Download Lecture 10 - Second order linear differential equations

Lecture 10 - Second order linear differential equations
In the first part of the course, we studied differential equations of the general form:
dy
= f (t, y)
dt
dy
dy
In other words,
is equal to some expression involving both y and t (example:
= y 2 + t2 ). This type
dt
dt
of differential equations are called first order differential equations because they provide an expression
for the first derivative of y in terms of y and t.
Here is a summary of the methods we studied so far:
dy
= p(t)y + g(t)
I. First order linear differential equations:
dt
dy
= ay − b
Form:
dt
Method: substitution rule of integration
dy
Form:
+ p(t)y = g(t)
dt
Method: product rule of derivatives
II. First order nonlinear differential equations
dy
= g(t) (separable equations)
Form: h(y)
dt
Method: chain rule of derivatives
y
Form: Equations that can be made separable by the substitution u(x) = x
Method: change of variable from y to u
Form: Equations that can be made separable by the substitution u(x) = ax + by
Method: change of variable from y to u
dy
Form:
+ p(t)y = g(t)y n (Bernoulli equations)
dt
Method: change of variable u(x) = y −(n−1)
In the second part of the course, we will study methods for solving second order differential equations.
A second order differential equation has the general form:
dy
d2 y
= f (t, y, )
dt
dt2
In other words,
d2 y
dy
is equal to some expression involving
, y, and t (example:
dt
dt2
d2 y
dy
=5
− 7y + sin(t)). We will focus mainly on second order linear differential equations because there
dt
dt2
are very few algorithms available for second order nonlinear differential equations. A second order linear
differential equation has the general form:
d2 y
dy
= q(t)
+ r(t)y + g(t)
dt
dt2
We will encounter most second order linear differential equations in the form:
p(t)y 00 + q(t)y 0 + r(t)y = g(t)
If g(t) = 0, then we call such an equation homogeneous.
If g(t) 6= 0, we call it nonhomogeneous.
(People who are familiar with Linear Algebra, take notice!)
We will first study homogeneous equations where the functions p(t), q(t), and r(t) are constant.
ay 00 + by 0 + cy = 0
Problem: Solve the differential equation y 00 − y = 0, y(0) = 2, y 0 (0) = −1.
Observations:
1. a = 1, b = −1, c = 0
1
2
2. We are now given two initial conditions: one for y and another for y 0 . Intuitively, this is because when
you take the antiderivative twice, you end up with two constants that need to be solved for.
Solution:
y 00 − y = 0
y 00 = y
We are looking for a function y that is equal to its second derivative!
First guess: y(t) = et .
Is y a solution?
1. Does y satisfy the differential equation? Yes!
y 0 (t) = et
y 00 (t) = et
y 00 = y
2. Does y satisfy the initial conditions? No!
y(0) = 1
Improved first guess: y(t) = cet .
Is y a solution?
1. Does y satisfy the differential equation? Yes!
y 0 (t) = cet
y 00 (t) = cet
y 00 = y
2. Can we find c so that y satisfies the initial conditions? No!
Solving for c:
y(0) = c = 2
y(t) = 2et
y 0 (t) = 2et
y 0 (0) = 2
Second guess: y(t) = de−t .
Is y a solution?
1. Does it satisfy the differential equation? Yes!
y 0 (t) = −de−t
y 00 (t) = de−t
y 00 = y
2. Can we find d so that y satisfies the initial conditions? No!
Solving for d:
y(0) = d = 2
y(t) = 2e−t
y 0 (t) = −2e−t
y 0 (0) = −2
So far we found two families y1 (t) = cet and y2 (t) = de−t (y1 (t) = cet is a family of functions because it
encompasses infinitely many functions corresponding to different values of c) both of which satisfy the
differential equation but neither of which contains a function that satisfies the initial conditions. Here
comes the key insight:
Add the two families: y(t) = cet + de−t
Is y a solution?
1. Does y satisfy the differential equation? Yes!
y 0 (t) = cet − de−t
y 00 (t) = cet + de−t
y 00 = y
2. Can we find c and d so that y satisfies the initial conditions? Yes!
Solving for c and d:
y(0) = c + d = 2
y 0 (0) = c − d = −1.
3
c + d = 2 and c − d = −1
c=2−d
2 − d − d = −1
d = 32
c = 2 − 32 = 21
1 t 3 −t
e + e
2
2
Fact 1: Suppose v(t) is a solution to p(t)y 00 + q(t)y 0 + r(t)y = 0, then so is w(t) = cv(t) for any constant c.
“A solution multiplied by a constant is also a solution.”
y(t) =
Proof:
Since v(t) is a solution to p(t)y 00 + q(t)y 0 + r(t)y = 0, we have:
p(t)v 00 + q(t)v 0 + r(t)v = 0
We need to verify that p(t)w00 + q(t)w0 + r(t)w = 0!
p(t)w00 + q(t)w0 + r(t)w =
p(t)(cv)00 + q(t)(cv)0 + r(t)(cv) =
p(t)cv 00 + q(t)cv 0 + r(t)cv =
c(p(t)v 00 + q(t)v 0 + r(t)v) =
c · 0 = 0!
Fact 2: Suppose v(t) and w(t) are solutions to p(t)y 00 + q(t)y 0 + r(t)y = 0, then so is u(t) = v(t) + w(t).
“The sum of two solutions is also a solution.”
Proof:
Since v(t) and w(t) are solutions to p(t)y 00 + q(t)y 0 + r(t)y = 0, we have:
p(t)v 00 + q(t)v 0 + r(t)v = 0
p(t)w00 + q(t)w0 + r(t)w = 0
We need to verify that p(t)u00 + q(t)u0 + r(t)u = 0!
p(t)u00 + q(t)u0 + r(t)u = 0
p(t)(v + w)00 + q(t)(v + w)0 + r(t)(v + w) =
p(t)(v 00 + w00 ) + q(t)(v 0 + w0 ) + r(t)(v + w) =
p(t)v 00 + p(t)w00 + q(t)v 0 + q(t)w0 + r(t)v + r(t)w =
p(t)v 00 + q(t)v 0 + r(t)v + p(t)w00 + q(t)w0 + r(t)w =
0 + 0 = 0!
Fact 3: Suppose y1 (t) and y2 (t) are solutions to p(t)y 00 + q(t)y 0 + r(t)y = 0, then so is
y(t) = c1 y1 (t) + c2 y2 (t).
“A linear combination of two solutions is also a solution!”
Proof: Combine facts 1 and 2.
The general strategy for solving second order linear equations of the form ay 00 + by 0 + cy = 0 will involve, as
above, finding two families of functions which satisfy the differential equation and adding them together
together to obtain the solution we are looking for.
Problem: Solve the differential equation y 00 + 5y 0 + 6y = 0, y(0) = 2, y 0 (0) = 3.
Solution:
Guess: y(t) = ert for some constant r
Solving for r:
y 0 (t) = rert
y 00 (t) = r2 ert
y 00 + 5y 0 + 6y = r2 ert + 5rert + 6ert = (r2 + 5r + 6)ert = 0
Therefore the differential equation will be satisfied exactly when r2 + 5r + 6 = 0!
The equation
r2 + 5r + 6 = 0
is called the characteristic equation of the differential equation y 00 + 5y 0 + 6y = 0!
4
To obtain r, we solve the characteristic equation:
r2 + 5r + 6 = (r + 2)(r + 3) = 0
So r = −2 and r = −3.
Therefore y1 (t) = c1 e−2t and y2 (t) = c2 e−3t are two families of solutions.
The general solution:
y(t) = c1 e−2t + c2 e−3t
It remains to solve for c1 and c2 :
y(0) = c1 + c2 = 2
y 0 (t) = −2c1 e−2t − 3c2 e−3t
y 0 (0) = −2c1 − 3c2 = 3
c1 + c2 = 2 and −2c1 − 3c2 = 3
c1 = 9
c2 = −7
y(t) = 9e−2t − 7e−3t
Problem: Solve the differential equation 4y 00 − 8y 0 + 3y = 0, y(0) = 2, y 0 (0) = 21 .
Solution:
Guess: y(t) = ert for some constant r
Solving for r:
y 0 (t) = rert
y 00 (t) = r2 ert
4y 00 − 8y 0 + 3y = 4r2 ert − 8rert + 3ert = (4r2 − 8r + 3)ert = 0
The characteristic
equation: 4r2 − 8r + 3 = 0
p
√
√
2 − 4ac
−b
±
b
64
−
4
·
4
·
3
8
±
16 = 8 ± 4
8
±
r=
=
=
2a
8
8
8
1
3
So r = 2 and r = 2 .
3
1
Therefore y1 (t) = c1 e 2 t and y2 (t) = c2 e 2 t are two families of solutions.
The general solution:
3
1
y(t) = c1 e 2 t + c2 e 2 t
It remains to solve for c1 and c2 :
3
1
y(0) = c1 + c2 = 2 y 0 (t) = 23 c1 e 2 t + 12 c2 e 2 t
y 0 (0) = 32 c1 + 12 c2 = 12
So c1 + c2 = 2 and 32 c1 + 12 c2 = 12 .
c1 = − 21
c2 = 52
1 3
5 1
y(t) = − e 2 t + e 2 t
2
2
General strategy for equations ay 00 + by 0 + cy = 0, y(t0 ) = m, y 0 (t0 ) = n.
(Characteristic equation with two real solutions.)
Step 1: Find two families of solutions.
Guess: y(t) = ert for some constant r
Solving for r:
y 0 (t) = rert
y 00 (t) = r2 ert
ar2 ert + brert + cert = 0
ert (ar2 + br + c) = 0
The characteristic equation ar2 + br + c = 0 has two solutions r1 6= r2 .
Therefore y1 (t) = er1 t and y2 (t) = er2 t are two families of solutions.
The general solution:
y(t) = c1 er1 t + c2 er2 t
Step 2: Solve for c1 and c2 .
y(t0 ) = c1 er1 t0 + c2 er2 t1 = m
y 0 (t) = r1 c1 er1 t + r2 c2 er2 t
5
y 0 (t0 ) = r1 c1 er1 t0 + r2 c2 er2 t0 = n
(1) c1 er1 t0 + c2 er2 t1 = m
(2) r1 c1 er1 t0 + r2 c2 er2 t0 = n
We solve for c1 using equation (1):
r2 t0
c1 = m − rc12te0
e
We substitute into equation (2):
r2 t0
r1 m − rc12te0
er1 t0 + r2 c2 er2 t0 = n
e
r1 (m − c2 er2 t0 ) + r2 c2 er2 t0 = n
r1 m − c2 r1 er2 t0 + r2 c2 er2 t0 = n
c2 er2 t0 (r2 − r1 ) = n − r1 m
c2 = r2nt0 − r1 m
(r2 − r1 )
e
The only way we would not be able to obtain a solution for c2 is if r2 − r1 = 0. But since r1 6= r2 , we have
r2 − r1 6= 0.
Thus, we can always find c1 and c2 to satisfy the initial conditions!
If we can choose the two solutions in such a way that we can satisfy ANY initial conditions, then we call
such solutions independent. Two independent solutions serve as building blocks for all other solutions!
Thus, once you find two independent solutions, you can build any other solution out of them by choosing
the appropriate constants. Two independent solutions are also called a fundamental set of solutions.
Questions:
1. What happens if the characteristic equation has one solution or no solutions?
2. From the general strategy, we learned that a differential equation ay 00 + by 0 + cy = 0 with initial
conditions y(t0 ) = m and y 0 (t0 ) = n always has a solution as long as the characteristic equation has two
solutions. Is this solution unique?