:
Z ln |I| =
M y − Nx
dx
N
Making our integrating factor:
My −Nx
N
R
dx
I(x) = e
Case 2: I is a function of y making Ix = 0. In this case the integrating factor is:
R
Nx −My
M
dy
I(y) = e
An Example: Solve:
(2y 7 + y 4 )dx + (6xy 6 − 3)dy = 0
M = 2y 7 + y 4
N = 6xy 6 − 3
Test for exactness:
My = 14y 6 + 4y 3
Nx = 6y 6
44
CHAPTER 1. FIRST ORDER EQUATIONS.
Since My 6= Nx our equation is not exact. We will look for an integrating factor. Since
−4y 3 (2y 3 + 1)
6y 6 − 14y 6 − 4y 3
−4
Nx − M y
=
=
=
7
4
4
3
M
2y + y
y (2y + 1)
y
is a function of only y our integrating factor is:
I=e
R
Nx −My
M
dy
=e
R
−4
y
= y −4
Multiplying both sides of the differential equation by y −4 gives:
(2y 3 + 1)dx + (6xy 2 − 3y −4 )dy = 0
M = 2y 3 + 1
N = 6xy 2 − 3y −4
Now
My = 6y 2 = Nx
Since My = Nx our differential equation is exact, making M = 2y 3 + 1 the partial derivative of the function F we are
solving for with respect to x. That is:
Fx = 2y 3 + 1
So
Z
F =
(2y 3 + 1)dx = 2xy 3 + x + g(y)
And Fy = N
Fy = 6xy 2 + g 0 (y) = 6xy 2 − 3y −4
g 0 (y) = −3y −4
g(y) = y −3
And our final solution is:
F (x, y) = 2xy 3 + x + y −3 = C
190.
Solve:
2 3
3 2
3y + 5x y + 4x dx + x + 3x y dy = 0
191.
Solve:
192.
Solve:
2x4 + x3 yexy − 2 dx + 2x3 y + x4 exy dy = 0
1.9. INTEGRATING FACTORS FOR NON-EXACT EQUATIONS
45
(2x2 + y)dx + (x2 y − x)dy = 0
y(1) = 1
193.
Solve:
(16x2 y 2 + 64x4 + 1)dx + (4xy 3 + 16x3 y) = 0
y(1) = 1
194.
Solve:
2
2
(xey )dx + (yey )dy = 0
y(0) = 0
195.
Solve:
(2xy 7 + y)dx + (3x2 y 6 − 3x)dy = 0
y(1) = 1
196.
Solve:
2x
y
2x
y
2e + e dx + 3e + 4xe dy = 0
197.
Solve:
6xy
3+
dx + 3dy = 0
1 + x2
198.
Solve:
2
2
2
2
ex (2x2 + 1) + e−y (2y 3 + 1) dx + 2xyex + e−y (6xy 2 + 1) dy = 0
199.
Solve:
3xydx + (x2 + 1)dy = 0
Sometimes we cannot find an integrating factor of just a single variable using the above formulas. In this case we
guess the form of the integrating factor and try to find constants that make it work.
200.
Solve by finding an integrating factor of the form I = xn y m :
3y 4 + 18y −1 dx + 5xy 3 + 2x−2 dy = 0
201.
Solve by finding an integrating factor of the form I = xn y m :
46
CHAPTER 1. FIRST ORDER EQUATIONS.
xy
y3
dx +
dy = 0
2
x(x + y)
(x + y)2
202.
Solve by finding an integrating factor of the form I = xn y m :
2 cos(x2 y 2 )(ydx + xdy) = 0
203.
If I(x) = x is an integrating factor for
f (x)
dy
+ x2 + y = 0
dx
Find all functions f (x)
204.
Solve by finding an integrating factor of the form I = ekx cos(y):
dy
= tan(y) − ex sec(y)
dx
205.
Solve by finding an integrating factor of the form I = sinn (x) cosm (y):
4 cos(x) + 3 cot(x) dx + − 2 sin(x) tan(y) − 3 tan(y) dy = 0
206.
Solve:
Show that I =
1
x2 +y 2
is an integrating factor for
2
2
y + xf (x + y ) dx + yf (x + y ) − x dy = 0
2
2
Use this result to solve
2
2 2
y + x(x + y )
2
2 2
dx + y(x + y ) − x dy = 0
207.
Show that
I=
1
Ax2 + Bxy + Cy 2
is an integrating factor for
xdy − ydx = 0
208.
Show that
G(x, y) = ln(x + y) −
1
x+y
1.9. INTEGRATING FACTORS FOR NON-EXACT EQUATIONS
is a solution to the exact equation
1
1
1
1
dx +
dy = 0
+
+
x + y (x + y)2
x + y (x + y)2
Now multiply both sides of this exact equation by (x + y)2 producing
(x + y + 1)dx + (x + y + 1)dy = 0
which has a solution
F (x, y) =
1 2 1 2
x + y + xy + x + y
2
2
What is the relationship between F and G
Now show in general that if F (x, y) = C is a solution to
Fx dx + Fy dy = 0
and G(x, y) = C is a solution to
I(x, y)Fx dx + I(x, y)Fy dy = 0
then
Fx Gy = Gx Fy
209.
In the study of first order linear differential equations:
dy
+ P (x)y = Q(x)
dx
we learned that multiplying by the integrating factor:
R
I(x) = e
P (x)dx
will transform the equation into a separable differential equation. Show that by multiplying both sides of
P (x)y − Q(x) dx + dy = 0
by the same integrating factor will transform the above equation into an exact equation.
Bernoulli’s Spread of Smallpox:
Let x(t) represent the population of all living people.
Let y(t) represent the population of all living people who have not contracted smallpox.
Let a > 0 be the rate at which population y(t) contracts smallpox.
Therefore
dy
= −ay
dt
Let 0 < b < 1 be the precentage of population y(t) that get and die from smallpox.
47
48
CHAPTER 1. FIRST ORDER EQUATIONS.
Therefore
dx
= −aby
dt
Let d(t) be the average death rate of populations x(t) and y(t) from causes other than smallpox.
Without smallpox we have
dx
= −d(t)x(t)
dt
dy
= −d(t)y(t)
dt
With smallpox we have
dx
= −d(t)x(t) − aby(t)
dt
dy
= −d(t)y(t) − ay(t)
dt
To eliminate the average death rate d(t) we multiply
y
dx
dt
by y(t) and
dy
dt
by x(t) and subtract
dx
dy
−x
= axy − aby 2
dt
dt
Recognizing the left hand is the numerator of
d x
dt y
we now divide both sides by y 2
dy
y dx
x
dt − x dt
= a − ab
2
y
y
This equation is now
x
d x
= a − ab
dt y
y
The substitution z =
x
y
yields the seperable equation
dz
= az − ab
dt
dz
= adt
z−b
ln(z − b) = at + C
z − b = Ceat
y = x(Ceat + b)
1.10
Orthogonal Trajectories
A common geometric problem in many applications involves finding a family of curves: Orthogonal Trajectories, that
intersect a given family of curves orthogonally at each point. If you are given a family of curves in the form F (x, y) = K
then the slope of this family of curves is given by the derivitive:
dy
Fx
=−
dx
Fy
1.10. ORTHOGONAL TRAJECTORIES
49
Since we are looking for an orthogonal family of curves, and in R2 orthogonal lines have negative reciprocal slopes,
we will be for a family of curves that satisfy the following differential equation:
Fy
Fx
y0 =
An Example:
Find the orthogonal trajectories for the circle:
x2 + y 2 = r 2
To find the orthogonal trajectories we take F (x, y) = x2 + y 2 we will need to solve the differential equation:
y0 =
y
2y
=
2x
x
This equation is separable
dx
dy
=
y
x
So
ln(y) = ln(x) + C
The family of curves orthogonal to the circle is:
y = kx
Conversely, the family of curves orthogonal to the lines y = kx is given by the circle x2 + y 2 = r2 .
An Example: Find the orthogonal trajectories for the circle:
x2 + y 2 = Cx
Here we will take F (x, y) = x2 + y 2 − Cx and we are looking for a family of curves that satisfy the following:
y0 =
2y
2x − C
The problem with this differential equation is that it involves the constant C. To eliminate this constant we will solve
for C in the original equation of the circle and insert it into the differential equation.
C=
x2 + y 2
x
Our differential equation becomes:
y0 =
2xy
− y2
x2
50
CHAPTER 1. FIRST ORDER EQUATIONS.
Recognizing that the sum of the exponents in each term in both the numerator and denominator add to the same
value of 2 we see that this is a homogenous equation. Remember, to solve the homogenous equation you must first write
it as:
dy
y
=f
dx
x
and make the substitution
z=
Multiplying both the numerator and denominator by
1
x2
y
x
we get the homogenous equation:
2 xy
dy
=
dx
1 − ( xy )2
Making the substitutions
z=
dz
dy
=x
+z
dx
dx
y
x
Our differential equation becomes:
x
dz
2z
+z =
dx
1 − z2
This equation is now separable:
x
dz
z3 + z
=
dx
1 − z2
1 − z2
dz =
z(z 2 + 1)
Z
Z
dx
x
The integral on the left requires partial fraction decomposition. After applying partial fraction we get:
Z Z
1
2z
dx
−
dz =
z
1 + z2
x
Integrating
ln(z) − ln(1 + z 2 ) = ln(x) + C
ln
z
1 + z2
= ln(x) + C
z
= Cx
1 + z2
z = Cx + Cxz 2
Expressing this quadratic with the coefficient of z 2 being 1 gives
z2 −
1
z+1=0
Cx
1.10. ORTHOGONAL TRAJECTORIES
51
To solve for z in this equation we must compleat the square by adding
z2 −
1
4C 2 x2
to both sides:
1
1
1
+1=
z+
Cx
4C 2 x2
4C 2 x2
Factoring the first 3 terms on the left we get
z−
1
2Cx
2
1
z−
=±
2Cx
1
z=
±
2Cx
Solving for y remembering that z =
y
x
1
−1
4C 2 x2
=
r
√
1 − 4C 2 x2
4C 2 x2
1 − 4C 2 x2
2Cx
gives:
y=x
1±
√
1 − 4C 2 x2
2Cx
The orthogonal family of curves we desire is:
y=
1±
√
1 − 4C 2 x2
2C
Sometimes we are concerned with finding a family of curves that make an angle of α 6= 90◦ with a given family of
curves. These curves are called Oblique Trajectories.
Given a family of curves: F (x, y) = K with its derivative (slope) given by:
dy
= f (x, y)
dx
Treating dy as the change in y and dx as the change in x and creating a triangle gives:
p
(dx)2 + (dy)2
dy
θ
dx
From the triangle we see, tan(θ) =
dy
dx
= f (x, y) so the tangent line has an angle of inclination of arctan(f (x, y)). So
the tangent line of an oblique trajectory that intersects this curve at an angle of α will have an angle of inclination of:
arctan(f (x, y)) + α
Making the slope of the oblique trajectory
tan
arctan(f (x, y)) + α
=
f (x, y) + tan(α)
1 − f (x, y) tan(α)
52
CHAPTER 1. FIRST ORDER EQUATIONS.
Thus the differential equation of this family of oblique trajectories is given by:
y0 =
f (x, y) + tan(α)
1 − f (x, y) tan(α)
An Example:
Find the family of oblique trajectories that intersect the family of straight lines y = Cx at an angle of 45◦ .
Here we take F (x, y) = y − Cx and compute the slope of the tangent line:
dy
Fx
C
=−
=
=C
dx
Fy
1
Solving for C we get
dy
y
= = f (x, y)
dx
x
Using this function for f (x, y) and α = 45◦ the differential equation of this family of oblique trajectories is
y0 =
y
x
+ tan(45◦ )
x+y
=
1 − xy tan(45◦ )
x−y
1
x
Multiplying both numerator and denominator by
produces
1+
dy
=
dx
1−
y
x
y
x
This is a homogenous equation so we make the following substitution
z=
dy
dz
=z+x
dx
dx
y
x
The differential equation becomes
z+x
dz
1+z
=
dx
1−z
This equation is now separable
dx
1−z
dz =
2
1+z
x
Integrating gives
1
ln(1 + z 2 ) = ln(x) + C
2
Treating C as ln(K) and using some properties of logs we get
arctan(z) −
2 arctan(z) = ln(K 2 x2 (1 + z 2 )
Converting back to x and y we get the family of oblique trajectories
y
2 arctan
= ln(K 2 (x2 + y 2 ))
x
1.10. ORTHOGONAL TRAJECTORIES
53
210.
Find the orthogonal trajectories for the family of straight lines.
y = mx + 1
211.
Find the orthogonal trajectories for each given family of curves.
y = Cx3
212.
Find the orthogonal trajectories for each given family of curves.
x2 + y 2 = Cx3
213.
Find the orthogonal trajectories for each given family of curves.
y = Ce2x
214.
Find the orthogonal trajectories for each given family of curves.
y = x − 1 + Ce−x
215.
Find the orthogonal trajectories for each given family of curves.
x − y = Cx2
216.
Find the value of n so that the curves xn + y n = C are the orthogonal trajectories of
y=
x
1 + Kx
217.
Show that the following family of curves is self orthogonal.
y 2 = 4C(x + C)
218.
Show that the following family of curves is self orthogonal.
y 2 = 2Cx + C 2
219.
Find a family of oblique trajectories that intersect the family of circles x2 + y 2 = r2 at an angle of 45◦
220.
54
CHAPTER 1. FIRST ORDER EQUATIONS.
Find a family of oblique trajectories that intersect the family of circles y 2 = Cx2 at an angle of 60◦
221.
Let O be the origin and P be a point on the curve P (x). Let N be the point on the x-axis where the normal line to
P (x) intersects the x-axis. If OP = ON what is the equation of P (x)?
Chapter 2
Second Order Equations.
2.1
Wronskian, Fundamental Sets and Able’s Theorem
In this section we will mostly be dealing with the second order linear differential equation:
y 00 + P (x)y 0 + Q(x)y = 0
If we want to find all solutions to this equation it can be shown that we are looking for two solutions y1 and y2 to
the equations with the one restriction that y1 and y2 cannot be scalar multiples of each other. But in order to expand
our knowledge to third order and higher order equations we replace the restriction that the solutions cannot be scalar
multiples of each other with the restriction that the set of solutions must be linearly independent.
The Set of functions{y1 , y2 , ..., yn }is Linearly Independent if the only solution to
C1 y1 + C2 y2 + ... + Cn yn = 0 is C1 = C2 = ... = Cn = 0
Although this is the formal definition of Linearly Independent sets we will not be using it. Instead we will be using
the Wronskian to determine if a set is linearly independent. Since we are considering only second order equations in this
chapter we will limit our study to the linear independence or dependence of two functions y1 and y2
The Wronskian of two functions y1 and y2 is given by the determinate:
y
1
W (y1 , y2 ) = 0
y1
y2 = y1 y20 − y2 y10
y20 If two functions y1 and y2 are linearly dependent then, from the formal definition, is possible to express one function
as a scalar multiple of the other. That is:
y2 = Cy1
Making the Wronskian:
55
56
CHAPTER 2. SECOND ORDER EQUATIONS.
Cy1 = Cy1 y10 − Cy1 y10 = 0
Cy10 y
1
W (y1 , y2 ) = W (y1 , Cy1 ) = 0
y1
So if two functions are linearly dependent their Wronskian is identically zero.
If y1 and y2 are both solutions to y 00 + P (x)y 0 + Q(x)y = 0 and {y1 , y2 } is linearly independent then {y1 , y2 } is a
Fundamental Solution Set of the differential equation.
222.
Show that the set is a fundamental solution set of the differential equation
y 00 − 6y 0 + 9y = 0
{y1 , y2 } = {e3x , xe3x }
223.
Show that the set is a fundamental solution set of the differential equation
{y1 , y2 } = {x2 , x3 }
x2 y 00 − 4xy 0 + 6y = 0
224.
Show that the set is a fundamental solution set of the differential equation
{y1 , y2 } = {sinh(x), cosh(x)}
y 00 − y = 0
Remember
sinh(x) =
ex − e−x
2
cosh(x) =
ex + e−x
2
225.
Show that the set is a fundamental solution set of the differential equation
{y1 , y2 , y3 } = {ex , e2x , e3x }
y 000 − 6y 00 + 11y 0 − 6y = 0
226.
Show that the set is a fundamental solution set of the differential equation
{y1 , y2 } = {eax sin(bx), eax cos(bx)}
y 00 − 2ay 0 + (a2 + b2 )y = 0
227.
Show that the two sets are both fundamental solution set of the differential equation. Which one would you rather
work with?
S1 = {y1 , y2 } = {ex , e2x }
S2 = {y1 , y2 } = {4ex , e2x − 6ex }
y 00 − 3y 0 + 2y = 0
228.
Show that if y1 and y2 both have a relative extrema at x = x0 then they cannot be a fundamental solution set to
y 00 + P (x)y 0 + Q(x)y = 0 on an interval containing x = x0
229.
Show that y1 = cos(2x) and y2 = cos2 (x) − sin2 (x) are Linearly Dependent
2.1. WRONSKIAN, FUNDAMENTAL SETS AND ABLE’S THEOREM
230.
Show for a, b Constants that
W (ay1 , by2 ) = abW (y1 , y2 )
231.
Show for f , g, h differentiable functions that
W (f g, f h) = f 2 W (g, h)
232.
Calculate and simplify the following
e
R
W
f
g
, fg dx
233.
Show
W (y1 + a, y2 + a) = W (y1 , y2 ) + a
d
(y2 − y1 )
dx
234.
Show
W (y1 , y2 ) =
y22
d y1
dx y2
235.
Show
y2
y1 · W
, y1
y1
y1
, y2
+ y2 · W
y2
=
d
(y1 · y2 )
dx
236.
If W (y1 , y2 ) = e4x and y1 = ex find y2 if y2 (0) = 2
237.
Show:
W (f, g + h) = W (f, g) + W (f, h)
238.
If
W (f, g) = e5x
Find
W (f + g, f − g)
239.
If
W (y, y 2 ) = e3x
y(0) = 1
57
58
CHAPTER 2. SECOND ORDER EQUATIONS.
Find y(x)
240.
Show
W 0 (f, g) = W (f, g 0 ) + W (f 0 , g)
241.
If W (f (x), g(x))|x=0 = 10 Find W (f (3x), g(3x))|x=0
242.
If
W (f, g) = W (g, h)
Show
g(x) = C f (x) + h(x)
243.
If
W (f, g) = f · g
Find
g
f
244.
Show:
W f (x) cos(x), f (x) sin(x)
=
245.
If
W (f (g), g) = 0
Show
f (x) = Cx
Interpret the results in terms of linear independence
246.
Solve for f (x) if
√
W (x, f ) =
247.
Find
1 + x2
x2
2
f (x)
2.1. WRONSKIAN, FUNDAMENTAL SETS AND ABLE’S THEOREM
R
f, f1
W
59
dx
e
248.
Show:
W (x · f, f ) ≤ 0
249.
Let y1 and y2 be solutions to
y 00 + P (x)y 0 + Q(x)y = 0
Show:
P (x) = −
y1 y200 − y2 y100
W (y1 , y2 )
and
Q(x) =
y10 y200 − y20 y100
W (y1 , y2 )
Use this result to show the differential equation with solutions y1 = xn and y2 = xm with n 6= m has the form
ax2 y 00 + bxy 0 + cy = 0
250.
Show the second order linear homogenous equation
y 00 + P (x)y 0 + Q(x)y = 0
with fundamental solution set {y1 , y2 } and Wronskian W can be written as
y
1 0
y
W y 00
y1
y10
y100
y2 y20 = 0
y 00 2
251.
Use the results of either of the previous problem to find a differential equation with the following Fundamental Solution
Set
{y1 , y2 } = {sin(kx), cos(kx)}
252.
Show
2
3
W 1, x, x , x , ..., x
n
= 1! · 2! · 3! · ... · n!
253.
The differential equation
y 00 + P (x)y 0 + Q(x)y = 0
can be converted into Normal Form:
60
CHAPTER 2. SECOND ORDER EQUATIONS.
u00 + f (x)u = 0
with the substitution
y(x) = u(x) · v(x)
v(x) = e
−1
2
R
P (x)dx
Use this to convert Bessel’s equation of order v to normal form
x y + xy + x − v y = 0
2 00
0
2
2
Another way of calculating the Wronskian of the two solutions:y1 and y2 of y 00 + P (x)y 0 + Q(x)y = 0 on (a, b) is to
use Abel’s Identity:
−
W (y1 , y2 ) = Ce
Rx
x0
P (t)dt
x0 ∈ (a, b)
P and Q continuous on (a, b)
This can be easily derived by noticing that if {y1 , y2 } is a fundamental solution set to y 00 + P (x)y 0 + Q(x)y = 0 on
(a, b) then:
y100 + P y10 + Qy1 = 0
y200 + P y20 + Qy2 = 0
Multiplying the first equation by y2 and the second by y1 gives:
y2 y100 + P y2 y10 + Qy2 y1 = 0
y1 y200 + P y1 y20 + Qy1 y2 = 0
Subtracting the second equation from the first gives:
y2 y100 − y1 y200 + P (y10 y2 − y2 y10 ) = 0
Remembering
W (y1 , y2 ) = y10 y2 − y2 y10 and calculating W 0 (y1 , y2 ) = y2 y100 − y1 y200
The above equation becomes:
W 0 (y1 , y2 ) + P W (y1 , y2 ) = 0 or
W 0 (y1 , y2 )
= −P
W (y1 , y2 )
Integrating and solving for W (y1 , y2 )
Z
x
ln |W (y1 , y2 )| = −
P (t)dt + C
x0
−
W (y1 , y2 ) = Ce
This result is known as Abel’s Identity
Rx
x0
P (t)dt
2.1. WRONSKIAN, FUNDAMENTAL SETS AND ABLE’S THEOREM
61
254.
Use Abel’s Identity to find the Wronskian up to a constant for
(1 + x2 )y 00 + 2xy 0 + y = 0
(−∞, ∞)
on
255.
Use Abel’s Identity to find the Wronskian up to a constant for
(x2 + 3x + 2)y 00 + y 0 + y = 0
(0, ∞)
on
256.
Use Abel’s Identity to find the Wronskian up to a constant for
00
2
cos (x)y +
3
0
cos (x) + 1 y + y = 0
on
−π π
,
4 4
257.
Use Abel’s Identity to find the Wronskian up to a constant for
xy 00 + (2x2 + 1)y 0 + xy = 0
(0, ∞)
on
258.
Use Abel’s Identity to find the Wronskian up to a constant for
6x
y0 + y = 0
y + 4
x + 5x2 + 4
00
on
− ∞, ∞
259.
Use Abel’s Identity to find the Wronskian up to a constant for
y 00 −
16x 0
16
y +
y=0
4x − 1
4x − 1
on
1
,∞
4
260.
Show y1 =
1
x−1
and y2 =
1
x+1
are solutions to the given differential equation
(x2 − 1)y 00 + 4xy 0 + 2y = 0
and calculate the Wronskian of y1 and y2 and then confirm your answer with Abel’s Identity.
261.
Use Abel’s Identity to find the Wronskian up to a constant for
(P (x)y 0 )0 + Q(x)y = 0
262.
If the Wronskian of the solutions to
y 00 + P (x)y 0 + Q(x)y = 0
is a constant, what does it say about P (x)
62
CHAPTER 2. SECOND ORDER EQUATIONS.
2.2
Reduction of Order
Question: given a second order differential equation of the form:
y 00 + P (x)y 0 + Q(x)y = 0
and one solution to the differential equation can we find a second solution? If y1 is a solution to y 00 +P (x)y +Q(x)y = 0
then we know y100 + P (x)y10 + Q(x)y1 = 0. Let us try to find a second solution of the form y2 = vy1 . Differentiating gives:
y20 = vy10 + v 0 y1
y200 = vy100 + 2v 0 y10 + v 00 y1
Substituting these into the original differential equation gives:
vy100 + 2v 0 y10 + v 00 y1 + P (x)(vy10 + v 0 y1 ) + Q(x)vy1 = 0
Which reduces to:
v(y100 + P (x)y10 + Q(x)y1 ) + 2v 0 y10 + v 00 y1 + P (x)v 0 y1 = 0
v 00 y1 + 2v 0 y10 + P (x)v 0 y1 = 0
v 00 y1 + v 0 (2y10 + P (x)y1 ) = 0
2y10
v 00
=
−
− P (x)
v0
y1
v 00 y1 = −v 0 (2y10 + P (x)y1 )
Integrating gives:
ln |v 0 | = −2 ln |y1 | −
Z
P (x)dx
Solving for v 0
R
ln |y1−2 |− P (x)dx
0
v =e
=
e−
R
P (x)dx
y12
Integrating again give the formula for v
Z
v=
e−
R
P (x)dx
y12
dx
An Example: Find a second linearly independent solution:
xy 00 − (x + 1)y 0 + y = 0
y1 = ex
Writing the equation is standard form gives:
00
y +
1 0 1
−1−
y + y=0
x
x
2.2. REDUCTION OF ORDER
63
R
Z
v=
−
1
−1− x
e
e2x
Z
v=
dx
Z
dx =
ex+ln(x)
dx
e2x
xe−x dx = −(x + 1)e−x
y2 = y1 v = ex (−(x + 1)e−x ) = −x − 1
Making the homogenous solution
yh = C1 y1 + C2 y2
yh = C1 ex + C2 (x + 1)
♠
263.
y1 = e2x is one solution to y 00 − 6y 0 + 8y = 0. Use reduction of order to find a second linearly independent solution.
264.
y1 = x−2 is one solution to x2 y 00 + 6xy 0 + 6y = 0. Use reduction of order to find a second linearly independent solution.
265.
y1 =
1
x
is one solution to xy 00 + (2x + 2)y 0 + 2y = 0. Use reduction of order to find a second linearly independent
solution.
266.
y1 = ex is one solution to xy 00 − (x + 1)y 0 + y = 0. Use reduction of order to find a second linearly independent
solution.
267.
y1 = ex is one solution to xy 00 + (1 − 2x)y 0 + (x − 1)y = 0. Use reduction of order to find a second linearly independent
solution.
268.
y1 = ex is one solution to (2x − 1)y 00 − (4x2 + 1)y 0 + (4x2 − 2x + 2)y = 0. Use reduction of order to find a second
linearly independent solution.
269.
y1 = ex is one solution to (sin(x) − cos(x))y 00 − 2 sin(x)y 0 + (sin(x) + cos(x))y = 0. Use reduction of order to find a
second linearly independent solution.
270.
y1 = x2 is one solution to (x2 − 2x)y 00 + (2 − x2 )y 0 + (2x − 2)y = 0. Use reduction of order to find a second linearly
independent solution.
271.
y1 = tan(x) is one solution to y 00 −tan(x)y 0 −sec2 (x)y = 0. Use reduction of order to find a second linearly independent
solution.
272.
y1 = x sin(x) is one solution to x2 y 00 −2xy 0 +(2+x2 )y = 0. Use reduction of order to find a second linearly independent
solution.
64
CHAPTER 2. SECOND ORDER EQUATIONS.
273.
y1 = x + 1 is one solution to (x2 + 2x − 1)y 00 − (2x + 2)y 0 + 2y = 0. Use reduction of order to find a second linearly
independent solution.
274.
y1 = x2 + 1 is one solution to y 00 −
2x
0
x2 −1 y
+
2
x2 −1 y
= 0. Use reduction of order to find a second linearly independent
solution.
275.
y1 = sin(x) is one solution to y 00 − 3 cot(x)y 0 +
3−2 sin2 (x)
y
sin2 (x)
= 0. Use reduction of order to find a second linearly
independent solution.
276.
y1 =
1
x−2
is one solution to (x2 − 4)y 00 + 4xy 0 + 2y = 0. Use reduction of order to find a second linearly independent
solution.
277.
y1 = x2 is one solution to
x3 − 3x + 1 0 2x3 − 2x2 + 2
y +
y=0
x3 − 3x
x4 − 3x3
Use reduction of order to find a second linearly independent solution.
y 00 −
278.
y1 = x is one solution to
1
x
y0 +
y=0
x−1
x−1
Use reduction of order to find a second linearly independent solution.
y 00 −
279.
y1 =
sin(x)
√
x
is one solution to the Bessel equation of order 12 :
1
y=0
x2 y 00 + xy 0 + x2 −
4
Use reduction of order to find a second linearly independent solution.
280.
y1 = ln(x) is one solution to
2
x2 ln(x) y 00 − 2x ln(x)y 0 + 2 + ln(x) y = 0
Use reduction of order to find a second linearly independent solution.
281.
The Hermite equation is an equation of the form:
y 00 − 2xy 0 + λy = 0
Find the homogenous solution for the given values of λ and y1
A) λ = 4 and y1 = 1 − 2x2
B) λ = 6 and y1 = 3x − 2x3
282.
x>0
2.3. EQUATIONS OF THE FORM Y”=F(X,Y’) AND Y”=F(Y,Y’)
65
The Legendre equation is an equation of the form:
(1 − x2 )y 00 − 2xy 0 + λ(λ + 1)y = 0
x ∈ (−1, 1)
Find the homogenous solution for the given values of λ and y1
A) λ = 1 and y1 = x
B) λ = 2 and y1 = 3x2 − 1
C) λ = 3 and y1 = x3 − 3x
283.
The Laguerre equation is an equation of the form:
xy 00 + (1 − x)y 0 + λy = 0
Find the homogenous solution for the given values of λ and y1
A) λ = 1 and y1 = x − 1
B) λ = 2 and y1 = x2 − 4x + 2
284.
First, use Abel’s Identity to find the Whronskian up to a constant for
y 00 −
16
16x 0
y +
y=0
4x − 1
4x − 1
on
1
,∞
4
Second, notice y1 = x is a solution and apply reduction of order to find y2
285.
The reduction of order algorithm can be applied to third order equations although the formula we derived for v above
will not work. y1 = ex is a solution to:
xy 000 − xy 00 + y 0 − y = 0
Use y2 = vy1 to reduce this equation to a second order equation by letting w = y 0
2.3
Equations of the form y”=f(x,y’) and y”=f(y,y’)
For a second order equation of the form:
y 00 = f (x, y 0 )
the substitution v = y 0 will transform the equation into first order equation.
An Example:
Solve:
xy 00 − y 0 = 3x2
Using the substitution v = y 0 , making v 0 = y 00 our equation becomes:
66
CHAPTER 2. SECOND ORDER EQUATIONS.
xv 0 − v = 3x2
v0 +
in standard form
−1
v = 3x
x
This equation is now first order linear. Creating an integrating factor
I(x) = e
−1
x
R
= x−1
And now the solution in terms of v
Z
x3xdx + C
v=x
v = x(x3 + C)
Converting back to y
y 0 = x4 + Cx
Integrating
y=
Cx2
x5
+
+K
5
2
If the differential equation is of the form:
y 00 = f (y, y 0 )
the substitution v = y 0 will transform the equation into a first order equation.
An Example from differential geometry:
The curvature of a circle of radius r is defined to be
1
r
and the curvature of a straight line defined to be zero. For
2
other equations in R the curvature is given by the second order equation:
K=
By replacing K with
1
r
|y 00 |
3
(1 + (y 0 )2 ) 2
and solving the second order equation we should obtain the equation of a circle of radius r.
For simplicity we will assume y 00 > 0 so we do not have to deal with the pesky absolute values.
1
y 00
=
3
r
(1 + (y 0 )2 ) 2
Substituting v = y 0 making v 0 = y 00 we get:
1
v0
=
3
r
(1 + (v)2 ) 2
or
dx
dv
=
3
r
(1 + (v)2 ) 2
This is now a separable and we now need to integrating both sides.
2.3. EQUATIONS OF THE FORM Y”=F(X,Y’) AND Y”=F(Y,Y’)
Z
Z
dx
=
r
67
dv
3
(1 + (v)2 ) 2
The integral on the left is easy while the integral on the right will require a trig substitution.
x
+C =
r
Z
dv
√
dv = sec2 θdθ
v = tan θ
3
(1 + (v)2 ) 2
1 + v2
v
θ
1
Under this substitution the integral becomes:
x
+C =
r
Z
sec2 θdθ
3
(1 + tan2 θ) 2
Z
x
+ C = cos θdθ
r
x
+ C = sin θ
r
Using the triangle to convert back to v:
x
v
+C = √
r
1 + v2
Converting back to y gives another separable equation:
x+C
y0
=p
r
1 + (y 0 )2
Solving for y 0
(x + C)(
p
1 + y 02 ) = ry 0
(x + C)2 (1 + (y 0 )2 ) = r2 (y 0 )2
(x + C)2 + (x + C)2 (y 0 )2 = r2 (y 0 )2
(y 0 )2 =
y0 = p
Z
(x + C)
r2
Substituting u = r2 − (x + C)2 ,
− (x +
−1
2 du
(x + C)2
r2 − (x + C)2
C)2
or
Z
dy =
(x + C)
p
r2
− (x + C)2
= (x + C)dx for the integral on the right.
dx
68
CHAPTER 2. SECOND ORDER EQUATIONS.
y+K =
y+K =−
−1
2
p
Z
u
−1
2
du
r2 − (x + C)2
Squaring both sides and rearranging terms gives the equation of a circle of radius r:
(y + K)2 + (x + C)2 = r2
286.
Solve:
(1 + x2 )y 00 = 2xy 0
287.
Solve:
y 00 +
1 0
4x
y = 0
x
y
y 0 (1) = 2
288.
Solve:
xy 00 − y 0 = 3x2
289.
Solve:
x2 y 00 = 2xy 0 + (y 0 )2
290.
Solve:
y 00 + x(y 0 )2 = 0
291.
Solve:
x2 y 00 = (y 0 )2
If the independent variable x is missing from the differential equation then you will have an equation of the form:
y 00 = f (y, y 0 )
and the substitution v = y 0 will transform the second order equation into a first order equation.
v=
dy
dx
d2 y
dv
dv dy
dv
=
=
·
=v
2
dx
dx
dy dx
dy
2.3. EQUATIONS OF THE FORM Y”=F(X,Y’) AND Y”=F(Y,Y’)
Using this substitution our equation becomes:
v
dv
= f (y, v)
dy
An Example:
Solve
y 00 + k 2 y = 0
Using the substitution above the second order equation becomes:
v
dv
+ k2 y = 0
dy
This equation is now separable:
vdv = −k 2 ydy
Integrating and solving for v:
−k 2 y 2
v2
=
+C
2
2
v=
p
C − k2 y2
or
p
dy
= ±k A2 − y 2
dx
This equation is now separable:
Z
dy
Z
= ±kdx
A2 − y 2
y
= ±kx + B
arcsin
A
p
y = A sin(±kx + B)
292.
Solve:
yy 00 + (y 0 )2 = 0
293.
Solve:
yy 00 = y 2 y 0 + (y 0 )2
294.
Solve:
y 00 + 2yy 0 = y
69
70
CHAPTER 2. SECOND ORDER EQUATIONS.
295.
Solve:
yy 00 = (y 0 )3
296.
Solve:
3
y 00 = 12y(y 0 ) 2
297.
If
{y1 , (y1 )2 }
are solutions to
y 00 − 3y 0 + ky = 0
Find y1
298.
Solve:
W (y, y 0 ) = 0
Interprit the results in terms of linear independce or depencence
2.4
Homogenous Linear Equations with Constant Coefficients.
Let us consider the second order equation:
ay 00 + by 0 + cy = 0
and look for a solution of the form y = erx . Differentiating twice give:
y = erx
y 0 = rerx
y 00 = r2 erx
Substituting these into the differential equation gives:
ar2 erx + brerx + cerx = 0
Dividing by erx gives the characteristic polynomial:
ar2 + br + c = 0
This quadratic equation can have 3 types of roots: real and distinct, complex or real and repeated.
2.4. HOMOGENOUS LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS.
71
If the two roots, r1 and r2 of the characteristic polynomial are real and distinct: r1 6= r2 then the solution to the
differential equation is
y = C1 er1 x + C2 er2 x
If the two roots, r1 and r2 of the characteristic polynomial are complex r1 = α + βı and r = α − βı then the solution
to the differential equation is
y = C1 e(α+βı)x + C2 e(α−βı)x or y = C1 eαx eβıx + C2 eαx e−βıx
After applying Euler’s equation: eıθ = cos(θ) + ı sin(θ) we get the solution to the differential equation to be:
y = C1 eαx cos(βx) + C2 eαx sin(βx)
If the two roots, r1 and r2 of the characteristic polynomial are real and repeated: r1 = r2 =
−b
2a
then one solution to
the differential equation is
y1 = er1 x
The second linearly independent solution comes by applying the reduction of order algorithm to the problem. If
−b
c
b
y1 = er1 x = e 2a x is a solution to y 00 + y 0 + y = 0
a
a
then the reduction of order algorithm gives:
Z
v=
e−
R
b
a dx
−b
(e 2a x )2
Z
dx =
e
−bx
a
e
−bx
a
dx = x
Making y2 = xy1 = xer1 x and the solution to the differential equation is
y = C1 er1 x + C2 xer2 x
299.
Solve
y 00 − y 0 − 2y = 0
y(0) = 2
y 0 (0) = 1
300.
Solve
y 00 − 12y 0 + 36y = 0
y(0) = 1
y 0 (0) = 1
y 00 − 2y 0 + 10y = 0
y(0) = 2
y 0 (0) = 1
301.
Solve
302.
Solve
72
CHAPTER 2. SECOND ORDER EQUATIONS.
y 00 − 8y 0 + 41y = 0
y(0) = 1
y 0 (0) = 1
y 00 − 16y 0 + 64y = 0
y(0) = 1
y 0 (0) = 9
303.
Solve
304.
Solve
y 000 − 6y 00 + 11y 0 − 6y = 0
y 0 (0) = 1
y(0) = 1
y 00 (0) = 1
305.
Solve
y 000 − 7y 00 + 15y 0 − 9y = 0
306.
Solve
y 0000 − 8y 000 + 24y 00 − 32y 0 + 16y = 0
307.
Solve
y 000 − 2y 00 + 4y 0 − 8y = 0
y(0) = 2
y 0 (0) = 4
y 00 (0) = 0
308.
Find a third order differential equation with the following solution
yh = C1 e3x + C2 x3 sin(2x) + C3 x3 cos(2x)
309.
Solve the differential equation for different values of k and sketch the solutions
y 00 + ky 0 + 6y = 0
k ∈ {−7, 5, 2, 0}
310.
Solve
y 00 − 2ıy 0 + 3y = 0
311.
Consider the differential equation:
ay 00 + by 0 + cy = 0
a, b, c ∈ R+
Show all solutions y tend to zero as x tends to infinity. Show this is not true if b = 0
2.4. HOMOGENOUS LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS.
73
312.
Consider the differential equation:
ky 00 + (k + 1)y 0 + (k + 2)y = 0
Find the values of k so that the characteristic ploynomial has real and distint roots, real and repeated roots and
complex roots. What values of k make the solution y tend to zero as x → ∞
313.
Consider the differential equation:
y 00 + ay 0 + a2 y = 0
Show the characteristic polynomial has complex roots for all a 6= 0
314.
Show that if y1 and y2 are solution to the second order linear equation
ay 00 + by 0 + cy = 0
a, b, c ∈ R
Then y = Cy1 + Ky2 is also a solution.
315.
Find a third order homogenous differential equation with constant coefficients with the given solution
y = C1 e2x + C2 xe2x + C3 x2 e2x
316.
If y = xex is a solution to
y 00 + ay 0 + by = 0
Find a and b
317.
Solve
dz
= e(a+bı)t
dt
by separating into real and imaginary parts. Then use the fact that
Z
eat cos(bt)dt =
eat
a
cos(bt)
+
b
sin(bt)
+C
a2 + b2
To calculate
Z
eat sin(bt)dt
318.
For the given differential equation:
y 00 + 2y 0 + Cy = 0
C∈R
74
CHAPTER 2. SECOND ORDER EQUATIONS.
A) For what values of C does the characteristic equation have two real distinct roots, real repeated roots and complex
roots.
B) For the case where you have two real distinct roots find the values of C that make the solution tend to zero as
x → ∞.
319.
Solve the differential equation with the discontinuous coefficient function
y 00 + sgn(x)y = 0
(
sgn(x) =
2.5
−1
x<0
1
0<x
The Method of Undetermined Coefficients
In this section we will study a way to solve the Linear Differential equation:
y 00 + ay 0 + by = g(x)
for different equations g(x).
For example if you were given the differential equation
y 00 − 3y 0 + 5y = 3ex
It would be reasonable to think the solution would involve the exponential function ex simply because exponential
functions differentiate into more exponential functions.
If you were given the differential equation
y 00 + y 0 + y = cos(x) + sin(x)
It would be reasonable to think the solution would involve the functions sin(x) and cos(x) since sin(x) and cos(x)
differentiate into each other.
So in the method of undetermined coefficients we guess the form of the solution and try to make the constants work.
For
y 00 − 3y 0 + 5y = 3ex
we would guess the form of the solution to be y = Aex and solve for the value of A that makes it a solution.
For
y 00 + y 0 + y = cos(x) + sin(x)
we would guess the form of the solution to be y = A sin(x) + B cos(x) and solve for the values of A and B that makes
it a solution.
One problem that can arise using this method is the problem of finding homogenous solution twice. If you were given
2.5. THE METHOD OF UNDETERMINED COEFFICIENTS
75
y 00 − 4y 0 + 4y = e2x
it would be reasonable to think the solution would be of the form y = Ae2x but we know the homogenous solution to
y 00 − 4y 0 + 4y = 0
is y = C1 e2x + C2 xe2x . So looking for a particular solution of the form y = Ae2x will again find the homogenous
solution, not the particular solution we are interested in. The solution to this problem is to multiply the particular solution
by xs so that it is linearly independent of the terms in the fundamental solution set. The nonnegative integer s should
be chosen to be the smallest nonnegative integer so that no term in the particular solution appears in the fundamental
solution set. In this problem the fundamental solution set is:
{e2x , xe2x }
So we would choose s = 2 and look for a particular solution of the form y = Ax2 e2x .
If you choose the wrong form of the particular solution you’re gonna have a bad time.
In general we choose the form of the particular solution based on the following table:
For y 00 + ay 0 + by = g(x)
g(x)
yp
n
s
pn = an x + ... + a1 x + a0
ae
s
n
x Pn = x (An x + ... + A1 x + A0 )
αx
Axs eαx
xs (A cos(βx) + B sin(βx))
a cos(βx) + b sin(βx)
pn eαx
xs (Pn eαx )
pn cos(βx) + pm sin(βx)
ae
αx
αx
cos(βx) + be
s
x (PN cos(βx) + QN sin(βx)) N = max(n, m)
xs (Aeαx cos(βx) + Beαx sin(βx))
sin(βx)
pn eαx cos(βx) + qm eαx sin(βx)
xs (PN eαx cos(βx) + QN eαx cos(βx)) N = max(n, m)
The integer s is chosen to be the smallest nonnegative integer so that no term in the particular solution appears in the
homogenous solution.
An Example: Solve:
y 00 − 4y 0 + 4y = 6e2x
First we need the homogenous solution, so we form the auxiliary equation:
r2 − 4r + 4 = 0
The homogenous solution is:
(r − 2)2 = 0
y1 = e2x
y2 = xe2x
76
CHAPTER 2. SECOND ORDER EQUATIONS.
yh = C1 e2x + C2 xe2x
The Fundamental Solution Set is:
F.S.S. = {e2x , xe2x }
Due to the right hand side of the differential equation we will choose a particular solution of the form yp = xs (Ae2x )
with s chosen to be the smallest positive integer so that no term in yp is in the fundamental solution set. So we choose
s = 2 making yp = Ax2 e2x
yp0 = e2x (2Ax + 2Ax2 )
Differentiating and simplifying gives:
yp00 = e2x (4Ax2 + 8Ax + 2A)
Substituting these derivatives into the original differential equation gives:
e2x (4Ax2 + 8Ax + 2A) − 4e2x (2Ax + 2Ax2 ) + 4Ax2 e2x = 6e2x
This simplifies nicely to:
2Ae2x = 6e2x
A=3
So our particular solution is:
yp = 3x2 e2x
The general solution to the differential equation is the sum of the homogenous solution and the particular solution:
y = C1 e2x + C2 xe2x + 3x2 e2x
320.
Solve
y 00 + 3y 0 + 2y = 18 sin(2x) + 26 cos(2x)
y(0) = 2
y 0 (0) = 4
321.
Solve
y 00 + 4y = (5x2 + 9x + 4)ex
322.
Solve:
y 00 − 5y 0 + 6y = ex (4 sin(x) − 2 cos(x))
323.
Solve
y(0) = 3
y 0 (0) = 3
2.5. THE METHOD OF UNDETERMINED COEFFICIENTS
y 00 − 4y 0 + 4y = 6xe2x
77
y(0) = 2
y 0 (0) = 4
324.
Solve
y 00 − 4y 0 + 5y = (2x + 2)ex
325.
Solve
y 00 + 4y = 14 cos(x) + 12x sin(x)
y(0) = 2
y 0 (0) = 4
y 00 − 8y 0 + 7y = −15xe2x − 32e2x
y(0) = 2
y 0 (0) = 4
326.
Solve
327.
Solve
y 00 − 3y 0 + 2y = ex sin(x)
y(0) = 2
y 0 (0) = 4
328.
Solve:
y 00 − 4y 0 + 4y = 8e2x
329.
Solve
y 00 − 2y 0 + y = (x + 3)ex
y(0) = 1
y 0 (0) = 2
330.
Solve
y 00 + 4y = (5x2 + 4x + 2)ex
y(0) = 4
y 0 (0) = 3
331.
Solve
y 00 − 3y 0 + 2y = 6x3 − 19x2 + 8x − 11
332.
Solve
y 00 − 3y 0 + 2y = 2x2 − 2x − 4 − ex
333.
y(0) = 2
y 0 (0) = 6
78
CHAPTER 2. SECOND ORDER EQUATIONS.
Solve
y 00 − 4y 0 + 4y = 16 sin(x) cos(x) + 16 cos2 (x) − 16 sin2 (x)
334.
Solve
y 00 + 4y = cos3 (x)
Hint
cos3 (x) =
3
1
cos(x) + cos(3x)
4
4
335.
Find the form of the solution to
y 00 + 6y 0 + 10y = x2 e−3x sin(x)
336.
Find the form of the solution to
y 000 − y 00 − y 0 + y = x3 (ex + e−x )
337.
Find the form of the solution to
y 000 − y 00 + 4y 0 − 4y = x2 (ex + sin(2x))
338.
Find a solution of the form y = Ax2 + Bx + C to solve the nonlinear differential equation
y 00 + (y 0 )2 + y = 5x2 + 5x + 3
339.
The form of the particular solution to the third order linear differential equation:
ay 000 + by 00 + cy 0 + d = g(x)
is
yp = (Ax4 + Bx3 + Cx2 )ex + Dxe2x
what does this tell you about the roots of the characteristic polynomial
ar3 + br2 + cr + d = 0
and the function g(x).
2.6. VARIATION OF PARAMETERS
2.6
79
Variation of Parameters
In the method of variation of parameters we will develop a solution to the second order linear differential equation:
y 00 + P (x)y 0 + Q(x)y = g(x)
As you can see the method of variation of parameters is a far more general method of solving differential equation than
the method of undetermined coefficients since variation of parameters allows for variable coefficients of y 0 and y whereas
the method of undetermined coefficients works only for linear equations with constant coefficients and the function g(x)
does not need to be listed in the table in the section on undetermined coefficients.
Given the fundamental solution set {y1 , y2 } for the homogenous equation:
y 00 + P (x)y 0 + Q(x)y = 0
we will look for a particular solution to the nonhomogeneous equation of the form
y = v1 y1 + v2 y2
Differentiating yields:
y 0 = v1 y10 + v10 y1 + v2 y20 + v20 y2
To avoid second derivatives of the unknown functions v1 and v2 we impose the condition:
v10 y1 + v20 y2 = 0
Making
y 0 = v1 y10 + v2 y20
Differentiating again gives:
y 00 = v1 y100 + v10 y10 + v2 y200 + v20 y20
Substituting y, y 0 and y 00 into the original differential equation gives:
v1 y100 + v10 y10 + v2 y200 + v20 y20 + P (x)(v1 y10 + v2 y20 ) + Q(x)(v1 y1 + v2 y2 ) = g(x)
This equation can be rewritten as:
v1 (y100 + P (x)y10 + Q(x)y1 ) + v2 (y200 + P (x)y20 + Q(x)y2 ) + y10 v10 + y20 v20 = g(x)
Since y1 and y2 are solutions of the homogenous equation
y 00 + P (x)y 0 + Q(x)y = 0
80
CHAPTER 2. SECOND ORDER EQUATIONS.
the above equation reduces to:
y10 v10 + y20 v20 = g(x)
This equation along with the restriction we made on y 0 : v10 y1 + v20 y2 = 0 gives us a system of two equations with two
unknown variables: v10 and v20 that we will solve using Cramer’s Rule. Writing the system as a matrix equation gives:
"
y1
y2
y10
y20
#" #
v10
v20
"
=
0
#
g(x)
Solving for the two unknown variables: v10 and v20 using Cramer’s Rule gives:
v10 = −
y2 g(x)
y2 g(x)
y1 g(x)
y1 g(x)
=−
and v20 =
=
y1 y20 − y2 y10
W (y1 , y2 )
y1 y20 − y2 y10
W (y1 , y2 )
Integrating gives:
Z
v1 = −
y2 g(x)
dx = −
y1 y20 − y2 y10
Z
y2 g(x)
dx and v2 =
W (y1 , y2 )
Z
y1 g(x)
dx =
y1 y20 − y2 y10
Z
y1 g(x)
dx
W (y1 , y2 )
An Example: Solve:
y 00 + y = sec(x)
First note that this equation cannot be solved by the method of undetermined coefficients due to the fact that the
right hand side: sec(x) is not in the table in the undetermined coefficients section.
First we need the homogenous solution, so we form the auxiliary equation:
r2 + 1 = 0
r = ±ı
y1 = cos(x)
y2 = sin(x)
The homogenous solution is:
yh = C1 cos(x) + C2 sin(x)
Our fundamental solution set is:
F.S.S = {cos(x), sin(x)}
The Wronskian of the fundamental solution set is:
cos(x)
W (cos(x), sin(x)) = − sin(x)
sin(x) = cos2 (x) + sin2 (x) = 1
cos(x)
Z
v1 = −
Z
sin(x) sec(x)dx = −
tan(x)dx = ln(cos(x))
Z
v2 =
cos(x) sec(x)dx = x
2.6. VARIATION OF PARAMETERS
81
yp = v1 y1 + v2 y2 = ln(cos(x)) cos(x) + x sin(x)
So the general solution is:
y = C1 cos(x) + C2 sin(x) + ln(cos(x)) cos(x) + x sin(x)
340.
Solve:
y 00 − 3y 0 + 2y = xe2x
341.
Solve:
y 00 + y = tan(x)
342.
Solve:
y 00 − 4y 0 + 5y = xe2x
343.
Solve:
y 00 − 2y 0 + y =
ex
1 + x2
344.
Solve:
y 00 + 4y = sin3 (2x)
345.
Solve:
y 00 − 2y 0 + y = ex arcsin(x)
346.
Solve:
y 00 + y 0 = ln(x)
347.
Solve:
y 00 + 4y = csc(2x)
82
CHAPTER 2. SECOND ORDER EQUATIONS.
348.
Solve:
y 00 + 4y = ln(x + 1)
349.
The fundamental solution set for the differential equation
xy 00 − y 0 − 4x3 y = x3 ex
2
is
2
2
{y1 , y2 } = {ex , e−x }
Use variation of parameters to solve the differential equation
350.
Show that y1 = tan(x) is a solution to the homogenous differential equation:
y 00 − tan(x)y 0 − sec2 (x)y = 0
Use reduction of order to find the second homogenous solution y2 and then use variation of parameters to find the
particular and general solution of:
y 00 − tan(x)y 0 − sec2 (x)y = sin(x)
351.
Show that y1 = x3 + x is a solution to the homogenous differential equation:
y 00 −
6x2 − 2
4x 0
y
+
y=0
x2 + 1
(x2 + 1)2
Use reduction of order to find the second homogenous solution y2 and then use variation of parameters to find the
particular and general solution of:
y 00 −
4x 0
6x2 − 2
2
y + 2
y=
2
+1
(x + 1)
1 + x2
x2
352.
Show that y1 = ex is a solution to the homogenous differential equation:
xy 00 − (x + 1)y 0 + y = 0
Use reduction of order to find the second homogenous solution y2 and then use variation of parameters to find the
particular and general solution of:
xy 00 − (x + 1)y 0 + y = x2 e2x
353.
Show that y1 = x2 + 1 is one solution to
2.6. VARIATION OF PARAMETERS
83
2
2x 0
y + 2
y = 0.
−1
x −1
Use reduction of order to find the second homogenous solution y2 and then use variation of parameters to find the
y 00 −
x2
particular and general solution of:
y 00 −
2x 0
2
y + 2
y = 2x.
x2 − 1
x −1
354.
Show that y1 = x is a solution to the homogenous differential equation:
(x + 1)y 00 + xy 0 − y = 0
Use reduction of order to find the second homogenous solution y2 and then use variation of parameters to find the
particular and general solution of:
(x + 1)y 00 + xy 0 − y = (x + 1)2
355.
Show that y1 = x is a solution to the homogenous differential equation:
(x2 − 1)y 00 − 2xy 0 + 2y = 0
Use reduction of order to find the second homogenous solution y2 and then use variation of parameters to find the
particular and general solution of:
(x2 − 1)y 00 − 2xy 0 + 2y = x2 − 1
356.
Show that y1 = 5x − 1 is a solution to the homogenous differential equation:
xy 00 + (5x − 1)y 0 − 5y = 0
Use reduction of order to find the second homogenous solution y2 and then use variation of parameters to find the
particular and general solution of:
xy 00 + (5x − 1)y 0 − 5y = x2 e−5x
357.
Show that y1 = x + 1 is a solution to the homogenous differential equation:
(x2 + 2x)y 00 − 2(x + 1)y 0 + 2y = 0
Use reduction of order to find the second homogenous solution y2 and then use variation of parameters to find the
particular and general solution of:
(x2 + 2x)y 00 − 2(x + 1)y 0 + 2y = (x + 2)2
358.
84
CHAPTER 2. SECOND ORDER EQUATIONS.
Show that y1 = sin(x2 ) is a solution to the homogenous differential equation:
xy 00 − y 0 + 4x3 y = 0
Use reduction of order to find the second homogenous solution y2 and then use variation of parameters to find the
particular and general solution of:
xy 00 − y 0 + 4x3 y = 2x3
359.
Show that y1 = sin(x) is a solution to the homogenous differential equation:
sin2 (x)y 00 − 2 sin(x) cos(x)y 0 + (1 + cos2 (x))y = 0
Use reduction of order to find the second homogenous solution y2 and then use variation of parameters to find the
particular and general solution of:
sin2 (x)y 00 − 2 sin(x) cos(x)y 0 + (1 + cos2 (x))y = sin3 (x)
360.
Show that y1 = sin(x) is a solution to the homogenous differential equation:
y 00 − 3 cot(x)y 0 +
3 − 2 sin2 (x)
y=0
sin2 (x)
Use reduction of order to find the second homogenous solution y2 and then use variation of parameters to find the
particular and general solution of:
y 00 − 3 cot(x)y 0 +
3 − 2 sin2 (x)
y = sin3 (x)
sin2 (x)
361.
Solve by first guessing a solution to the homogenous equation then using reduction of order to find the second
homogenous solution y2 and then use variation of parameters to find the particular and general solution of:
y 00 −
2
2x 0
y +
y = 1 + x2
2
1+x
1 + x2
362.
The Bessel Equation of order one half is:
1
x2 y 00 + xy 0 + x2 −
y=0
4
and has solutions y1 =
cos(x)
√
x
and y2 =
sin(x)
√ .
x
Use variation of parameters to solve
5
1
x2 y 00 + xy 0 + x2 −
y = x2
4
363.
One solution to the equation:
y 00 + P (x)y 0 + Q(x)y = 0
2.7. CAUCHY EULER EQUATION
85
is (1 + x)2 , and the Wronskian of the two solutions is 1. Find the general solution to:
y 00 + P (x)y 0 + Q(x)y = 1 + x
2.7
Cauchy Euler Equation
:
The second order Cauchy Euler Equation is:
dy
d2 y
+ bx
+ cy = 0
2
dx
dx
This equation can be transformed into a second order liner differential equation with constant coefficients with use of
ax2
the substitution:
x = et
Differentiating with respect to t gives:
dx
= et
dt
So
dx
=x
dt
dy
In the Cauchy Euler equation we see the term x dx
which we need to substitute for, so I will multiply both sides of
dx
dt
= x by
dy
dx :
dy
dx dy
=x
dt dx
dx
Differentiating
dy
dt
This simplifies to
dy
dy
=x
dt
dx
dy
= x dx
with respect to x gives:
d dy
dy
d2 y
=
+x 2
dx dt
dx
dx
2
d y
In the Cauchy Euler equation we see the term x2 dx
2 which we need to substitute for, so I will multiply the left side
of the above equation by
dx
dt
and the right hand side by x (Remember they are equal):
d dy dx
=
dx dt dt
dy
Replacing x dx
with
dy
dt
dy
d2 y
+x 2 x
dx
dx
This simplifies to
2
d2 y
dy
2d y
=
x
+
x
dt2
dx
dx2
2
d y
and solving for x2 dx
2 gives:
d2 y
d2 y dy
=
−
dx2
dt2
dt
So under this substitution the Cauchy Euler equation becomes:
x2
2
d y dy
dy
a
−
+b
+ cy = 0
2
dt
dt
dt
This simplifies to the second order linear equation with constant coefficients:
a
d2 y
dy
+ (b − a)
+ cy = 0
dt2
dt
86
CHAPTER 2. SECOND ORDER EQUATIONS.
Which we can solve by finding the roots of the characteristic polynomial:
ar2 + (b − a)r + c = 0
Again there are three cases: the roots are real and distinct, the roots are real and repeated or the roots are complex.
Case 1: we have two real and distinct roots r1 and r2 . Then the solutions to the differential equation are:
y 1 = e r1 t
and
y2 = er2 t
)
and
y2 = er2 ln(x) = eln(x
y1 = xr1
and
y2 = xr2
Since x = et , t = ln(x) Making the solution:
y1 = er1 ln(x) = eln(x
r1
r2
)
So
Case 2: The roots are real and repeated r1 = r2 . Then the solutions to the differential equation are:
y1 = er1 t
and
y2 = ter1 t
Since x = et , t = ln(x) Making the solution:
y1 = er1 ln(x) = eln(x
r1
)
y2 = ln(x)er1 ln(x) = ln(x)eln(x
and
r1
)
So
y 1 = x r1
and
y2 = ln(x)xr1
Case 3: The roots are complex r1 = α + βı and r2 = α − βı. Then the solutions to the differential equation are:
y1 = eαt cos(βt)
and
y2 = eαt sin(βt)
and
y2 = xα sin(β ln(x))
Since x = et , t = ln(x) Making the solution:
y1 = xα cos(β ln(x))
An Example: Solve:
x2 y 00 − 5xy 0 + 13y = 0
Forming the characteristic polynomial
r2 + (−5 − 1)r + 13 = 0
r2 − 6r + 13 = 0
(r − 3)2 = −4
So we have complex roots so the solution is:
yh = C1 e3t cos(2t) + C2 e3t sin(2t)
Converting back to x gives:
r = 3 ± 2ı
2.7. CAUCHY EULER EQUATION
87
yh = C1 x3 cos(2 ln(x)) + C2 x3 sin(2 ln(x))
An Example: Solve:
dy
d2 y
+ tan(x)
− k 2 cos2 (x)y = 0
dx2
dx
Although this is not the Cauchy-Euler equation we can transform it into one using the following substitution:
sin2 (x)
u = sin(x)
dy du
dy
dy
=
·
= cos(x)
dx
du dx
du
Using the product rule gives
d2 y
d
=
2
dx
dx
d dy
dy
+ cos(x)
·
du
dx du
d2 y
dy
d dy du
= − sin(x)
+ cos(x)
dx2
du
dx du du
dy
cos(x)
du
= − sin(x)
dy
d2 y du
d2 y
=
−
sin(x)
+
cos(x)
·
dx2
du
du2 dx
d2 y
d2 y
dy
2
=
−
sin(x)
+
cos
(x)
dx2
du
du2
Substituting these expressions into the differential equation gives
dy
d2 y
dy
sin2 (x) − sin(x)
+ cos2 (x) 2 + tan(x) cos(x)
− k 2 cos2 (x)y = 0
du
du
du
Rearranging the terms gives
d2 y
sin cos (x) 2 +
du
dy
− k 2 cos2 (x)y = 0
sin(x) − sin (x)
du
d2 y
dy
− k 2 cos2 (x)y = 0
sin2 cos2 (x) 2 + sin(x) 1 − sin2 (x)
du
du
2
2
3
After a trig identity we see a cos2 (x) in each term that we can eliminate yielding
d2 y
dy
+ sin(x)
− k2 y = 0
2
du
du
Since u = sin(x) we get the Cauchy-Euler equation
sin2
u2
d2 y
dy
+u
− k2 y = 0
du2
du
Now the Cauchy-Euler substitution
u = et
u
dy
dy
=
du
dt
u2
d2 y
d2 y dy
= 2 −
2
du
dt
dt
88
CHAPTER 2. SECOND ORDER EQUATIONS.
Our differential equation now becomes
d2 y
− k2 y = 0
dt2
which has the characteristic equation
r2 − k2 = 0
r = ±k
So the solution is
y = C1 ekt + C2 e−kt
Converting back to the variable u
y = C1 uk + C2 u−k
Converting back to the variable x gives the final solution
y = C1 sink (x) + C2 sin−k (x)
364.
Solve:
x2
dy
d2 y
+ 8y = 0
+ 7x
2
dx
dx
365.
Solve:
x2
dy
d2 y
+ 9x
+ 12y = 0
2
dx
dx
x2
d2 y
dy
− 3x
+ 20y = 0
dx2
dx
366.
Solve:
367.
Solve:
x2
d2 y
dy
− 11x
+ 36y = 0
dx2
dx
368.
Solve:
x2
369.
Solve:
d2 y
dy
− 4x
+ 6y = x3 ln(x)
dx2
dx
2.7. CAUCHY EULER EQUATION
89
x2
dy
d2 y
+ 3x
− 8y = (ln(x))3 − ln(x)
2
dx
dx
370.
Solve:
x2
dy
d2 y
+x
+y =x
dx2
dx
371.
Solve:
x2
dy
d2 y
− 5x
+ 8y = x3 arctan(x)
2
dx
dx
372.
Solve:
x2
dy
1
d2 y
+x
+ 4y =
dx2
dx
x
x2
dy
d2 y
+ 3x
+ y = 8x
2
dx
dx
373.
Solve:
374.
Solve:
x2
d2 y
dy
− 7x
+ 16y = x3
dx2
dx
375.
Solve:
x2
p
d2 y
dy
− 4x
+ 6y = 1 − x2
2
dx
dx
376.
Solve:
x2
d2 y
dy
+x
+ 4y = 8
dx2
dx
377.
Solve:
x3
d2 y
dy
+ x2
+ xy = 0
2
dx
dx
378.
Find a Cauchy Euler equation with the following solution
y=
379.
C1 x2 + C2
x
90
CHAPTER 2. SECOND ORDER EQUATIONS.
Find a Cauchy Euler equation with the following solution
y = C1 x2 cos(4 ln(x)) + C2 x2 sin(4 ln(x))
380.
Solve:
xy 00 + y 0 = 0
381.
Solve:
2(x − 4)2
dy
d2 y
+ 5(x − 4)
− 2y = 0
dx2
dx
382.
Solve:
x3
d3 y
d2 y
dy
− 4x2 2 + 8x
− 8y = 4 ln(x)
3
dx
dx
dx
383.
Find the values of α that make the solution: y, to the given equation tend to zero as x → ∞
x2
1
d2 y
dy
+ y=0
+ αx
dx2
dx 4
384.
Find the value of α that makes y → 0 as x → 0
x2
d2 y
− 6y = 0
dx2
y(1) = 1
y 0 (1) = α
385.
The nonlinear equation
y 0 + y 2 + P (x)y + Q(x) = 0
is an example of a Riccati Equation. Show the substitution y =
z0
z
transform the equation to
z 00 + P (x)z 0 + Q(x)z = 0
386.
Use the results of the previous problem to solve
y0 + y2 −
387.
Use the results of the previous problem to solve
4
6
y+ 2 =0
x
x
2.8. EVERYONE LOVES A SLINKY: SPRINGS
91
y0 + y2 −
388.
Use the substitution x =
√
4
12
y+ 2 =0
x
x
t to transform the differential equation to an equation with constant coefficients and then
solve.
1 0
y + 4x2 y = 0
x
y 00 −
389.
Show that if y(x) is a solution to the Cauchy Euler Equation for x > 0 then y(−x) is a solution for x < 0
2.8
Everyone Loves a Slinky: Springs
The goal of this section is to develop a differential equation that governs the motion of a mass connected to an ideal
spring. We will first study the theoretical case of a spring with no damping: (internal resistance of the spring, air friction
ect.). We will also study springs with damping and then with a forcing function attached to the mass.
Newton’s Second law states that force equals mass times acceleration: F = ma. So if y(t) represents the position of
d2 y
dt2 .
a moving mass on a spring then its acceleration is a =
Now consider a mass spring system with a mass m attached and stretched so that the mass is still. This unmoving
system is said to be in equilibrium. We measure the distance, y, to be the displacement of the mass from equilibrium.
When the mass is displaced from equilibrium , the spring is stretched or compressed and it exerts a force in the opposite
direction of the displacement. The force exerted by the spring is given by Hooke’s Law:
Fspring = −ky
k>0
where k is a constant dependent on the stiffness of the spring and y is the displacement of the mass from equilibrium.
All mass spring systems experience some form of internal resistance known as damping which is proportional to the
velocity of the mass. Since the mass has a velocity v =
dy
dt
the damping force is given by:
Fdamping = −b
dy
dy
b>0
where b is the damping coefficient.
All other forces on the system are external making the differential equation governing the mass spring system:
m
d2 y
dy
= −b
− ky + Fexternal (t)
2
dx
dy
Or
m
d2 y
dy
+b
+ ky = Fexternal (t)
dx2
dy
In the absence of damping and an external force: b = 0, Fexternal (t) = 0 the differential equation becomes:
m
d2 y
+ ky = 0
dx2
92
CHAPTER 2. SECOND ORDER EQUATIONS.
Which has an auxiliary mr2 + k = 0 which has purely imaginary roots r = ±ωı making the solution:
r
y = C1 cos(ωt) + C2 sin(ωt)
ω=
k
m
This solution can be written in the form:
y = A sin(ωt + φ)
by first applying a trig identity to sin(ωt + φ).
y = A sin(ωt + φ) = A(sin(ωt) cos(φ) + cos(ωt) sin(φ))
These two solution are equal if
A sin(φ) = C1
and
A cos(φ) = C2
We see the amplitude of the solution: A is given by:
A=
q
C12 + C22
and
tan(φ) =
We see the solution is a sinusoid with angular frequency ω =
q
k
m
C1
C2
and Period T =
2π
ω .
All springs experience some form of damping. To explore the nature of this damping let us consider the equation:
my 00 + by 0 + ky = 0
The auxiliary equation is:
mr2 + br + k = 0
with roots:
√
−b
1 p 2
b2 − 4mk
=
±
b − 4mk
2m
2m 2m
The nature of the solution depends on the discriminate b2 − 4mk.
r=
−b ±
If b2 − 4mk < 0 the roots will be complex and we say the spring system has Underdamped Motion.
Letting α be the real part and β the imaginary part of the roots we have:
α=
−b
2m
and
β=
1 p
4mk − b2
2m
The solution is:
y = eαt (C1 cos(βt) + C2 sin(βt))
We can express this solution in a alternate form as we did earlier:
y = Aeαt sin(βt + φ)
A=
q
C12 + C22
and
tan(φ) =
C1
C2
So our solution is the product of a sinusoid: sin(βt+φ) and an exponential damping factor: Aeαt . As t → ∞ Aeαt → 0
and our solution also tends to zero. Further as b → 0 α = α =
−b
2m
→ 0 and the solution tends to the sinusoid:
2.8. EVERYONE LOVES A SLINKY: SPRINGS
93
y = A sin(βt + φ)
Going back to the discriminate: b2 − 4mk. If b2 − 4mk > 0 the roots will real and distinct and we say the spring
system has Overdamped Motion. The roots to the auxiliary equation are:
r1 =
−b
1 p 2
+
b − 4mk
2m 2m
−b
1 p 2
−
b − 4mk
2m 2m
r2 =
And the solution is:
y = C1 er1 t + C2 er2 t
It is clear that r2 is negative; r1 is negative as well since b2 > b2 − 4mk making b >
√
−b
1
b2 − 4mk < 0.
this inequality and dividing by 2m gives r1 = 2m
+ 2m
√
b2 − 4mk. Subtracting b from
Since both r1 and r2 are negative the solution y will approach zero as t → ∞.
Going back to the discriminate: b2 − 4mk. If b2 − 4mk = 0 the roots will real and repeated and we say the spring
system has Critically Damped Motion. The roots to the auxiliary equation are:
r1 = r2 =
−b
2m
And the solution is:
−b
y = e 2m (C1 + C2 t)
As t → ∞ our solution will tend to zero.
An Example:
A 1kg mass hangs from a spring stretching it .392m from equilibrium. The mass is then pulled down .5m and released.
Find the equation of the motion of the mass if:
1) damping constant b = 0
2) damping constant b = 8
3) damping constant b = 10
4) damping constant b = 26
Solution: We already know mass m and damping constant b so all we need is the spring constant k. Using Hook’s
Law:
9.8 = k(.392)
k = 25
The differential equation governing the system is:
y 00 + by 0 + 25y = 0
y(0) = −.392
1) If there is no damping then b = 0 and our equation becomes:
y 00 + 25y = 0
y 0 (0) = 0
94
CHAPTER 2. SECOND ORDER EQUATIONS.
The auxiliary equation and roots are:
r2 + 25 = 0
r = ±5ı
The solution is:
y = C1 cos(5t) + C2 sin(5t)
After the laborious task of applying the initial conditions we get:
y = −.392 cos(5t)
We see the Amplitude is A = .392 and φ =
π
2
The solution can be written in the form:
π
y = −.392 sin 5t +
2
2) The damping constant is b = 8. Our equation becomes:
y 00 + 8y 0 + 25y = 0
The auxiliary equation and roots are:
r2 + 8r + 25 = 0
r = −4 ± 3ı
The solution is:
y = C1 e−4t cos(3t) + C2 e−4t sin(3t)
After the laborious task of applying the initial conditions we get:
y = −.392e−4t cos(3t) − .52267e−4t sin(3t)
We see the Amplitude is A = .653336 and φ = .643498. Since the initial displacement is down (against the force of
the spring) we will use A = −.653336. The solution can be written in the form:
y = −.65336e−4t sin(3t + .643498)
3) The damping constant is b = 10. Our equation becomes:
y 00 + 10y 0 + 25y = 0
The auxiliary equation and roots are:
r2 + 10r + 25 = 0
r = −5 is repeated root
The solution is:
y = C1 e−5t + C2 te−5t
After the laborious task of applying the initial conditions we get:
2.8. EVERYONE LOVES A SLINKY: SPRINGS
95
y = −.392e−5t − 1.96te−5t
4) The damping constant is b = 26. Our equation becomes:
y 00 + 26y 0 + 25y = 0
The auxiliary equation and roots are:
r2 + 26r + 25 = 0
r = −1, −25
The solution is:
y = C1 e−t + C2 e−26t
After the laborious task of applying the initial conditions we get:
y = .40768e−t − .01568e−26t
390.
A 3kg mass is attached to a spring with stiffness k = 48N/m. The mass is displaced 1/2m to the left of equilibrium
point and given a velocity of 2m/s to the right. The damping constant is 0. Find the equation of motion of the mass
along with the amplitude, period, and frequency.
391.
A 1/8kg mass is attached to a spring with stiffness k = 16N/m. The mass is displaced 3/4m to the left of equilibrium
point and given a velocity of 2m/s to the left. The damping constant is 2Ns/m. Find the equation of motion of the mass.
392.
Consider the following differential equation
y 00 + ty 0 + y = 0
Although we cannot solve this differential equation we can determine the nature of the solution for large t by thinking
of the equation in terms of a mass spring system.
Find
lim y(t)
t→∞
Now let us consider the mass spring system with a forcing function applied to the system. The differential equation
governing this system is:
my 00 + by 0 + ky = F0 cos(ωt)
F0 > 0
ω>0
Let us first explore the underdamped case (0 < b2 < 4mk). From previous discussion we know the homogenous
solution is:
yh = Ae
b
t
− 2m
√
sin
4mk − b2
t+φ
2m
96
CHAPTER 2. SECOND ORDER EQUATIONS.
With
C1
C2
To find the particular solution we apply the method of undetermined coefficients. We choose the form of the particular
A=
q
C12 + C22
tan(φ) =
solution to be:
yp = A1 cos(ωt) + A2 sin(ωt)
Making:
yp0 = −A1 ω sin(ωt) + A2 ω cos(ωt)
yp00 = −A1 ω 2 cos(ωt) − A2 ω 2 sin(ωt)
Substituting this into the differential equation and simplifying gives:
(k − mω 2 )A1 + A2 bω cos(ωt) + (k − mω 2 )A2 + A1 bω sin(ωt) = F0 cos(ωt)
Equating corresponding coefficients gives:
(k − mω 2 )A1 + A2 bω = F0
(k − mω 2 )A2 + A1 bω = 0
Solving this system of equations gives:
A1 =
F0 (k − mω 2 )
(k − mω 2 )2 + b2 ω 2
A2 =
F0 bω
(k − mω 2 )2 + b2 ω 2
Making the solution :
yp =
F0
2
(k
−
mω
)
cos(ωt)
+
bω
sin(ωt)
(k − mω 2 )2 + b2 ω 2
Let
tan(θ) =
Drawing a triangle for tan(θ) = yp =
p
(k − mω 2 )2 + b2 ω 2
A1
A2
k−mω 2
bω
k − mω 2
θ
bω
From the triangle we see:
sin(θ) = p
k − mω 2
(k −
mω 2 )2
+
b2 ω 2
cos(θ) = p
bω
(k − mω 2 )2 + b2 ω 2
Making:
k − mω 2 =
Now our solution is:
p
(k − mω 2 )2 + b2 ω 2 sin(θ)
bω =
p
(k − mω 2 )2 + b2 ω 2 cos(θ)
2.8. EVERYONE LOVES A SLINKY: SPRINGS
97
p
p
F0
2
2
2
2
2
2
2
2
(k − mω ) + b ω sin(θ) cos(ωt) + (k − mω ) + b ω cos(θ) sin(ωt)
yp =
(k − mω 2 )2 + b2 ω 2
yp = p
F0
(k − mω 2 )2 + b2 ω 2
(sin(θ) cos(ωt) + cos(θ) sin(ωt))
After a trig identity we get:
F0
yp = p
sin(ωt + θ)
(k − mω 2 )2 + b2 ω 2
So the general solution is:
y = yh + yp = Ae
b
− 2m
t
√
sin
4mk − b2
F0
sin(ωt + θ)
t+φ + p
2m
(k − mω 2 )2 + b2 ω 2
The first term in the solution yh tends to zero as t tends to infinity. So we refer to this term as the transient solution.
As t gets large and yh approaches zero the solution approaches the particular solution yp . Hence we call this term the
steady state solution.
The factor:
1
p
(k − mω 2 )2 + b2 ω 2
in the particular solution represents the ratio of the magnitude of the forcing function F0 to the magnitude of the
sinusoidal response to the input force so we call it: the frequency gain and has units length/force.
An Example:
An 8-kg mass is attached to a spring hanging from the ceiling causing the spring to stretch 1.96m upon coming to
rest at equilibrium. At t = 0 the forcing function F (t) = cos(2t) is applied to the system. The damping constant for the
system is 3 N-sec/m. Find the steady state solution and the frequency gain.
Solution:
At t = 0 the system is equilibrium so it has an initial position y(0) = 0 and initial velocity y 0 (0) = 0
First we need the spring constant k. Since the 8-kg mass stretched the spring 1.96m Hooks Law gives:
8(9.8) = k(1.96)
k = 40
The differential equation governing the system is:
8y 00 + 2y 0 + 40y = cos(2t)
We could use the method of undetermined coefficients to find the steady state solution but we have already derived
equations for it:
F0
yp = p
sin(ωt + θ)
(k − mω 2 )2 + b2 ω 2
tan(θ) =
Plugging in the values for m, b, k, F0 , θ and ω gives:
tan(θ) =
And the steady state solution is:
1
2
θ = .463648
k − mω 2
bω
98
CHAPTER 2. SECOND ORDER EQUATIONS.
1
yp = √ sin(2t + .463648)
80
The frequency gain is:
1
p
(k − mω 2 )2 + b2 ω 2
=
1
80
393.
An 8-kg mass is attached to a spring hanging from the ceiling causing the spring to stretch 7.84m upon coming to
rest at equilibrium. At t = 0 the forcing function F (t) = 2 cos(2t) is applied to the system. The damping constant for
the system is 1 N-sec/m. Find the steady state solution and the frequency gain.
394.
An 2-kg mass is attached to a spring hanging from the ceiling causing the spring to stretch .2m upon coming to rest
at equilibrium. At t = 0 the mass is displaced .005m below equilibrium and released. At t = 0 the forcing function
F (t) = .3 cos(t) is applied to the system. The damping constant for the system is 5 N-sec/m. Find the steady state
solution and the frequency gain.
395.
Although we cannot solve the following differential equation
y 00 + et y 0 + y = 0
we can determine the limiting behavior of the solution by thinking about the differential equation in terms of a spring
equation.
Find
lim y(t)
t→∞
For a given mass spring system with known mass: m, damping constant: b and spring constant k with a forcing
function F0 cos(ωt) we know the frequency gain: M (ω), will be a be a function of the variable ω and is given by:
1
M (ω) = p
(k − mω 2 )2 + b2 ω 2
Often times we want to know the value of ω that maximizes the frequency gain. Differentiating gives:
M 0 (ω) =
−3
−1
((k − mω 2 )2 + b2 ω 2 ) 2 (2(k − mω 2 )(−2mω) + 2b2 ω)
2
Simplifying:
0
M (ω) = ω
−2m2 ω 2 − b2 + 2km
3
(k − mω 2 )2 + b2 ω 2 ) 2
So M has critical numbers:
ω = 0 in this case the forcing function is constant
The critical number we care about is:
2.9. CIRCUITS
99
r
ω=
k
b2
−
m 2m2
And the maximum frequency gain occurs when:
r
k
b2
−
m 2m2
The maximum Amplitude of the steady state response is:
ω=
F0 M (ω) = F0 p
1
(k − mω 2 )2 + b2 ω 2
396.
Let the following differential equation govern the motion of a mass on a spring.
1 00
y + by 0 + 10y = 3 cos(2t)
2
Find the value of b that maximizes the amplitude of the steady state response.
2.9
Circuits
We now turn our study to the circuit consisting of a resister whose letter representation is R and is measured in Ohms, a
capacitor whose letter representation is C and the inductor whose letter representation is L connected to a voltage source
whose letter representation is E.
This circuit will be governed by Kirchhoff’s loop rules. Here they are:
1. The sum of the currents flowing into any junction point are zero.
2 The sum of the voltage around any closed loop is zero.
From physics we can find the voltage drop by the resistor, capacitor and the inductor. Here they are
1. The voltage drop across a resistor is given by
ER = IR
where I is the current passing through the resistor
2. The voltage drop across a capacitor is
100
CHAPTER 2. SECOND ORDER EQUATIONS.
1
Q
C
3 The voltage drop across an inductor is
EC =
where Q is the charge on the capacitor
dI
dt
If a voltage source is connected to the circuit an adds voltage at a level of E(t) then Kirchhoff’s voltage law gives:
EL = L
EL + ER + EC = E(t)
Or
1
dI
+ RI + Q = E(t)
dt
C
Since the change in charge is the current we have
L
dQ
=I
dt
Making
d2 Q
dI
= 2
dt
dt
Now our differential equation becomes
d2 Q
dQ
1
+R
+ Q = E(t)
2
dt
dt
C
Sometimes we want to determine the current I(t) in the circuit so we differentiate the above equation to get:
L
L
d2 I
dI
1
+R
+ I = E 0 (t)
2
dt
dt
C
397.
A RLC circuit has a voltage source given by E(t) = 40 cos(2t)V a resistor of 2 ohms, a inductor of .25 henrys and a
capacitor of
1
13
farads. If the initial current is zero and the initial charge on the capacitor is 3.5 coulombs, determine the
charge on the capacitor as a function of t.
398.
A RLC circuit with no voltage source has a resistor of 20 ohms, a inductor of .1 henrys and a capacitor of
1
25
farads.
If the initial current is zero and the initial charge on the capacitor is 10 coulombs, determine the charge on the capacitor
as a function of t.
399.
A RLC circuit has a voltage source given by E(t) = 40V a resistor of 10 ohms, a inductor of .2 henrys and a capacitor
of
1
13
of t.
farads. If the initial current is zero and the initial charge on the capacitor is 0, determine the current as a function
Chapter 3
Series Solution
In calculus we learned that all continuously differentiable function can be represented by a Taylor series. The Taylor
series for a function centered at x = c is:
f (x) = f (c) + f 0 (c)(x − c) +
f 000 (c)(x − c)3
f (4) (c)(x − c)4
f 00 (c)(x − c)2
+
+
+ ...
2!
3!
4!
If the series is centered at zero then we call it a Mclauren series.
Some common power series are:
ex = 1 + x +
sin(x) = x −
∞
X
x2
x3
x4
xn
+
+
+ ... =
2!
3!
4!
n!
n=0
∞
X
x5
x7
(−1)n x2n+1
x3
+
−
+ ... =
3!
5!
7!
(2n + 1)!
n=0
cos(x) = 1 −
∞
X
x2
x4
x6
(−1)n x2n
+
−
+ ... =
2!
4!
6!
(2n)!
n=0
∞
X
1
= 1 + x + x2 + x3 + ... =
xn
1−x
n=0
arctan(x) = x −
3.1
∞
X
x3
x5
x7
(−1)n x2n+1
+
−
=
3
5
7
2n + 1
n=0
Series Solutions Around Ordinary Points
In this chapter we will not be looking for some equation f (x) that is the solution to a differential equation, instead we
will be looking for its Power Series (normally a Taylor Series). For the case of the Taylor Series we will look for a solution
to the differential equation of the form:
y=
∞
X
an xn
n=0
101
102
CHAPTER 3.
SERIES SOLUTION
with the sequence an to be determined by substituting y, y 0 and y 00 into the differential equation and developing a
recurrence relation for an and solving the recurrence relation for a formula for an .
An Example:
Find at least the first seven terms in the power series that is a solution to the differential equation
y 00 + 3xy 0 + 2y = 0
We will look for a solution of the form:
y=
∞
X
an xn
making
y0 =
n=0
0
∞
X
nan xn−1
y 00 =
n=1
∞
X
n(n − 1)an xn−2
n=2
00
Substituting y, y and y into the differential equation produces:
∞
X
n(n − 1)an xn−2 + 3
n=2
∞
X
n(n − 1)an xn + 2
n=1
∞
X
an xn = 0
n=0
I will now shift the first series in the above expression so that it too has an xn factor:
∞
X
n
(n + 2)(n + 1)an+2 x + 3
n=0
∞
X
n
nan x + 2
n=1
∞
X
a n xn = 0
n=0
Now I will add the zero term in the first and third series so that all indexes will be n = 1
2a2 + 2a0 +
∞
X
(n + 2)(n + 1)an+2 xn + 3
∞
X
nan xn + 2
an xn = 0
n=1
n=1
n=1
∞
X
n
Now that all three series have an x factor and all 3 indexes are the same: n = 1, we can combine the three series
into one series:
∞ X
2a2 + 2a0 +
(n + 2)(n + 1)an+2 + (3n + 2)an xn = 0
n=1
Setting the constant term on the left: 2a2 + 2a0 equal to the constant term on the right: 0 and setting the coefficient
on xn on the left: (n + 2)(n + 1)an+2 + (3n + 2)an equal to the coefficient of xn on the right: 0 gives:
2a2 + 2a0 = 0
(n + 2)(n + 1)an+2 + (3n + 2)an = 0
a2 = −a0 and our recurrence relation is: an+2 =
−(3n + 2)an
(n + 2)(n + 1)
Substituting n = 1, 2, ...5 in to the recurrence relation gives:
a3 =
−5
a1
6
a4 =
2
a0
3
a5 =
11
a1
24
a6 =
−14
a0
45
The solution is
5
2
11
14
y = a0 + a1 x − a0 x2 − a1 x3 + a0 x4 + a1 x5 − a0 x6
6
3
24
45
Separating this solution into two linearly independent solutions:
3.1. SERIES SOLUTIONS AROUND ORDINARY POINTS
103
2
14
5
11
y = a0 1 − x2 + x4 − x6 + a1 x − x3 + x5
3
45
6
24
Since a pattern in the terms cannot be found we shall leave the solution as a sixth degree polynomial.
Sometimes a pattern can be found and the solution can be written much more concisely. Consider the following
example:
Example: Solve using power series about x = 0:
(x + 1)y 00 − 2xy 0 − 4y = 0
First lets note that y = e2x is a solution. We will try to find this solution using power series. The series expansion for
e2x that we are trying to obtain is:
e2x =
∞
X
2n xn
4
2
4
= 1 + 2x + 2x2 + x3 + x4 + x5 ...
n!
3
3
15
n=0
We will look for a solution of the form:
y=
∞
X
an xn
making
y0 =
n=0
0
∞
X
nan xn−1
y 00 =
n=1
∞
X
n(n − 1)an xn−2
n=2
00
Substituting y, y and y into the differential equation produces:
∞
X
n(n − 1)an xn−1 +
∞
X
n(n − 1)an xn−2 − 2
nan xn − 4
n=1
n=2
n=2
∞
X
∞
X
an xn = 0
n=0
n
Shifting the first and second series so that they each have an x factor.
∞
X
(n + 1)(n)an+1 xn +
n=1
∞
X
(n + 2)(n + 1)an+2 xn − 2
n=0
∞
X
n=1
nan xn − 4
∞
X
an xn = 0
n=0
Adding the n = 0 term in he second and fourth series so that all series start with an index of n = 1 and condensing
into one series give:
∞ X
2a2 − 4a0 +
(n + 1)(n)an+1 + (n + 2)(n + 1)an+2 − 2nan − 4an xn = 0
n=1
2a2 − 4a0 = 0
(n + 1)(n)an+1 + (n + 2)(n + 1)an+2 − 2nan − 4an = 0
a2 = 2a0
an+2 =
2(n + 2)an − n(n + 1)an+1
(n + 2)(n + 1)
So our recurrence relation simplifies to
an+2 =
2
n
an −
an+1
n+1
n+2
104
CHAPTER 3.
1
2
a3 = a1 − a2 = a1 − a0
3
3
a4 =
2
1
5
1
a2 − a3 = a0 − a1
3
2
3
2
4
a5 = a1 − a0
3
a6 =
14
4
a0 − a1
9
5
SERIES SOLUTION
As of now our solution is:
1
12
5
4
14
2
3
4
5
a0 − a1 x + a1 − a0 x +
a0 − a1 x6
y = a0 + a1 x + 2a0 x + a1 − a0 x +
3
3
2
3
9
15
2
Or
2
5
4
14
1
4
y = a0 1 + 2x2 − x3 + x4 − x5 + x6 + a1 x + x3 − x4 + x5 − x6
3
3
3
9
2
5
This solution does not look like the series expansion for e2x that we are expecting. Noticing the first and third terms
in the first linearly independent solution: 1 + 2x2 match the first and third terms in the series but we seem to be missing
the second term: 2x. I will rewrite the linear term in the second linearly independent solution so that it has a 2x term
in it. I will do this by letting
a1 = 2a0 + K
Making
1
4
a3 = a1 − a2 = a0 + K
3
3
a4 =
2
1
2
1
a2 − a3 = a0 − K
3
2
3
2
4
4
4
a5 = a1 − a0 =
a0 + K
3
15
5
Now our solution is
2
y = a0 + (2a0 + K)x + 2a0 x +
2
4
1
4
4
3
4
a0 + K x +
a0 − K x +
a 0 + K x5
3
3
2
15
5
Separating this into two linearly independent solutions gives
4 3 2 4
4 5
1 4 4 5
2
3
y = a0 1 + 2x + 2x + x + x + x + K x + x − x + x
3
3
15
2
5
Now the first linearly independent solution has the expansion for e2x so our solution simplifies to
y = a0 e
2x
1
4
+ K x + x − x4 + x5
2
5
3
Now that we have one solution we normally would apply the reduction of order algorithm to find the second solution
but the integral it produces is quite unpleasant so we must accept the series expansion as our second linearly independent
solution.
400.
Find the first six terms of the solution to the differential equation:
y 00 + xy 0 + y = 0
3.1. SERIES SOLUTIONS AROUND ORDINARY POINTS
401.
Find the first six terms of the solution to the differential equation:
(x + 1)y 00 − xy 0 − y = 0
402.
Find the first six terms of the solution to the differential equation:
y 00 + (3x − 1)y 0 − y = 0
403.
Find the first six terms of the solution to the differential equation:
(2x + 3)y 00 − (4x + 8)y 0 + 4y = 0
404.
Find the first six terms of the solution to the differential equation:
(3x + 4)y 00 − (3x + 4)y 0 + (2x + 3)y = 0
405.
Find the first six terms of the solution to the differential equation:
(x2 − 2x)y 00 + (2 − x2 )y 0 + (2x − 2)y = 0
406.
Find the first six terms of the solution to the differential equation:
y 00 −
407.
x
1
y0 +
y=0
x−1
x−1
Find the first six terms of the solution to the differential equation:
4xy 00 + 3y 0 + xy = 0
408.
Find the first six terms of the solution to the differential equation:
(x2 + 1)y 00 + y = 0
409.
Find the first six terms of the solution to the differential equation:
(2x + 1)y 00 − 2y 0 − (2x + 3)y = 0
410.
Find the first six terms of the solution to the differential equation:
(x2 + 2)y 00 + xy 0 − y = 0
105
106
CHAPTER 3.
SERIES SOLUTION
411.
Find the first 5 terms in y1 in the series expansion about x = 0 for a solution to:
y 00 − 2xy 0 − 2y = 0
412.
Find the first 5 terms in y1 in the series expansion about x = 0 for a solution to:
y 00 + xy 0 − (4 + 2x)y = 0
413.
Show that the solutions to
y 00 = y 0 + y
is given by
∞
X
Fn xn
n!
n=0
where Fn is the nth Fibonacci number
414.
Use the power series for sin(x), cos(x) and ex to derive Euler’s Equation:
eix = cos(x) + i sin(x)
415.
Use your knowledge of power series to evaluate
∞
X
(−1)n
2n + 1
n=0
3.2
Method of Frobenius:
We will now try to find a power series solution to the differential equation:
y 00 + P (x)y 0 + Q(x)y = 0
centered at a point where either P (x) or Q(x) is not analytic. To motivate our procedure lets reconsider the Cauchy
Euler equation.
x2
d2 y
dy
+ ax
+ by = 0
2
dx
dx
After dividing by x2 we get
d2 y
a dy
b
+
+ y=0
dx2
x dx x2
So both
3.2. METHOD OF FROBENIUS:
107
P (x) =
a
x
and
Q(x) =
b
x2
are not analytic at x = 0.
The Cauchy Euler equation has a characteristic equation
r(r − 1) + ar + b = 0
which can be created by the following equation
r(r − 1) + xP (x) + x2 Q(x) = 0
This equation is quite similar to the characteristic equation we use in the method of Frobenous.
First a definition:
x = x0 is a Singular Point of the above differential equation if either P (x) or Q(x) is not analytic at x = x0 . x = x0
is a Regular Singular Point if the following limits exist:
q0 = lim (x − x0 )2 Q(x)
p0 = lim (x − x0 )P (x)
x→x0
x→x0
To find a series solution about a regular singular point we use the Method of Frobenius. In the Method of Frobenius
we look for a series solution of the form:
y=
∞
X
an x
n+r
making
0
y =
∞
X
(n + r)an x
n+r−1
00
y =
(n + r)(n + r − 1)an xn+r−2
n=0
n=0
n=0
∞
X
The values of r that we will use are the roots of the Indicial Equation:
r(r − 1) + p0 r + q0 = 0
There will be three cases we will have to consider based on the roots: r1 and r2 to the indical equation. Note: we
always take r1 to be the larger of the two roots. The first case we will consider is the case when the roots differ by a non
integer. That is:
r1 − r2 6∈ Z
An Example:
Find the first 6 terms in the series expansion for the solution to the following differential equation centered at x = 0.
2x(x − 1)y 00 + 3(x − 1)y 0 − y = 0
In standard form this equation is:
y 00 +
p0 = lim x
x→0
The indical equation is:
−1
3 0
y +
y=0
2x
2x(x − 1)
3
3
=
2x
2
q0 = lim x2
x→0
−1
=0
2x(x − 1)
108
CHAPTER 3.
3
r(r − 1) + r = 0
2
The roots are r1 = 0 and r2 =
SERIES SOLUTION
1
=0
r r+
2
or
−1
2 .
Inserting:
y=
∞
X
an xn+r
making
y0 =
n=0
∞
X
y 00 =
(n + r)an xn+r−1
n=0
∞
X
(n + r)(n + r − 1)an xn+r−2
n=0
into the original differential equation gives:
2
∞
X
(n+r)(n+r−1)an xn+r −2
n=0
∞
X
(n+r)(n+r−1)an xn+r−1 +3
n=0
∞
X
(n+r)an xn+r −3
n=0
∞
X
(n+r)an xn+r−1 −
n=0
∞
X
an xn+r = 0
n=0
Shifting the first, third and fifth summation so that all series will have an xn+r−1 factor:
2
∞
X
(n + r − 1)(n + r − 2)an−1 xn+r−1 − 2
n=1
3
∞
X
∞
X
(n + r)(n + r − 1)an xn+r−1 +
n=0
(n + r − 1)an−1 xn+r−1 − 3
n=1
∞
X
(n + r)an xn+r−1 −
n=0
∞
X
an−1 xn+r−1 = 0
n=1
Adding the n = 0 terms in the second and forth series and condensing into a single series gives:
(−2r(r−1)a0 −3ra0 )xr−1 +
∞ X
2(n+r−1)(n+r−2)an−1 −2(n+r)(n+r−1)an +3(n+r−1)an−1 −3(n+r)an −an−1 xn+r−1 = 0
n=1
Setting the coefficients of the polynomial on the left hand side equal to zero: the coefficients of the polynomial on the
right hand side. We see that the coefficient of xr−1 is zero at both r1 = 0 and r2 =
zero. We also get our recurrence relation from setting the coefficient of x
n+r−1
−1
2
so a0 is a free is not necessarily
to zero.
2(n + r − 1)(n + r − 2)an−1 − 2(n + r)(n + r − 1)an + 3(n + r − 1)an−1 − 3(n + r)an − an−1 = 0
Solving for an
an = an−1
1 − 3(n + r − 1) − 2(n + r − 1)(n + r − 2)
−2(n + r)(n + r − 1) − 3(n + r)
We will find two linearly independent solutions by substituting our two values of r into the recurrence relation. For
r1 = 0
an = an−1
1 − 3(n − 1) − 2(n − 1)(n − 2)
−2n(n − 1) − 3n
Which simplifies to:
an =
2n − 3
an−1
2n + 1
To obtain the first 6 terms in the solution we will find the first 3 terms in y1 using the above recurrence relation
created by using r1 = 0 and the first 3 terms in y2 using the using a recurrence relation created by using r2 =
first 3 terms in this recurrence relation will have coefficients a0 , a1 and a2 . They are:
−1
2 .
The
3.2. METHOD OF FROBENIUS:
109
a1 =
−1
a0
3
1
1 −1
−1
a1 =
a0 =
a0
5
5 3
15
a2 =
Using r1 = 0 the first 3 terms in y1 are:
y1 = a0 + a1 x + a2 x2
y1 = a0
x x2
1− −
3
15
Since this is a second order differential equation we would expect the form of the solution to be:
y = C1 y1 + C2 y2
In our answer C1 is the constant a0 .
To find y2 we insert r2 =
−1
2
into the recurrence relation:
an =
1 − 3(n − 32 ) − 2(n − 32 )(n − 25 )
−2(n − 12 )(n − 23 ) − 3(n − 12 )
After a bit of simplifying we arrive at our recurrence relation:
an =
2n − 1
an−1
n
The first three terms in this recurrence relation will have coefficients a0 , a1 and a2 . They are:
a1 = a0
a2 =
y2 = a0 x
y2 = a0 x
−1
2
−1
2
3
3
a1 = a0
2
2
1
3
+ a1 x 2 + a2 x 2
1
3
3
+ a0 x 2 + a0 x 2
2
−1
1
3 3
y2 = a0 x 2 + x 2 + x 2
2
An Example:
Solve the following differential equation by method of Forbenious.
4xy 00 + 2y 0 + y = 0
In standard form our equation is:
y 00 +
1
1 0
y +
y=0
2x
4x
p0 = lim x
x→0
1
1
=
2x
2
110
CHAPTER 3.
SERIES SOLUTION
1
=0
4x
q0 = lim x2
x→0
Our indical equation becomes:
1
r(r − 1) + r = 0
2
Which has roots r1 =
1
2
and r2 = 0
Assume a solution to the differential equation of the form:
y=
∞
X
an xn+r
n=0
Therefore:
y0 =
∞
X
∞
X
y 00 =
(n + r)an xn+r−1
n=0
(n + r)(n + r − 1)an xn+r−2
n=0
Inserting the series for y, y’ and y” into our differential equation yields the following:
∞
X
4(n + r)(n + r − 1)an xn+r−1 +
n=0
∞
X
2(n + r)an xn+r−1 +
n=0
∞
X
an xn+r = 0
n=0
Shifting the last series on the LHS of the above equation yields:
∞
X
4(n + r)(n + r − 1)an x
n+r−1
+
∞
X
2(n + r)an x
n+r−1
+
an−1 xn+r−1 = 0
n=1
n=0
n=0
∞
X
Adding the n = 0 terms in the first two series and then combining the rest into one series yields:
(4r(r − 1)a0 + 2ra0 )xr−1 +
∞
X
[4(n + r)(n + r − 1)an + 2(n + r)an + an−1 ]xn+r−1 = 0
n=1
This gives the recurrence relation:
4(n + r)(n + r − 1)an + 2(n + r)an + an−1 = 0
Which simplifies to:
an =
For r1 =
1
2
−an−1
(2n + 2r)(2n + 2r − 1)
our recurrence relation becomes:
an =
−an−1
(2n + 1)(2n)
Substituting values of n into our recurrence relation yields:
a1 =
−a0
3·2
a3 =
a2 =
−a1
a0
=
5·4
5·4·3·2
−a2
−a0
=
7·6
7·6·5·4·3·2
3.2. METHOD OF FROBENIUS:
111
The solution to this recurrence relation is:
an =
(−1)n · a0
(2n + 1)!
Making the solution:
y1 =
∞
X
√
(−1)n n+1/2
x
= sin( x)
(2n + 1)!
n=0
For r2 = 0 our recurrence relation becomes:
an =
−an−1
(2n)(2n − 1)
Substituting values of n into our recurrence relation yields:
a1 =
−a0
2·1
a3 =
a2 =
−a1
a0
=
4·3
4·3·2
−a0
−a2
=
6·5
6·5·4·3·2
The solution to this recurrence relation is:
an =
(−1)n · a0
(2n)!
Making the solution:
y2 =
∞
X
√
(−1)n n
x = cos( x)
(2n)!
n=0
416.
Determain if x = 0 is a regular or iregular singular point. If x = 0 is a regular singular point find the indicial equation
and its roots.
00
y +
e2x − 2x − 1 0
sin(4x)
y ++
y=0
xex − x
(x + sin(x))2
417.
Determain if x = 0 is a regular or iregular singular point. If x = 0 is a regular singular point find the indicial equation
and its roots.
2 0
arctan(x)
y ++
y=0
y + x sin
x
x
00
418.
Determain if x = 0 is a regular or iregular singular point. If x = 0 is a regular singular point find the indicial equation
and its roots.
y 00 +
1
1
arcsin(x)
−
y0 + +
y=0
x sin(x)
ln(x + 1)
419.
Solve by method of Frobenious. You should be able to recognise a pattern and identify the power series as a known
function.
112
CHAPTER 3.
SERIES SOLUTION
2xy 00 + (x + 1)y 0 + y = 0
420.
Solve by method of Frobenious. You should be able to recognise a pattern and identify the power series as a known
function.
y 00 +
2 0
y +y =0
x
421.
Find the first 5 terms in y1 in the series expansion about x = 0 for a solution to:
x2 y 00 − xy 0 + (1 − x)y = 0
422.
Find the first 5 terms in y1 in the series expansion about x = 0 for a solution to:
(x2 − 2x)y 00 + (2 − x2 )y 0 + (2x − 2)y = 0
423.
Find the first 5 terms in y1 in the series expansion about x = 0 for a solution to:
(6x2 + 2x)y 00 + (x + 1)y 0 − y = 0
424.
Find the first 5 terms in y1 in the series expansion about x = 0 for a solution to:
xy 00 + y 0 + xy = 0
425.
Find the first 5 terms in y1 in the series expansion about x = 0 for a solution to:
xy 00 + y 0 − 4y = 0
426.
Find the first 5 terms in y1 in the series expansion about x = 0 for a solution to:
xy 00 + 2y 0 + xy = 0
427.
Show that one solution to
x2 y 00 = (a2 − a + bx)y
is
y =x 1+
a
bx
(bx)2
(bx)3
+
+
+ ...
1! · 2a 2! · (2a)(2a + 1) 3! · (2a)(2a + 1)(2a + 2)
3.3. THE GAMMA FUNCTION
3.3
113
The Gamma Function
As you have seen many of our series solutions involve factorials. The problem factorials is that they are only defined for non
negative integers. The gamma function extend the idea of factorials to all positive real numbers. The Gamma Function
is given by the improper integral:
∞
Z
e−t tx−1 dt
Γ(x) =
x>0
0
The first important property that we will prove of the gamma function shows its similarities to the factorial function:
Thm:
Γ(x + 1) = xΓ(x)
Proof:
Z
∞
e−t tx dt
Γ(x + 1) =
0
Using integration by parts
u = tx
dv = e−t dt
v = −e−t
du = xtx−1 dt
b
Z
Γ(x + 1) = lim −t e + x
b→∞
b
x −t 0
tx−1 e−t dt
0
The first term is zero at both zero and infinity so we hvae
Z
Γ(x + 1) = x
∞
e−t tx−1 dt = xΓ(x)
0
This theorem demonstrates the relationship with the factorial function. The next result is a direct corollary to the
above theorem and is the most well known property of the gamma function.
If n is a positive integer
Γ(n) = (n − 1)!
Example:
Calculate Γ( 12 )
Z ∞
Z ∞ −t
−1
1
e
√ dt
Γ
=
e−t t 2 dt =
2
t
0
0
Substituting
u=
√
t
dt
2du = √
t
when t = 0 u = 0
Z ∞
2
1
Γ
=2
e−u du
2
0
as t → ∞ u → ∞
114
CHAPTER 3.
SERIES SOLUTION
2
Although we cannot find an antiderivative for e−u we can evaluate this integral with a clever trick
Z ∞
Z ∞
2
1
−u2
=2
e
du = 2
e−v dv
Γ
2
0
0
So
Z ∞
2 Z ∞
Z ∞Z ∞
2
2
1
−u2
−v 2
= 2
e
e−u −v dudv
du · 2
e dv = 4
Γ
2
0
0
0
0
Converting to polar coordinates
2
Z π2 Z ∞
2
1
e−r rdrdθ
Γ
=4
2
0
0
b
2
Z π2
Z π2
1
−r 2 Γ
= −2
lim e dθ = 2
dθ = π
2
0 b→∞
0
0
So
√
1
Γ
= π
2
428.
Find
3
Γ
2
429.
Show
lim Γ(x) = ∞
x→0+
430.
Use the principle of mathematical induction to prove:
√
1
(2n)! π
Γ n+
=
2
4n n!
n = 0, 1, 2, ...
431.
Show that the definition of the Gamma function is equivalent to Euler’s original definition:
Z
Γ(x) =
0
1
x−1
1
ln
dt
t
432.
Another important function related to the Gamma function is Euler’s Psi function: the derivative of the logarithm of
the Gamma function.
d
Γ0 (x)
ψ(x) =
ln Γ(x) =
dx
Γ(x)
3.4. BESSEL’S EQUATION
115
Show the following property of the Psi function
ψ(x + 1) =
1
+ ψ(x)
x
Use the above results to also show for positive integers n
ψ(n + 1) = ψ(1) +
n
X
1
k
k=1
3.4
Bessel’s Equation
We will now consider a differential equation of great importance in applied mathematics. The Bessel Equation of order
v is:
x2 y 00 + xy 0 + (x2 − v 2 )y = 0
or in standard form:
y 00 +
1 0
v2
y + 1− 2 y =0
x
x
We see that x = 0 is a singular point of the Bessel Equation. Since both
P0 = lim x
x→0
1
=1
x
v2
Q0 = lim x2 1 − 2 = −v 2
x→0
x
And
exist x = 0 is a regular singular point and our indical equation is
r(r − 1) + r − v 2 = 0
Or
r2 − v2 = 0
Which has roots
r = ±v
If the difference between the roots: 2n is not an integer then the method of Frobenius gives two linearly independent
solutions:
Jv =
∞
X
(−1)n x2n+v
22n+v n!Γ(1 + v + n)
n=0
and
J−v =
∞
X
n=0
(−1)n x2n−v
− v + n)
22n−v n!Γ(1
An Example:
Solve the Bessel equation of order
1
2
1
2
x y + xy + x −
=0
4
2 00
0
116
CHAPTER 3.
SERIES SOLUTION
In standard form we have
1 0
1
y + y + 1− 2 y =0
x
4x
00
We see that x = 0 is a singular point. To show it is a regular singular point we compute:
1
P0 = lim x = 1
x→0 x
Q0 = lim x
2
x→0
1
1− 2
4x
=
−1
4
Since both limits exist we see that x = 0 is a regular singular point and we can use the method of Frobenius to find a
series solution about x = 0. The indical equation and roots are:
r(r − 1) + r −
y=
∞
X
y0 =
an xn+r
n=0
∞
X
1
=0
4
(n + r)an xn+r−1
r=±
y=
n=0
0
∞
X
1
2
(n + r)(n + r − 1)an xn+r−2
n=0
00
Inserting the series for y, y and y into the original differential equation gives:
∞
X
(n + r)(n + r − 1)an xn+r +
n=0
∞
X
(n + r)an xn+r +
n=0
∞
X
an xn+r+2 −
∞
X
1
an xn+r = 0
4
n=0
an−2 xn+r −
∞
X
1
an xn+r = 0
4
n=0
n=0
Shifting the third series so that it too has an exponent of n + r
∞
X
(n + r)(n + r − 1)an xn+r +
∞
X
(n + r)an xn+r +
n=2
n=0
n=0
∞
X
Adding the first and second terms in the first, second and fourth series will produce four series all with an index
starting at n = 2 that can be simplified into a single series.
∞ X
1
1
1
r
1+r
(n + r)(n + r − 1) + (n + r) −
r(r − 1) + r −
a0 x + (1 + r)r + (1 + r) −
a1 x
+
an + an−2 xn+r = 0
4
4
4
n=2
The coefficient of a0 xr is identical to the indical equation and is zero at both values of r making a0 not zero whereas
the coefficient of a1 x1+r is not zero at both values of r making a1 = 0. Our recurrence relation is:
(n + r)(n + r − 1) + (n + r) −
an =
For the larger root r =
1
2
−an−2
(n + r)2 −
1
an + an−2 = 0
4
1
4
our recurrence relation simplifies to:
an =
−an−2
n(n + 1)
Producing a sequence of terms in an
−a0
−a0
−a1
=
a3 =
=0
2·3
3!
4·3
Since a5 depends on a3 it too must be zero. In fact an = 0 for all odd values of n
a2 =
3.4. BESSEL’S EQUATION
117
a4 =
−a2
a0
a0
=
=
5·4
5·4·3·2
5!
a6 =
−a4
−a0
=
7·6
7!
Recognizing the pattern we see
a2n =
(−1)n a0
(2n + 1)!
And the first solution is
y1 =
∞
∞
∞
X
X
(−1)n a0 x2n+1
(−1)n a0 2n+ 1
1 X (−1)n a0 2n+1
2 =
√
x
=√
x
(2n + 1)!
(2n + 1)!
x
x n=0 (2n + 1)!
n=0
n=0
Recognizing the series gives:
cos(x)
y1 = J −1 (x) = a0 √
2
x
Since r1 − r2 is an integer the second linearly independent solution would be quite difficult to find using series, but
we were able to find a closed form expression for y1 =
cos(x)
√
x
so we can use the method of reduction of order form chapter
2 to find the second linearly independent solution.
y 00 +
Z
v=
e−
R
1 0
1
y + 1− 2 y =0
x
4x
1
x dx
cos2 (x)
x
Z
dx =
sec2 (x)dx = tan(x)
Our second linearly independent solution is
y2 = y1 v =
cos(x)
√ tan(x)
x
sin(x)
y2 = J 12 (x) = √
x
And the homogenous solution is:
cos(x)
sin(x)
y = C1 √
+ C2 √
x
x
A few important properties of the solutions to the Bessel equation are:
d
v
x Jv (x) = xv Jv−1 (x)
dx
Jv+1 (x) =
2v
Jv (x) − Jv−1 (x)
x
d
−v
x Jv (x) = −x−v Jv+1 (x)
dx
Jv+1 (x) = Jv−1 (x) − 2Jv0 (x)
433.
The Bessel functions of order n +
1
2
for integer n are related to the spherical Bessel functions. Use one of the above
properties of the Bessel function and the results of the example to calculate J 23 and J 25
118
CHAPTER 3.
434.
Show the orthogonality of the Bessel functions by showing
1
Z
0
435.
Show that the substitution z =
√
xJ −1 J 12 dx = 0
2
xy transforms the Bessel equation to normal form:
1 − 4v 2
z=0
z + 1+
4x2
00
436.
The parametric Bessel equation is:
x2 y 00 + xy 0 + (λ2 x2 − v 2 )y = 0
x>0
Show
y = C1 Jv (λx) + C2 J−v (λx)
v∈
/Z
is the solution
437.
Use the results from the previous problem and to find the solution to
1
x2 y 00 + xy 0 + 25x2 −
y=0
4
x>0
438.
Use ideas from the previous two problems and the fact ı2 x2 = −x2 to solve
1
y=0
x2 y 00 + xy 0 − x2 +
4
439.
Use the change of variable y = x
−1
2
v(x) to solve
x2 y 00 + 2xy 0 + λ2 x2 y = 0
x>0
SERIES SOLUTION
Chapter 4
Laplace Transform
4.1
Calculating Laplace and Inverse Laplace Transforms
The Laplace Transform of a function f (x) will be a function of s given by the improper integral:
Z
L(f ) =
∞
f (x)e−sx dx
0
provided the integral converges. Functions that outgrow ekx , for constant k, do not have a Laplace Transform since
n
the integral diverges. So f (x) = ex will not have a Laplace Transform for n > 1. Let us now try to find the Laplace
Transform of f (x) = eax .
ax
Z
L(e ) =
∞
ax −sx
e e
Z
dx =
0
∞
e(a−s)x dx
0
b
1
1
e(a−s)b
e(a−s)x −
=
= lim
b→∞ a − s
b→∞ a − s 0
a−s
s−a
= lim
For s > a.
Here is a basic table of laplace transforms:
119
120
CHAPTER 4.
f (x)
eax
sin(ax)
cos(ax)
n
x
n∈N
ax n
e x
n∈N
eax − ebx
eax sin(bx)
eax cos(bx)
x sin(ax)
x cos(ax)
sinh(ax)
cosh(ax)
r
x r ∈√
R r > −1
x
√1
x
1
LAPLACE TRANSFORM
F(s) = L(f (x))
1
s−a
a
s2 +a2
s
s2 +a2
n!
sn+1
n!
(s−a)n+1
a−b
(s−a)(s−b)
b
(s−a)2 +b2
s−a
(s−a)2 +b2
2as
(s2 +a2 )2
2
s −a2
(s2 +a2 )2
a
s2 −a2
s
s2 −a2
Γ(r+1)
r+1
s√
π
3
2s
√2
π
√
s
√
1·3·5...(2n−1) π
xn− 2 n ∈ N
1
2n sn+ 2
4a3
sin(ax) cosh(ax) − cos(ax) sinh(ax)
s4 +4aa
2a2 s
sin(ax) sinh(ax)
s4 +4a4
2a3
sinh(ax) − sin(ax)
s4 −a4
2a2 s
cosh(ax) − cos(ax)
s4 −a4
2a3
sin(ax) − ax cos(ax)
(s2 +a2 )2
2as2
sin(ax) + ax cos(ax)
(s2 +a2 )2
−2a(a2 −3s2 )
x2 sin(ax)
(s2 +a2 )3
2s(s2 −3a2 )
2
x cos(ax)
(s2 +a2 )3
ax
e g(x)
G(s − a) G(s)
= L(g(x))
xn g(x)
n
d
(−1)n ds
L(g(x))
n
Notice there is not a Laplace Transform for any function that outgrows f (x) = eax . This is because the improper
integral from the definition will diverge.
One key property of the Laplace Transform is that it is a Linear Transformation, meaning:
L(af (x) + bg(x)) = aL(f (x)) + bL(g(x))
An Example: Find the Laplace Transform of:
f (x) = 4x sin(3x) + x2 e6x
2 6x
Using the table we see that the Laplace Transform of 4x sin(3x) is 4 (s22·3s
is
+32 )2 and the Laplace Transform of x e
2
(s−6)3 .
So:
4.1. CALCULATING LAPLACE AND INVERSE LAPLACE TRANSFORMS
L(f (x)) =
121
24s
2
+
2
+ 9)
(s − 6)3
(s2
An Example: Find the inverse Laplace Transform of:
F (s) =
s3 − 8s2 + 23s − 7
(s2 − 4s + 20)(s − 3)2
Looking at the table of Laplace Transforms we notice that there is not an inverse Laplace Transform for a function
with a repeated linear factor multiplied by a quadratic factor so we must use partial fraction decomposition to separate
this function into a sum of functions that do have an inverse Laplace Transform. By partial fractions
F (s) =
s−3
1
+
s2 − 4s + 20 (s − 3)2
Looking at the table we notice all terms with a quadratic denominator are written in the completed square form.
So we will complete the square on the quadratic denominator and do nothing to the second term since we can find the
inverse Laplace Transform it as it is:
F (s) =
s−3
1
+
(s − 2)2 + 16 (s − 3)2
To find the inverse Laplace of a term with a denominator: (s − 2)2 + 42 we need to have either an s − 2 or a 4 in the
numerator so I will rewrite the s − 3 term as s − 2 − 1 and separate the fraction into two fractions:
F (s) =
I will now multiply the second term by
s−2
1
1
−
+
2
2
2
2
(s − 2) + 4
(s − 2) + 4
(s − 3)2
4
4
to make it fit the table:
1
4
s−2
1
−
F (s) =
+
(s − 2)2 + 42
4 (s − 2)2 + 42
(s − 3)2
Using the table to find the inverse Laplace Transform gives:
1
f (x) = e2x cos(4x) − e2x sin(4x) + xe3x
4
440.
Find the Laplace Transform of:
f (x) = x3 − 3x cos(4x)
441.
Find the Laplace Transform of:
f (x) = x2 sinh(3x) + e2x sin(3x)
442.
Find the Laplace Transform of:
f (x) = x4 e5x − sin(2x) sinh(2x)
443.
Use trig identities and the table in this section to find the Laplace transform of:
122
CHAPTER 4.
f (x) = sin(Ax + B)
444.
g(x) = cos(Ax + B)
Use the definition to find the Laplace Transform of:


 1
f (x) =
x

 2
x
445.
x<2
2≤x<6
x≥6
Find the Laplace Transform of:
3
5
f (x) = x 2 − x 2
446.
Find the Inverse Laplace Transform of:
F (s) =
447.
3s
12
+
s2 + 4 s2 − 10s + 34
Find the Inverse Laplace Transform of:
F (s) =
448.
Find the Inverse Laplace Transform of:
F (s) =
449.
4s3 + 4s2 − 27s − 18
s4 − 3s3 + 2s2
Find the Inverse Laplace Transform of:
F (s) =
450.
3s4 + s3 − 5s2 − 10s − 25
s5 + 2s4 + 5s3
Find the Inverse Laplace Transform of:
F (s) =
451.
500 + 40s
(s2 + 25)2
9s2 − 30s + 49
(s − 3)2 (s2 + 1)
Find the Inverse Laplace Transform of:
F (s) =
3s3 − 6s2 + 39s + 54
(s2 − 4s + 13)(s2 + 9)
452.
Find the Inverse Laplace Transform of:
F (s) =
∞
X
1
sn
n=1
You should be able to recognize the power series.
453.
Find the Inverse Laplace Transform of:
F (s) =
You should be able to recognize the power series.
∞
X
(−1)n
s2n+2
n=0
LAPLACE TRANSFORM
4.1. CALCULATING LAPLACE AND INVERSE LAPLACE TRANSFORMS
454.
Find the Inverse Laplace Transform of:
F (s) =
∞
X
(−1)n (2n)!
s2n+2
n=0
You should be able to recognize the power series.
455.
Use the series expansion for ln(1 + x) to find
L(ln(1 + x))
456.
2
Show L(ex ) does not exist.
457.
Show L
1
x2
does not exist.
458.
Use
cosh(x) =
ex + e−x
2
sinh(x) =
ex − e−x
2
to derive formulas for
L{cosh(ax + b)}
L{sinh(ax + b)}
and
L{cosh2 (x)}
L{sinh2 (x)}
Then find the inverse Laplace transform of
L{cosh2 (x)} − L{sinh2 (x)}
What can you conclude about the value of
cosh2 (x) − sinh2 (x)
459.
Calculate:
L(x2 · y 0 )
460.
Let
y=
n
X
i=0
and
(1 + t)n
123
124
CHAPTER 4.
F (s) = L(y)
Show that the coefficients of both
1
sn
and
1
sn+1
in F (s) are both n!
461.
Use the following property:
Z ∞
f (x)
L
F (x)dx
=
x
s
to find
L
sinh(x)
x
462.
Use a power series expansion to show:
−1
L
1
e− s
√
s
√
cos(2 t)
= √
πt
463.
Recall the Gamma function from the last chapter is given by
Z
∞
Γ(x) =
tx−1 e−t dt
0
Show that
L(xr ) =
Γ(r + 1)
sr+1
r > −1
464.
Show
L{y · y 0 } =
1
sL{y 2 } − y 2 (0)
2
465.
Another interesting inverse Laplace Transform is:
−1 −1
L−1 F (s) =
L
F 0 (s)
x
Use this transform to calculate the inverse Laplace Transform of
F (s) = arctan
1
s
466.
Use the previous problem to find the inverse Laplace Transform of
F (s) = ln(1 + s2 )
LAPLACE TRANSFORM
4.2. SOLVING INITIAL VALUE PROBLEMS
4.2
125
Solving Initial Value Problems
What is the use of Laplace Transforms in differential equations? The answer comes from the next derivation of the
Laplace Transform of y 0 (x):
Z
0
L(y (x)) =
∞
y 0 (x)e−sx dx
using integration by parts
0
u = e−sx
0
dv = y 0 (x)dx
du = −se−sx dx
∞
Z
0
L(y (x)) =
y (x)e
−sx
dx = lim e
b→∞
0
−sx
v = y(x)
b Z
y(x) −
0
b
−sy(x)e−sx dx
0
= lim e−sb y(b) − y(0) + sL(y(x))
b→∞
= sL(y(x)) − y(0)
In a similar way we can find the Laplace Transform of y 00 , y 000 ...
L(y 0 (x)) = sL(y(x)) − y(0)
L(y 00 (x)) = s2 L(y(x)) − sy(0) − y 0 (0)
L(y 000 (x)) = s3 L(y(x)) − s2 y(0) − sy 0 (0) − y 00 (0)
From the above equations it should be clear that you must have initial conditions evaluated at x = 0 to use Laplace
Transforms to solve a differential equation.
The procedure to solve a differential equation using Laplace Transforms is: First, using the above tables take the
Laplace Transform of both sides of the differential equation; Second, Solve for L(y(x)) as a function of s; Third, using
the above tables calculate the inverse Laplace of both sides of the equation.
A Example:
Solve:
y 00 + 4y = 20e t cos(t)
y(0) = 7
y 0 (0) = 10
Taking the Laplace transform of both sides gives:
s2 L(y) − 7s − 10 + 4L(y) = 20
s−1
(s − 1)2 + 1
Solving for L(y):
L(y)(s2 + 4) = 7s + 10 + 20
s−1
(s − 1)2 + 1
126
CHAPTER 4.
L(y) =
LAPLACE TRANSFORM
7s + 10
s−1
+ 20
2
2
s +4
((s − 1) + 1)(s2 + 4)
Applying Partial Fractions to the second fraction on the right hand side and adding the results to the first fraction
yields
L(y) =
4s − 2
3s + 4
+
s2 − 2s + 2 s2 + 4
Completing the square on the first fraction
L(y) =
4s − 2
3s + 4
+
(s − 1)2 + 1 s2 + 4
Manipulating the right hand side into our table of Laplace Transforms
L(y) = 4
1
s
2
s−1
+2
+3 2
+2 2
2
2
(s − 1) + 1
(s − 1) + 1
s +4
s +4
Taking the inverse Laplace of each side
y = 4et cos(t) + 2et sin(t) + 3 cos(2t) + 2 sin(2t)
An Example: Solve:
y 00 − 2y 0 − y = e2x − ex
y(0) = 1
y 0 (0) = 3
Taking the Laplace transform of both sides gives:
s L(y) − s − 3 − 2 sL(y) − 1 − L(y) =
2
1
1
−
s−2 s−1
Solving for L(y):
1
1
−
+s+1
s−2 s−1
√
√
Solving for L(y) and factoring s2 − 2s − 1 = (s − (1 + 2))(s − (1 − 2))
L(y)(s2 − 2s − 1) =
L(y) =
1
s+1
1
√
√
√
√
√
√
−
+
(s − 2)(s − (1 + 2))(s − (1 − 2)) (s − 1)(s − (1 + 2))(s − (1 − 2)) (s − (1 + 2))(s − (1 − 2))
Expanding with partial fractions gives:
L(y) =
√
√
3( 2 + 1)
1
3( 2 − 1)
1
1 1
1
√
√ +
−
−
4
4
s − ( 2 + 1)
s − (1 − 2) 2 s − 1 s − 2
Calculating the inverse Laplace Transform
√
√
3( 2 + 1) (√2+1)x 3( 2 − 1) (1−√2)x 1 x
y=
e
−
e
+ e − e2x
4
4
2
The Laplace Transform can also be used to solve linear differential equations without constant coefficients. To do so
we would like to have nice formula for:
4.2. SOLVING INITIAL VALUE PROBLEMS
127
L(f (x)y 0 )
L(f (x)y)
L(f (x)y 00 )
Unfortunately, these formulas either do not exist or are well beyond the scope of this book, so instead we will find
formulas for the specific case f (x) = x. Lets now calculate L(xy)
We know
L(xf (x)) = −
d
L(f (x))
ds
So
L(xy) = −
d
L(y)
ds
And
L(xy 0 ) = −
d
d
d
L(y 0 ) = −
sL(y) − y(0) = −L(y) − s L(y)
ds
ds
ds
And
d
d
d
00
2
0
s L(y) − sy(0) − y (0) = −2sL(y) − s2 L(y) + y(0)
L(xy ) = − L(y ) = −
ds
ds
ds
00
Using the notation L0 (y) =
d
ds L(y)
L(xy) = −L0 (y)
our formulas become:
L(xy 0 ) = −L(y) − sL0 (y)
L(xy 00 ) = −2sL(y) − s2 L0 (y) + y(0)
A Example:
Solve:
xy 00 + 2(x − 1)y 0 − 2y = 0
y(0) = 0
Taking the Laplace Transform of each side gives:
−2sL(y) − s2 L0 (y) + y(0) − 2L(y) − 2sL0 (y) − sL(y) + y(0) − 2L(y) = 0
−2sL(y) − s2 L0 (y) − 2L(y) − 2sL(y) − sL0 (y) − 2L(y) = 0
Which simplifies to:
(s2 + 2s)L0 (y) + (4s + 4)L(y) = 0
This equation is now separable
L0 (y)
4s + 4
=−
L(y)
s(s + 2)
Applying partial fraction to the right hand side gives
128
CHAPTER 4.
LAPLACE TRANSFORM
L0 (y)
2
2
=− −
L(y)
s s+2
Integrating gives
ln L(y) = −2 ln(s) − 2 ln(s + 2) + C
Treating C as ln(K) and using some log rules we get:
ln(s2 (s + 2)2 L(y)) = ln(K)
So
L(y) =
K
s2 (s + 2)2
Applying partial fraction to the right hand side gives
−1 1 1 1
1
1
1
1
· + · 2+ ·
+ ·
L(y) = K
4 s 4 s
4 s + 2 4 (s + 2)2
Absorbing the
1
4
into the constant and calculating the inverse Laplace Transform gives:
y=K
− 1 + x + e−2x + xe−2x
We see the first initial condition y(0) = 0 gives us no information about the constant K but our second initial condition
y(1) = 2e−2 does. Applying this condition gives
2e−2 = K
− 1 + 1 + e−2 + e−2
So K = 1 and the solution is
y=
467.
−1+x+e
Solve:
y(0) = 1
Solve:
y 00 + 6y 0 + 5y = 12ex
470.
y(0) = 1
y 0 + 2y = sin(2x)
469.
+ xe
−2x
Solve:
y 0 + y = ex
468.
−2x
y(0) = −1
y 0 (0) = 7
Solve:
y 00 − 2y 0 + y = 6tet
y(0) = 1
y 0 (0) = 2
4.2. SOLVING INITIAL VALUE PROBLEMS
471.
129
Solve:
y 00 − 2y 0 − 3y = 8 − 16x − 12x2
472.
Solve:
y 00 + y 0 − 2y = 3ex
473.
Solve:
Solve:
Solve:
Solve:
y(0) = 3
y 0 (0) = 5
Solve:
y 00 − 4y 0 + 3y = −ex (3 sin(x) + cos(x))
478.
y 0 (0) = 4
y(0) = 1
y 00 − 3y 0 + 2y = (1 − 2x)ex
477.
y 0 (0) = 0
y(0) = 1
y 00 − 4y 0 + 5y = te2t
476.
y 0 (0) = 5
y(0) = 3
y 00 − 2y 0 + y = (x + 2)e2x
475.
y 0 (0) = 8
y(0) = 5
y 00 + y = ex (3 cos(x) − sin(x))
474.
y 0 (0) = 2
y(0) = 2
y 0 (0) = 4
y(0) = 1
Solve:
y 00 + y = 2 sin(x) + 4x cos(x)
y(0) = 3
y 0 (0) = 2
479.
Solve:
y 00 + 4y = sin(2t)
480.
y(0) = 10
Solve:
y 00 − 7y 0 + 10y = 9 cos(x) + 7 sin(x)
481.
482.
483.
y 0 (0) = 0
y(0) = 5
y 0 (0) = −4
Solve:
y 00 + 4y = 4x2 − 4x + 10
y(0) = 0
y 0 (0) = 3
y 00 + 4y 0 + 5y = e2x sin(x)
y(0) = 1
y 0 (0) = −6
Solve:
Solve:
130
CHAPTER 4.
y 00 − 2y 0 + y = 18xex
484.
486.
487.
488.
Solve:
y 00 + y = e2x (4x3 + 12x2 + 6x)
y(0) = 1
y 0 (0) = 1
y 00 − y 0 − 2y = e2x (x2 + 6x + 2)
y(0) = 2
y 0 (0) = 1
y 00 − 2y 0 + 10y = −12 cos(3x) + x sin(x) + sin(x) − 6x cos(x)
y(0) = 1
Solve:
Solve:
y(0) = 2
y(0) = 0
y 0 (0) = 0
y 0 (0) = 2
y(0) = 0
y 00 (0) = −4
y(0) = 4
y 0 (0) = 8
Solve:
y(0) = 2
y 0 (0) = 5
Solve:
y 00 − y =
∞
X
enx
y(0) = 0
y 0 (0) = 0
n=0
494.
Solve:
xy 00 − y 0 = x2
495.
y 00 (0) = 0
Solve:
y 000 − 8y 00 + 21y 0 − 18y = 2e3x
493.
y 00 (0) = 4
Solve:
y 000 − 2y 00 + 4y 0 − 8y = 8e2x cos(2x) − 16e2x sin(2x)
492.
y 0 (0) = 4
Solve:
y 000 + y 00 + 3y 0 − 5y = 16e−x
491.
y 0 (0) = 13
Solve:
y 000 − 6y 00 + 11y 0 − 6y = et + e2t + e3t
490.
y 0 (0) = 3
y(0) = 3
Solve:
y 000 − y 00 + y 0 − y = −2et
489.
y 0 (0) = 3
y(0) = 1
y 00 − 5y 0 + 6y = ex (4 sin(x) − 2 cos(x))
485.
LAPLACE TRANSFORM
y(0) = 0
y(1) = 2
Solve:
xy 00 + y 0 − 2xy = ex
y(0) = 1
y(1) = 2
y 00 (0) = 15
y 00 (0) = 0
4.3. UNIT STEP FUNCTION
496.
131
Solve:
xy 00 + 2xy 0 + 2y = 0
497.
y(0) = 0
y(1) = 2
Solve:
xy 00 + (x + 2)y 0 + y = −1
y(0) = 0
y(1) = 2
498.
Use the formula
L(y 00 (x)) = s2 L(y(x)) − sy(0) − y 0 (0)
To find the Laplace transform of
y(x) = sin(ax)
499.
Convert the Cauchy Euler equation to the variable t and then use Laplace transforms to solve it.
x2
4.3
d2 y
dy
+x
− 4y = sin(ln(x))
dx2
dx
y(1) = 1
y 0 (1) = 3
Unit Step Function
So far we have learned how to use Laplace Transforms to solve differential equations involving continuous functions. We
will now learn how to solve differential equations involving piecewise functions. First a definition.
The Unit Step Function is defined as:
(
uc (x) =
0
x<c
1
x≥c
It can be shown that:
L(uc (x)f (x)) = e−cs L(f (x + c))
L−1 (e−cs F (s)) = uc (x)f (x − c)
An Example: Solve:
y 00 + y = g(x)
y(0) = 0
(
g(x) =
x
x<2
4
2<x
y 0 (0) = 1
Start by writing the equation using the unit step function:
y 00 + y = x + (4 − x)u2 (x)
132
CHAPTER 4.
LAPLACE TRANSFORM
Take the Laplace Transform of both sides:
s2 L(y) − 1 + L(y) =
1
+ e−2s L(4 − (x + 2))
s2
1
1
−2s 2
L(y)(s + 1) = 2 + e
+1
−
s
s s2
2
L(y) =
1
s2 (s2 + 1)
+
1
1
2
−2s
+
e
−
(s2 + 1)
s(s2 + 1) s2 (s2 + 1)
Applying partial fraction to the first term and on the terms multiplied by e−2s gives:
1
1
1
−2s
1
2
1
−2s
L(y) = 2 − 2
+
+e
+
+ −
s
s + 1 (s2 + 1)
s2 + 1 s2 + 1 s s2
L(y) =
1
−2s
1
2
1
−2s
+
e
+
+
−
s2
s2 + 1 s2 + 1 s s2
Calculating the inverse Laplace Transform:
y = x + u2 (x) − 2 cos(x) + sin(x) + 2 − x x−2
y = x + u2 (x) − 2 cos(x − 2) + sin(x − 2) + 2 − (x − 2)
y = x + u2 (x) − 2 cos(x − 2) + sin(x − 2) − x
500.
Express the piecewise function using the unit step function and then find its Laplace Transform:
(
f (x) =
501.
x
e
x<4
3x
x>4
Express the piecewise function using the unit step function and then find its Laplace Transform:


x<3
 sin(2x)
4x
f (x) =
xe
3<x<6


cosh(2x)
6<x
502.
Find the Inverse Laplace Transform of:
F (s) =
503.
e−5s
(s + 4)
Find the Inverse Laplace Transform of:
F (s) =
504.
Solve:
e−3s (s − 5)
(s + 1)(s + 2)
4.3. UNIT STEP FUNCTION
133
y 0 − 4y = 1 + uπ (x)x sin(x)
y(0) = 1
505.
Solve:
y 00 + 3y 0 + 2y = g(x)
(
g(x) =
y 0 (0) = −1
y(0) = 2
x2
1
x<3
3<x
506.
Solve:
y 00 + y = g(x)
y(0) = 2
(
g(x) =
x
x
y 0 (0) = −1
x<1
2
x>1
507.
Solve:
y 00 + 5y 0 + 6y = g(x)


 0
g(x) =
4x


8
y(0) = −1
y 0 (0) = 0
x<1
1<x<5
5<x
508.
Solve:
y 00 − 2y 0 + y = g(x)


 2x
g(x) =
x2


1
y(0) = 1
y 0 (0) = 3
x<2
2<x<5
5<x
509.
Solve:
y 00 + 8y 0 + 17y = g(x)


 x
g(x) =
x2


x
510.
y(0) = 1
x<1
1<x<5
5<x
y 0 (0) = 0
134
CHAPTER 4.
LAPLACE TRANSFORM
Solve:
y 00 + 4y = g(x)
y 0 (0) = 0
y(0) = 0


 x
g(x) =
x2


1
x<1
1<x<2
2<x
511.
Solve:
y 00 + y = g(x)
(
g(x) =
y 0 (0) = 1
y(0) = 0
cos(2x)
x<
sin(2x)
π
2
π
2
<x
512.
Solve:
y 00 + 4y = g(x)
(
g(x) =
y(0) = −3
| sin(x)|
0
y 0 (0) = 1
x < 2π
x > 2π
513.
Show: if
Y (s) =
∞
X
e
n=0
−ns
1
n
+
2
s
s
Then


t




2t



−1
3t
L
Y =


..


.




nt
x<1
1<x<2
2<x<3
n−1<x<n
514.
Express the floor function as a sum of unit step functions and find its Laplace Transform.
4.4
Convolution
If you are asked to find the inverse Laplace Transform of F (s)G(s) and you know the inverse Laplace Transform of both
F (s) and G(s) what is the relationship between the Laplace Transform of F (x)G(s) and Laplace Transform of F (s) and
G(s)? The answer is a new operation called Convolution.
4.4. CONVOLUTION
135
The Convolution of f (x) and g(x) is denoted f ∗ g and is given by the integral:
x
Z
f ∗g =
f (x − v)g(v)dv
0
For example the Convolution of x2 and x3 is:
x
Z
x2 ∗ x3 =
(x − v)2 v 3 dv =
Z
0
Z
x
(x2 − 2xv + v 2 )v 3 dv =
0
x
(x2 v 3 − 2xv 4 + v 5 )dv =
0
x
x6
x2 v 4
2xv 5
v6 2x6
x6
x6
−
+ =
−
+
=
4
5
6 0
4
5
6
60
Some basic properties of Convolution are:
f ∗g =g∗f
f ∗ (g + h) = f ∗ g + f ∗ h
(f ∗ g) ∗ h = f ∗ (g ∗ h)
f ∗0=0
It is also true that
f ∗ 1 6= f
since
Z
x
f ∗1=
f (x − v)dv 6= f
0
What makes Convolution useful in differential equations are the following properties involving the Laplace Transform:
(L)−1 (F (s)G(s)) = ((L)−1 (F (s))) ∗ ((L)−1 (G(s)))
L(f ∗ g) = L(f ) · L(g)
To show the first property:
L(f ∗ g) = L(f ) · L(g)
holds we let
∞
Z
−sx
F (s) = L(f ) =
f (x)e
Z
0
Z
L(f ∗ g) =
∞
−st
Z
e
0
t
∞
G(s) = L(f ) =
dx
g(y)e−sy dy
0
Z
f (t − v)g(v)dv dt =
0
∞
e
−st
Z
0
∞
ut−v (t)f (t − v)g(v)dv dt
0
Remember ut−v (t) will be zero if v > t
Reversing the order if integration gives
∞
Z
L(f ∗ g) =
Z
g(v)
0
The integral in parentheses is equal to e
−sv
Z
L(f ∗ g) =
e−st ut−v (t)f (t − v)dt dv
0
F (s) so we have:
∞
g(v)e
0
∞
−sv
Z
F (s)dv = F (s)
0
∞
g(v)e−sv dv
136
CHAPTER 4.
LAPLACE TRANSFORM
So
L(f ∗ g) = F (s) · G(s)
An Example: Solve:
y0 − 2
Z
x
ex−v y(v)dv = x
y(0) = 2
0
Rewriting the integral as convolution
y 0 − 2ex ∗ y = x
y(0) = 2
Taking the Laplace Transform of each side gives:
1
1
L(y) = 2
sL(y) − 2 − 2
s−1
s
Solving for L(y)
(s − 1)sL(y) − 2L(y) =
L(y)(s2 − s − 2) =
L(y) =
L(y) =
s−1
+ 2(s − 1)
s2
s−1
+ 2(s − 1)
s2
s−1
2(s − 1)
+ 2
− s − 2) s − s − 2
s2 (s2
s2 (s
s−1
2(s − 1)
+
− 2)(s + 1) (s − 2)(s + 1)
After partial fractions we have:
L(y) =
3 1
31 1 1
2
+
−
+
s + 1 4 s − 2 4 s 2 s2
Taking the inverse Laplace Transform
3
3 1
y = 2e−x + e2x − + x
4
4 2
Convolution can also be used to find the output response y(x) of a physical system for some input function f (x).
Consider the physical system governing the motion of a mass connected to a spring:
my 00 + by 0 + ky = f (x)
For Simplicity lets assume
y(0) = y 0 (0) = 0
4.4. CONVOLUTION
137
Taking the Laplace transform of both sides of the differential equation gives
ms2 L(y) + bsL(y) + kL(y) = L(f (x))
L(f (x))
ms2 + bs + k
L(y) =
The function
W (s) =
ms2
1
+ bs + k
is called the Transfer Function for the physical system.
L(y) = L(f (x))W (s)
The function
w(x) = L
−1
W (s)
is called the Weight Function of the system.
Through convolution we see the solution to the differentia equations is
x
Z
w(u)f (x − u)du
y(x) =
0
This equation reduces solving the same mass spring system for different input functions into solving a definite integral
for each input function.
515.
Find:
x ∗ ex
516.
Find:
sin(x) ∗ sin(2x)
517.
If y(0) = 0 show:
2 ∗ (y(x) · y 0 (x)) = y 2 (x)
518.
Find the Laplace Transform of:
Z
f (x) =
0
519.
Find the Laplace Transform of:
x
(x − v)2 e5v dv
138
CHAPTER 4.
x
Z
cos(x − v) · sin(4v)dv
f (x) =
0
520.
Find the inverse Laplace Transform of the following function by using Convolution:
1
F (s) =
s2 (s2
+ 1)
521.
Find the inverse Laplace Transform of the following function by using Convolution:
F (s) =
(s2
s
+ 1)2
522.
Solve:
y0 +
Z
x
(x − v)y(v)dv = t
y(0) = 0
0
523.
Solve:
y0 + y −
Z
x
sin(x − v)y(v)dv = − sin(x)
y(0) = 1
0
524.
Solve:
y 0 = 1 − sin(x)
Z
x
y(v)dv
y(0) = 0
0
525.
Solve:
y 00 + y = 2x ∗ ex
y(0) = y 0 (0) = 0
526.
Solve:
y 00 − y = 2x ∗ sin(x)
y(0) = 4
y 0 (0) = 2
527.
Solve using convolution. Leave your answer in terms of an integral involving g(x).
y 00 + y = g(x)
y(0) = 1
528.
Use convolution to evaluate the following integral
Z
0
t
(t − u)5 u8 du
y 0 (0) = 1
LAPLACE TRANSFORM
4.4. CONVOLUTION
139
529.
Use convolution to evaluate the following integral
t
Z
(t − u)n um du
0
530.
If f and g have the following properties
f (x) ∗ f (x) = xf (x)
f (0) = 4
Find f (x)
531.
Without using the definition of convolution calculate
f (t) = u1 (t)t ∗ u2 (t)t2
By calculating the Laplace Transform of f (t) simplifying and then calculating the inverse Laplace transform to find
f (t)
532.
If
f (x) ∗ f 0 (x) =
x2
2
f (0) = f 0 (0) = 1
Find f (x)
533.
Prove The following with mathamatical induction
n-times
}|
{
z
x ∗ x ∗ x ∗ ... ∗ x =
x2n−1
(2n − 1)!
534.
Show the folowing property holds
t ∗ tn =
tn+2
(n + 1)(n + 2)
535.
If f (x) = (1 + x)2
f (x) ∗ g(x) = f (x) · g(x)
Find g(x)
536.
Show
f ∗ g0 = f 0 ∗ g
g(0) = e−1
140
CHAPTER 4.
4.5
LAPLACE TRANSFORM
Delta Function
The Dirac Delta Function δ(x) is defined as
(
δ(x) =
x 6= 0
0
∞
x=0
And has the property:
Z
∞
f (x)δ(x)dx = f (0)
−∞
L(δ(x − c)) = e−cs
The delta function shows up in science when considering the impulse of a force over a short interval. If a force F (t)
on the time interval t0 to t1 then the impulse due to the force is:
Z
t1
Impulse =
F (t)dt
t0
By Newton’s second law
Z
t1
Z
t1
F (t)dt =
t0
m
t0
dv
dt = mv(t1 ) − mv(t0 )
dt
where m is the mass and v is the velocity. Since an objects momentum is the product of mass and velocity we see
that the impulse is equal to the change in momentum.
An Example: Solve:
y 00 + y = δ(x − π)
y(0) = 0
Taking the Laplace Transform of both sides gives:
s2 L(y) + L(y) = e−πs
L(y)(s2 + 1) = e−πs
L(y) = e−πs
1
s2 + 1
Finding the inverse Laplace Transform:
y = uπ (x) sin(x)
x−π
y = uπ (x) sin(x − π)
After a trig identity
y = −uπ (x) sin(x)
y 0 (0) = 0
4.5. DELTA FUNCTION
141
537.
Find:
3
Z
2
ex δ(x − 1)dx
0
538.
Find the value of k so that
Z
0
1
1
3
sin (π(x − k))δ x −
dx =
2
4
2
539.
Find
L{δ(sin(πx))}
540.
541.
542.
Solve
y 0 + y = δ(x − 1)
y(0) = 2
y 00 − 4y 0 + 4y = e2x + δ(x − 2)
y(0) = 1
Solve
Solve
y 00 + 2y 0 + 2y = δ(x − π)
543.
y(0) = 1
546.
Solve
Solve
Solve
548.
Solve
π
y +y =δ x−
2
3π
−δ x−
2
0
x
y(0) = 0
Solve
y 00 + y =
∞ X
δ x − nπ
y 0 (0) = 0
y(0) = 1
y 0 (0) = 1
y −y =x∗e +δ x−1
549.
y 0 (0) = 3
y(0) = 2
y 00 − 4y 0 + 13y = xu2 (x) + δ x − 2
Solve
y 0 (0) = 3
y(0) = 0
y 00 − 3y 0 + 2y = δ x − 1 − δ x − 2
00
547.
y 0 (0) = −2
y(0) = 2
y 00 − y 0 − 2y = 3δ(x − 1) + ex
545.
y 0 (0) = 1
Solve
y 00 + 2y 0 − 3y = δ(x − 1) − δ(x − 2)
544.
y 0 (0) = 3
y(0) = 1
y(0) = 0
y 0 (0) = 0
n=0
550.
Show
L{δ(x − n) ∗ δ(x − m)} = L{δ(x − m − n)}
142
CHAPTER 4.
LAPLACE TRANSFORM
Chapter 5
First Order Systems of Differential
Equations
In this chapter we will study systems of differential equation of the form:
dx
= F (t, x, y)
dt
dy
= G(t, x, y)
dt
Where the dependent variables x and y are linked together by the independent variable t.
5.1
Linear Systems
The theory of first order linear systems is very similar to the theory of second order equation that we studied in chapter
2. Consider the following system of equations:
dy
= 2x + y
dt
dx
= 4x − y;
dt
This system can be transformed into a second order equation by solving for y in the first equation and substituting it
into the second:
dx
y = 4x −
dt
d
dx
dx
4x −
= 2x + 4x −
dt
dt
dt
This produces the second order equation:
4
dx d2 x
dx
− 2 = 2x + 4x −
dt
dt
dt
Which simplifies to
d2 x
dx
−5
+ 6x = 0
2
dt
dt
This has the characteristic equation:
143
144
CHAPTER 5. FIRST ORDER SYSTEMS OF DIFFERENTIAL EQUATIONS
r2 − 5r + 6 = 0
(r − 3)(r − 2) = 0
Giving the solution:
x1 = e2t
Substituting these equations into y = 4x −
dx
dt
x2 = e3t
gives two solution for y
y1 = 2e2t
y2 = e3t
Making the general solution:
x = C1 e3t + C2 e2t
y = C1 e3t + 2C2 e2t
In chapter 2 we learned that if the Wronskian of the two solutions to a second order equation is nonzero on an interval
then the two solutions are linearly independent and form a Fundamental Solution Set. For systems of equations the
Wronskian is:
x
1
W (t) = y1
x2 = x1 y2 − x2 y1
y2 It can be shown that if the homogenous system:
dy
= a2 (t)x(t) + b2 y(t)
dt
dx
= a1 (t)x(t) + b1 y(t)
dt
has solutions:
x = x1 (t)
y = y1 (t)
and
x = x2 (t)
y = y2 (t)
x = x2 (t)
y = y2 (t)
and the Wronskian is non zero on the interval [a, b] then:
x = x1 (t)
y = y1 (t)
and
is the general solution to the system of differential equations on the interval [a, b]. We see that the Wronskian for the
last problem we solved is:
e3t
W (t) = 3t
e
e2t = e5t 6= 0
2e2t So the general solution is indeed:
x = C1 e3t + C2 e2t
y = C1 e3t + 2C2 e2t
You can also solve the same system of differential equations using matrices. Let us first write the system of equations
as a matrix equation of the form:
"
Ax = x’
where
x=
x
y
#
5.1. LINEAR SYSTEMS
145
dy
= 2x + y
dt
dx
= 4x − y;
dt
The system in matrix form is:
"
4
−1
2
1
#"
x
#
"
=
y
x0
#
y0
As we earlier showed the system of equations can be reduced to a second order equation with constant coefficients
whose solutions were of the form: ert . It is reasonable to assume the above matrix equation will have a solution of the
form:
"
x(t) =
x
y
#
= ert u
Where r is a constant and u is a constant nonzero vector. Substituting x(t) = ert u into Ax = x’ gives:
rert u = Aert u
Dividing by the nonzero factor ert and rearranging the terms gives:
A − rI u = 0
The values of r and u that satisfy the above equation are the eigenvalues and eigenvectors of the matrix A. To find
the eigenvalues r of a matrix we take the determinant of both sides to the above equation:
A − rI u = |0| = 0
Since u is nonzero the solutions: r, to the above equation come from the solutions to:
A − rI = 0
This equation is a polynomial with the variable r. This equation is called the Characteristic Equation whose roots
are the eigenvalues of the matrix A. Back to the problem we were solving. Our characteristic equation is:
4−r
2
−1 = (4 − r)(1 − r) + 2 = 0
1−r Factoring the characteristic equation and solving gives:
(r − 2)(r − 3) = 0
r1 = 2 r2 = 3
NOTE: this is the same characteristic equation we got in our first solution to this problem.
Now that we have the eigenvalues for A we need the eigenvectors u. To find the eigenvector for each eigenvalue we
must solve the following equation for u.
A − rI u = 0
For r1 = 2 we have:
146
CHAPTER 5. FIRST ORDER SYSTEMS OF DIFFERENTIAL EQUATIONS
"
2
−1
2
−1
#"
#
x
=0
y
We see the solution to this matrix equation is y = 2x. Taking x to be 1 we get our eigenvector:
"
#
1
u1 =
2
For r2 = 3 we have:
"
1
−1
2
−2
#"
#
x
=0
y
We see the solution to this matrix equation is y = x. Taking x to be 1 we get our eigenvector:
"
u2 =
1
#
1
Making the solution to the differential equation:
"
x
y
#
"
= C1 e
2t
1
#
"
3t
+ C2 e
2
1
#
1
Making:
x = C1 e3t + C2 e2t
y = C1 e3t + 2C2 e2t
The same solution we obtained earlier.
The matrix containing the two eigenvectors as its two columns is called the Fundamental Matrix. The fundamental
matrix for our problem is:
"
X(t) =
1
1
2
1
#
Some times the matrix has complex eigenvalues: r = α ± βı and eigenvectors a ± bı the solution becomes:
x(t) = C1 eαt cos(βt)a − eαt sin(βt)b + C2 eαt cos(βt)b + eαt sin(βt)a
Example:
Solve the system of differential equations:
x0 = −x − 2y
y 0 = 8x − y
Writing the system as a matrix equation gives:
"
Find the characteristic equation:
−1
−2
8
−1
#"
x
y
#
"
=
x0
y0
#
5.1. LINEAR SYSTEMS
147
= (−1 − r)2 + 16 = 0
−1 − r −1 − r
8
−2
Which has roots:
r = −1 ± 4ı
Using r = −1 + 4ı the eigenvector is:
"
−4ı
−2
8
−4ı
#"
x
y
#
=0
This has the solution:
y = −2xı
taking x = 1 the eigenvector is
"
u1 =
1
#
"
=
−2ı
1
#
"
+ı
0
#
0
−2
We could find the eigenvector corresponding to r = −1 − 4ı in the same we found the first eigenvector but we do not
need to. It is true that if u1 = a + ıb is a eigenvector for eigenvalue r1 = α + βı then u2 = a − ıb is a eigenvector for
r1 = α − βı. This means the eigenvector for r = −1 − 4ı is
"
u2 =
1
#
2ı
"
=
1
0
#
"
+ı
0
#
2
Making the solution:
"
#
"
#
"
#
"
#
1
0
0
1
−t
−t
−t
−t
x(t) = C1 e cos(4t)
− e sin(4t)
+ C2 e cos(4t)
+ e sin(4t)
0
2
2
0
551.
Solve:
x0 = 2x + y
y 0 = −x + 4y
552.
Solve:
x0 = −6x + 4y
y 0 = −5x + 3y
553.
Solve:
x0 = 2x + 4y
554.
Solve:
y 0 = −5x − 2y
148
CHAPTER 5. FIRST ORDER SYSTEMS OF DIFFERENTIAL EQUATIONS
x0 = −x − 2y
y 0 = 13x + y
x0 = −7x − 2y
y 0 = 5x − y
555.
Solve:
556.
Solve:
x0 = x + 3y
y 0 = 12x + y
557.
Solve:
x0 = x + 2y + 2z
y 0 = 2x + 3y
z 0 = 2x + 3y
558.
Solve:
x0 = 2x − 4y
y 0 = 2x − 2y
559.
Solve:
x0 = −2x − 5y
y 0 = x + 2y
560.
Solve:
x0 = 4x − y
y 0 = x + 2y
561.
Solve:
x0 = −x + 5y
y 0 = −x + 3y
562.
Convert the following differential equation governing the motion of a mass attached to a spring into a system of
equations
my 00 + by 0 + ky = 0
563.
Solve:
"
2
−3
1
−2
#"
x
y
#
"
=
x0
y0
#
5.2. LOCALLY LINEAR EQUATIONS
149
564.
Solve:
"
−1
2
−1
−3
#"
#
x
"
=
y
#
x0
y0
565.
Use the definition of the Wronskian in this section to prove Abel’s Theorem for systems: If
{y1 , y2 }
are solutions to
x0 = Ax
Then
Rt
W (y1 , y2 ) = Ce
5.2
t0
tr(A)du
Locally Linear Equations
We will now discuss nonlinear systems around their critical or equilibrium points. The general system of equations we
will be studying are systems of the form:
x0 = F (x, y)
y 0 = G(x, y)
The Critical or Equilibrium Points of this system are the values (x0 , y0 ) such that
x0 = F (x0 , y0 ) = 0
y 0 = G(x0 , y0 ) = 0
and
If both x0 > 0 and y0 > 0 then the critical point is defined to be a Positive Equilibrium Point and will be discussed
when we consider the competing species problem.
The system is said to be Locally Linear around a critical or equilibrium point if both F and G have continuous first
and second order partial derivatives at the critical point (x0 , y0 ).
For the general system:
x0 = F (x, y)
y 0 = G(x, y)
with equilibrium point (x0 , y0 ) we can form the locally linear system in matrix form:
"
Fx (x0 , y0 )
Fy (x0 , y0 )
Gx (x0 , y0 ) Gy (x0 , y0 )
#"
x
y
#
"
=
y0
With the matrix:
"
J=
called the Jacobian Matrix
Fx (x0 , y0 )
Fy (x0 , y0 )
Gx (x0 , y0 ) Gy (x0 , y0 )
x0
#
#
150
CHAPTER 5. FIRST ORDER SYSTEMS OF DIFFERENTIAL EQUATIONS
An Example
Find the positive equilibrium solution to the system and discuss the stability near this point.
dx
= x 4 − 2y
dt
dy
=y −2+x
dt
We see that the positive equilibrium solutions come from the equations :
4 − 2y = 0
−2+x=0
Giving the positive equilibrium solution (2, 2). The Jacobian Matrix for this problem is:
"
J=
#
4 − 2y
−2x
y
x − 2y
The Jacobian Matrix at the equilibrium point (2, 2) is:
"
J=
0
−4
2
−2
#
Producing the locally linear system:
"
0
−4
2
−2
#"
x
#
y
"
=
x0
#
y0
Now we need the eigenvalues and eigenvector of our Jacobian Matrix evaluated at the equilibrium points. The
characteristic equation is:
0−r
J =
2
= −r(−2 − r) + 8 = 0
−2 − r −4
r2 + 2r + 8 = 0
Which has roots:
r1,2 = −1 ±
√
7ı
I have no desire to find the eigenvectors so, instead of the two eigenvectors I present you with one rabbit with two
pancakes on its head.
5.2. LOCALLY LINEAR EQUATIONS
151
566.
Find all equilibrium solutions and discuss the stability of each point and interpret the results of the competing species
modle.
dx
=x 1−x−y
dt
dy
= y 2 − y − 3x
dt
567.
Find all equilibrium solutions and discuss the stability of each point and interpret the results of the competing species
modle.
dx
=x 1−x−y
dt
dy
= y 3 − 2x − 4y
dt
568.
Find all equilibrium solutions and discuss the stability of each point and interpret the results of the competing species
modle.
dx
=x 2−x−y
dt
dy
= y 3 − y − 2x
dt
569.
Find all equilibrium solutions and discuss the stability of each point and interpret the results of the competing species
modle.
dx
= x 2 − 2x − y
dt
dy
= y 3 − 2x − 2y
dt
152
CHAPTER 5. FIRST ORDER SYSTEMS OF DIFFERENTIAL EQUATIONS
570.
Find all equilibrium solutions and discuss the stability of each point and interpret the results of the competing species
modle
dx
=x 3−x−y
dt
dy
= y 4 − x − 2y
dt
571.
Find all equilibrium solutions and discuss the stability of each point and interpret the results of the competing species
modle
dx
1
=x 2−x− y
dt
2
dy
=y 3−x−y
dt
572.
Find all equilibrium solutions and discuss the stability of each point and interpret the results of the competing species
modle
x0 = x(2 − x − y)
y 0 = y(3 − 2y − x)
573.
Find all equilibrium solutions and discuss the stability of each point and interpret the results of the competing species
modle
x0 = x(2 − x − 2y)
y 0 = y(7 − 2y − 6x)
574.
Find all equilibrium solutions and discuss the stability of each point and interpret the results of the competing species
modle
x0 = x(4 − 2x − 2y)
y 0 = y(5 − y − 4x)
575.
Find all equilibrium solutions and discuss the stability of each point and interpret the results of the competing species
modle
x0 = x(4 − x − 2y)
y 0 = y(3 − y − x)
576.
Find the positive equilibrium solution to the system and discuss the stability near this point.
dx
=x 1−x−y
dt
dy
= y 3 − x − 2y
dt
577.
Find the positive equilibrium solution to the competing species system of equations and discuss the stability near this
point.
dx
x
4y
=x 2− −
dt
3 x+4
dy
x
1
=y
−
dt
x+4 2
5.3. LINEAR SYSTEMS AND THE LAPLACE TRANSFORM
153
578.
Find the equilibrium solution to the system and sketch phase plane diagram.
dx
= −y(y − 2)
dy
dy
= (x − 2)(y − 2)
dt
579.
Solve the system of equations by converting it into a first order differential equation
dx
x3
= xy −
dy
y
dy
= 3x2 − y 2
dt
580.
Solve the system of equations by converting it into a first order differential equation
dx
=x
dt
5.3
dy
= 2y − x3 y 2
dt
Linear Systems and the Laplace Transform
We can also use Laplace Transforms to solve linear systems with initial conditions. The procedure will be quite similar to
the procedure used to solve initial value problems in chapter 4. Given a system of equations we will start by taking the
Laplace Transform of each side of the equation, solving for the Laplace Transform of both x(t) and y(t) and then using
a table to find the inverse.
An Example: Solve:
dx
= 2y + 4t
dt
x(0) = 4
dy
= 4x − 2y − 4t − 2
dt
y(0) = −5
Taking the Laplace Transform of both sides of the differential equation involving x0 gives:
L(x0 ) = L(y) + L(4t)
Using a table we get:
sL(x) − x(0) = 2L(y) +
sL(x) − 4 = 2L(y) +
4
s2
4
s2
Taking the Laplace Transform of both sides of the differential equation involving y 0 gives:
L(y 0 ) = 4L(x) − 2L(y) − L(4t) − L(2)
sL(y) − y(0) = 4L(x) − 2L(y) −
sL(y) + 5 = 4L(x) − 2L(y) −
4
2
−
2
s
s
4
2
−
s2
s
154
CHAPTER 5. FIRST ORDER SYSTEMS OF DIFFERENTIAL EQUATIONS
We now have two equations with two desired variables: L(x) and L(y) that we will need to solve for. In our first
equation I will solve for L(x) and insert it into the second.
sL(x) − 4 = 2L(y) +
4
s2
2
4
4
L(y) + 3 +
s
s
s
Inserting L(x) into the second of our two equations gives:
L(x) =
sL(y) + 5 = 4L(x) − 2L(y) −
2
4
−
s2
s
2
4
2
4
4
sL(y) + 5 = 4 L(y) + 3 +
− 2L(y) − 2 −
s
s
s
s
s
Multiplying both sides by s3 and simplifying the result gives:
L(y)(s4 + 2s3 − 8s2 ) = 16 − 4s + 14s2 − 5s3
L(y) =
16 − 4s + 14s2 − 5s3
(s4 + 2s3 − 8s2 )
Applying partial fractions to the right hand side gives:
L(y) =
−6
1
2
+
−
(x + 4) (s − 2) s2
Using a table to find the inverse Laplace gives:
y = −6e−4t + e2t − 2t
Now that we have y we can substitute its equation into the original equation for x0 :
dx
= 2y + 4t
dt
becomes:
dx
= 2(−6e−4t + e2t − 2t) + 4t
dt
dx
= −12e−4t + 2e2t
dt
Integrating gives:
x = 3e−4t + e2t + C
Applying the initial condition shows C = 0 and our solutions are:
x = 3e−4t + e2t
581.
y = −6e−4t + e2t − 2t
5.3. LINEAR SYSTEMS AND THE LAPLACE TRANSFORM
155
Solve:
x0 = x + y
x(0) = 0
y 0 = x + y − tet
x0 = y − x + 4t2 + 1
x(0) = 1
y 0 = x − y + 2t2 + 2t − 1
y(0) = 1
582.
Solve:
y(0) = 0
583.
Solve:
x0 = 2x + y − t − 1
x(0) = 1
y0 = y − t
y(0) = 1
584.
Solve:
x0 = x + y − t sin(t) + t cos(t)
x(0) = 0
y 0 = x − t sin(t) + cos(t)
x0 = 2x − y + 1 − 2t
x(0) = 1
y 0 = 2y − x + t
y(0) = 0
585.
Solve:
y(0) = 1
586.
Solve:
x0 = x − y − t3 + 4t2 + 1
y 0 = 2x − y − 2t3 + t2 + t + 1
x(0) = 0
y(0) = 0
587.
Solve:
x0 = x + y + 2tet
y 0 = x0 − t2 et − 3tet
x(0) = 0
y(0) = 1
588.
Solve:
x0 = x + y
x(0) = 0
y 0 = x0 − x
y(0) = 1
589.
Solve:
x0 = y − 3t
x(0) = 0
y 0 = x0 + 6x − 6t3 − 3t2 + 2
y(0) = 1