Download second-order differential equations

SECOND-ORDER
DIFFERENTIAL EQUATIONS
OVERVIEW In this chapter we extend oUI study of differential equations to those of second
order. Second-order differential equations arise in many applications in the sciences and
engineering. For instance, they can be applied to the study of vibrating springs and electric
circuits. You will learn how to solve such differential equations by several methods in this
chapter.
Second-Order Linear Equations
An equation of the form
P(x)y"(x)
+ Q(x)y'(x) + R(x)y(x)
= G(x),
(1)
which is linear in y and its derivatives, is called a second-order linear differential equation. We assume that the functions P, Q, R, and G are continuous throughout some open
interval 1. If G(x) is identically zero on I, the equation is said to be homogeneous; otherwise it is called nonhomogeneous. Therefore, the form of a second-order linear homogeneous differential equation is
P(x)y"
+
Q(x)y'
+ R(x)y
=
O.
(2)
We also assume thatP(x) is never zero for any x EI.
Two fundamental results are important to solving Equation (2). The first of these says
that if we know two solutions Yl and Y2 of the linear homogeneous equation, then any
linear combination y = ClYl + C2Y, is also a solution for any constants Cl and C2.
THEOREM i-The Superposition Prindple
If Yl(X) and y,(x) are two solutions
Cl and C2, the
to the linear homogeneous equation (2), then for any constants
function
y(x) = cm(x)
+
C2Y2(X)
is also a solution to Equation (2).
17-1
17 -2
Chapter 17: Second-Order Differential Equations
Proof Substituting Y into Equation (2), we have
P(x)y"
+ Q(x)y' + R(x)y
= P(X)(CIYI + C2Y2)" + Q(X)(CIYI + C2Y2)' + R(X)(CIYI + C2Y2)
= P(X)(CIY!" + C2Y2") + Q(X)(CIYI' + C2Y2') + R(X)(CIYI + C2YV
= CI(P(X)YI" + Q(X)YI' + R(X)YI) + C2(P(X)Y2" + Q(X)Y2' + R(X)Y2)
=
=
CI(O)
Therefore, Y = CIYI
0, YI is a solution
=
0, Y2 is a solution
+ C2(0) = O.
+ C2Y2 is a solution of Equation (2).
•
Theorem I immediately establishes the following facta concerning solutions tu the
linear homogeneous equation.
1.
A sum of two solutions YI
+ Y2
tu Equation (2) is also a solution. (Choose CI =
C2 = \.)
2.
3.
A constant multiple kYI of any solution YI to Equation (2) is also a solution. (Choose
CI = k and C2 = 0.)
The trivial solution y(x) = 0 is always a solution tu the linear homogeneous equation. (Choose CI = C2 = 0.)
The second fundamental result about solutions to the linear homogeneous equation
concerns its general solution or solution containing all solutions. This result says that
there are two solutions YI and Y2 such that any solution is some linear combination of them
for suitable values of the constants CI and C2. However, not just any pair of solutions will
do. The solutions must be linearly independent, which means that neither YI nor Y2 is a
constant multiple of the other. For example, the functions f(x) = eX and g(x) = xe x are
linearly independent, whereas f(x) = X2 and g(x) = 7X2 are not (so they are linearly dependent). These results on linear independence and the following theorem are proved in
more advanced courses.
THEOREM 2
If P, Q, and R are continuous over the open interval I and P(x) is
never zero on I, then the linear homogeneous equation (2) has two linearly
independent solutions YI and Y2 on 1. Moreover, if YI and Y2 are any two linearly
independent solutions of Equation (2), then the general solution is given by
y(x) = CIYI(X) + C2Y2(X),
where CI and C2 are arbitrary constants.
We now turn our attention tu rmding two linearly independent solutions tu the special
case of Equation (2), where P, Q, andR are constant functions.
Constant-Coefficient Homogeneous Equations
Suppose we wish to solve the second-order homogeneous differential equation
ay"+by'+cy=O,
(3)
17-3
17.1 Second-Order Linear Equations
where a, b, and c are constants. To solve Equation (3), we seek a function which when
multiplied by a constant and added to a constant times its first derivative plus a constant
times its second derivative sums identically to zero. One function that behaves this W<rj is
the exponential function y = e"', when r is a constant. Two differentiations of this exponential function give y' = re'" and y' = r 2e"', which are just constant multiples of the
original exponential. Ifwe substitute y = e'" into Equation (3), we obtain
ar2e n: + bre1"X + ce rx
=
o.
Since the exponential function is never zero, we can divide this last equation through by
e"'. Thus, y = e'" is a solution to Equation (3) if and ouly if r is a solution to the algebraic
equation
ar2 + br + c
=
O.
(4)
Equation (4) is called the auxiliary equation (or charaeteristic equation) of the differential equation ay" + by' + cy = O. The auxiliary equation is a quadratic equation with
roots
-b
+ Vb 2 - 4ac
2a
and
'2 =
There are three cases to consider which depend on the value of the discriminant b 2
-
4ac.
Case 1: b2 - 4ac > O. In this case the auxiliary equation has two real and unequal roots
rl and r2. Then YI = e'" and Y2 = e"" are two linearly independent solutions to Equation
(3) because e"" is not a constant multiple of e'" (see Exercise 61). From Theorem 2 we
conclude the following result.
THEOREM 3
equation ar2
If rl and r2 are two real and unequal roots to the auxiliary
= 0, then
+ br + c
is the general solution to ay"
EXAMPLE 1
+ by' + cy
= O.
Find the general solution of the differential equation
y" - y' - 6y
=
O.
Solution Substitution of y = e'" into the differential equation yields the auxiliary
equation
which factors as
(r - 3)(r
+ 2)
=
O.
The roots are r1 = 3 and r2 = - 2. Thus, the general solution is
y = cle 3x
+ C2e- 2x•
•
17-4
Chapter 17: Second-Order Differential Equations
Case 2: b2 - 4ac = O. In this case rl = r2 = -bI2a. To simplify the notation, let
r = -bI2a. Then we have one solution YI = en< with 2ar + b = O. Since multiplication
of e'" by a constant fails to produce a second linearly independent solution, suppose we try
multiplying by afonction instead. The simplest such function would be u(x) = x, so let's
see if Y2 = xe'" is also a solution. Substituting Y2 into the differential equation gives
aY2" + by,' + CY2
=
a(2re'" + r 2xe"') + b(e'" + rxe"') + cxe'"
= (2ar + b)e'" + (ar2 + br + c)xe'"
= O(e"') + (O)xe'"
=
O.
The first term is zero because r = -bI2a; the second term is zero because r solves the
auxiliary equation. The functions YI = e'" and Y2 = xe'" are linearly independent (see
Exercise 62). From Theorem 2 we conclude the following result.
THEOREM 4
ar2
+ br + c
=
If r is the only (repeated) real root to the auxiliary equation
0, then
is the general solution to ay"
EXAMPLE 2
+ by' + cy
=
o.
Find the general solution to
y" + 4y' + 4y
=
o.
Salution The auxiliary equation is
r2+4r+4=0,
which factors into
(7 + 2)2
=
o.
Thus, r = -2 is a double root. Therefore, the general solution is
Y
=
cle-2x +
C2 xe - 2x•
•
2
Case 3: b - 4ac < O. In this case the auxiliary equation has two complex roots
rl = a + i{3 and r2 = a - i{3, where a and {3 are real numbers and i 2 = -1. (These real
numbers are a = -bl2a and {3 = Y4ac - b 2/2a.) These two complex roots then give
rise to two linearly independent solutions
Yl = e(a+ip)< = eax(cos {3x
+ i sin {3x)
and Y2 = e(a-ip)< = eax(cos {3x - i sin {3x).
(The expressions involving the sine and cosine terms follow from Euler's identity in Section 9.9.) However, the solutions YI and Y2 are complex valued rather than real valued
Nevertheless, because of the superposition principle (Theorem 1), we can obtain from
them the two real-valued solutions
Y3
=
I
+1
ZYI
ZY2
=
Q
e ax cos,...x
and
_I
I
_ax."
Y4 - 2i YI - 2i Y2 - e sm ,...x.
The functions Y3 and Y4 are linearly independent (see Exercise 63). From Theorem 2 we
conclude the following result.
17-5
17.1 Second-Order Linear Equations
If rl = a + if3 and r2 = a - if3 are two complex roots to the
auxiliary equation ar2 + br + c = 0, then
THEOREM 5
y = e""(cI cos f3x
+ by' + cy
is the general solution to ay"
EXAMPLE 3
+
=
C2Sin f3x)
o.
Find the general solution to the differential equation
y" - 4y' + 5y
=
O.
Solution The auxiliary equation is
r2-4r+5=0.
The roots are the complex parr r = (4 ± V16 - 20)/2 or
Thus, a = 2 and f3 = 1 give the general solution
rl =
2 + i and r2
=
2 - i.
•
Initial Value and Boundary Value Problems
To determine a unique solution to a {lISt-order linear differential equation, it was sufficient
to specify the value of the solution at a single point. Since the general solution to a secondorder equation contains two arbitrary constants, it is necessary to specify two conditions.
One way of doing this is to specify the value of the solution function and the value of its
derivative at a single point: y(xo) = Yo and y'(xo) = YI. These conditions are called initial
conditions. The following result is proved in more advanced texts and guarantees the existence of a unique solution for both homogeneous and nonhomogeneous second-order
linear initial value problems.
If P, Q, R, and G are continuous throughout an open interval f,
then there exists one and only one function y(x) satisfying both the differential
equation
THEOREM 6
P(x)y"(x)
+ Q(x)y'(x) + R(x)y(x)
=
G(x)
on the interval f, and the initial conditions
y(xo)
=
Yo
and
y'(xo)
=
Yl
at the specified point Xo E f.
It is important to realize that any real values can be assigned to Yo and YI and Theorem 6
applies. Here is an example of an initial value problem for a bomogeneous equation.
17-6
Chapter 17: Second-Order Differential Equations
EXAMPLE 4
Find the particular solution to the initial value problem
y"-2y'+y=0,
y(O)
= 1,
y'(0)
= -I.
Solution The auxiliary equation is
+ I
r2 - 2r
y
= (r -
1)2 = O.
The repeated real root is r = I, giving the general solution
Y
= CteX
+
C2 xex .
Then,
y' = cle x
-4
+ C2(X +
I)e x •
From the initial conditions we have
-6
-8
FIGURE 17"1
for Example 4.
and
Thus, CI = I and C2 = -2. The unique solution satisfying the initial conditions is
y
Particular solution curve
=
eX - be x .
•
The solution curve is shown in Figure 17.1.
Another approach to determine the values of the two arbitrary constants in the general
solution to a second-order differential equation is to specify the values of the solution
function at two different points in the interval!. That is, we solve the differential equation
subject to the boundary values
and
where Xl and X2 both belong to I. Here again the values for YI and Y2 can be any real
numbers. The differential equation together with specified boundary values is called a
boundary value problem. Unlike the result stated in Theorem 6, boundary value problems do not always possess a solution or more than one solution may exist (see Exercise
65). These problems are studied in more advanced texts, but here is an example for which
there is a unique solution.
EXAMPLE 5
Solve the boundary value problem
y"
+ 4y
=
0,
y(O)
= 0,
y(;;)
= I.
Solution The auxiliary equation is r2 + 4
= 0, which has the complex roots r = ±2i.
The general solution to the differential equation is
y = CI cos 2x + C2 sin 2x.
The boundary conditions are satisfied if
y(0)=cI"I+C2"0=0
It follows that CI = 0 and C2 = 2. The solution to the boundary value problem is
y
= 2 sin2x.
•
17-7
17.1 Second-Order Linear Equations
EXERCISES 17.1
In Exercises 1-30, fmd the general solution of the given equation.
a
4y ~ a
1. y' - y' - 12y
3. y'
+ 3y'
-
~
2. 3y" - y'
5.y·-4y~0
~
41. y" - 2y' - 3y
43. 4y"
6.y"-64y~0
45. 4y'
~
0
12y
11.y·+9y~0
12. y"
+ 4y' + 5y
13.y·+25y~0
14. y"
+Y = 0
+
19. y'
+ 5y
+ 2y' + 4y
+ 4y' + 9y
21. y"
=
a
a
~ a
0
49. 4y'
42.6y·-y'-y~0
+
lOy'
+y
+ 4y' + 5y
~
~
44. 9y'
53. 9y"
a
+
24y'
+ 9y ~
+ 16y ~
55. 6y" - 5y' - 4y
~
48. 6y'
0
0
51. 16y" - 24y'
0
+
12y'
+ 4y
+ \3y' - 5y
a
a
52.
+ 16y' + 52y
18. y" -2y'
+3y~0
In Exercises 56-60, solve the initial value problem.
20. 4y" - 4y'
+ \3y ~ 0
+ 16y ~ 0
56. y" - 2y'
d'y
dy
d'y
dy
d'y
23. tb;2+4tb;+4y~0
25. tb;2+6tb;+9y~0
d'y
dy
d'y
dy
dy
dy
26. 4tb;2 - 12tb;
58.
d'y
27.4tb;2+4tb;+y~0
59.
+ 9y
~ 0
dy
28.4tb;2-4tb;+y~0
d'y
29. 9tb;2+6tb;+y~0
dy
30. 9 tb;2 - 12 tb;
+ 4y
~
0
In Exercises 31-40, fmd the uoique solution of the second-order
initial value problero.
31. y" + 6y' + 5y ~ 0, y(O) ~ 0, y'(O) ~ 3
32. y" + 16y ~ 0, y(0) ~ 2, y'(O) ~ -2
33. y" + 12y ~ 0, y(0) ~ 0, y'(O) ~ I
34. 12y' + 5y' - 2y ~ 0, y(O) ~ I, y'(O) ~ -I
35. y" + 8y ~ 0, y(0) ~ -I, y'(O) ~ 2
36. y"
+ 4y' + 4y
~
+ 4y ~
38. 4y" - 4y' + y ~
37. y" - 4y'
d'y
39.4 tb;2
d'y
+
dy
12tb;
dy
40.9 tb;2 - 12tb;
+
0, y(O)
0, y(O)
0, y(O)
~
~
~
0, y'(O)
I, y'(O)
4, y'(O)
~
I
~
a
~
0, y(0)
~
0, y'(O)
~
60.
~
0, y(0)
~
2
61. Prove that the two solution fimctions in Theorem 3 are linearly independent.
62. Prove that the two solution fimctions in Theorem 4 are linearly independent.
63. Prove that the two solution fimctions in Theorem 5 are linearly independent.
64. Prove that if Yl and Y2 are linearly independent solutions to the
homogeneous equation (2), then the fimctions Y3 ~ Yl + Y2 and
y, ~ Yl - Y2 are also linearly independent solutions.
65. •• Show that there is no solution to the boundary value problem
~
0, y(O)
~
0, y(1T)
~
I.
b. Show that there are infmitely many solutions to the bouodary
value problero
y" + 4y
~
0, y(O)
~
0, y(1T)
~
~
O.
66. Show that if a, b, and c are positive constants, then all solutions of
the homogeneous differential equation
ay"+by'+cy=O
+ 4y
a
+ 2y' + y ~ 0, y(0) ~ I, y'(O) ~ I
4y" - 4y' + y ~ 0, y(O) ~ -I, y'(O) ~ 2
3y" + y' - 14y ~ 0, y(O) ~ 2, y'(O) ~ -I
4y" + 4y' + 5y ~ 0, y('IJ') ~ I, Y'(1T) ~ a
y" + 4y
~ 4
dy
9y ~ 0, y(0) ~ 2, tb;(0) ~ I
dy
-I, tb;(0)
~
57. y"
24. tb;2-6tb;+9y~0
d'y
+ 2y
0
a
16. y"
~
+ 8y'
~
6y'-5y'-6y~0
54. 4y'
~
22. y"
0
50.y·+4y'+6y~0
17. y'
16y
~
46.y·+2y'+2y~0
15. y' - 2y'
0
~
~
~
0
~
+ 4y' + y ~ a
+ 20y ~ 0
47. 25y"
+
10. 3y' - 2Oy'
0
In Exercises 41-55, fmd the general solution.
4.y"-9y=O
8. 9y" - y
7.2y·-y'-3y~0
9. 8y" - lOy' - 3y
0
=
I
approach zero as x ~
00.
17-8
Chapter 17: Second-Order Differential Equations
Nonh
Linear
In this section we study two methods for solving second-order linear noohomogeneous
differential equations with constant coefficients. These are the methods of undetermined
coefficients and variation ofparameters. We begin by considering the form of the general
solution.
Form of the General Solution
Suppose we wish to solve the noohomogeneous equation
ay"
+ by' + ey
(1)
= G(x),
where a, b, and c are constants and G is continuous over some open interval l. Let
Ye = ClYl + C2Y2 be the general solution to the associated complementary equation
ay"
+ by' + cy
=
o.
(2)
(We learned how to find Ye in Section 17.1.) Now suppose we could somehow come up
with a particular function YP that solves the noohomogeneous equation (I). Then the sum
Y = Ye
+ YP
(3)
also solves the noohomogeneous equation (I) because
a(ye
+ Yp)" + b(ye + Yp)' + c(ye + Yp)
= (aye" + bye' + eye) + (ayp' + byp' + eyp)
y, solves Eq. (2) and YP solves Eq. (\)
= 0 + G(x)
= G(x).
Moreover, if Y = y(x) is the general solution to the noohomogeneous equation (I), it must
have the form of Equation (3). The reason for this last statement follows from the observation that for any function YP satisfying Equation (I), we have
a(y - Yp)"
+ b(y
= (ay'
+ c(y
+ by' + ey)
- Yp)'
= G(x) - G(x) =
- Yp)
- (ayp"
+ byp' + eyp)
o.
Thus, Ye = Y - YP is the general solution to the homogeneous equation (2). We have
established the following result.
THEOREM 7
The general solution y = y(x) to the noohomogeneous differential equation (I) has the form
y = Ye
+ YP'
where the complementary solution Ye is the general solution to the associated
homogeneous equation (2) and Yp is any particular solution to the noohomogeneous equation (I).
17.2
Nonhomogeneous Linear Equations
17-9
The Method of Undetermined Coeffidents
This method for finding a particular solution YP to the nonhomogeneous equation (1) applies to special cases for which G(x) is a sum of terms of various polynomials p(x) multiplying an exponential with possibly sine or cosine factors. That is, G(x) is a sum of terms
of the following forms:
P2(x)e UX cos {3x,
P3(x)eax sin {3x.
For instance, 1 - x, e2x , xe x, cos x, and Sex - sin 2x represent functions in this category.
(Essentially these are functions solving homogeneous linear clifferential equations with
constant coefficients, but the equations may be of order higher than two.) We now present
several examples illustrating the method.
EXAMPLE 1
Solve the nonhomogeneous equation y" - 2y' - 3y = 1 - x 2.
Solution The auxiliary equation for the complementary equation y" - 2y' - 3y = 0 is
r2 - 2r - 3 = (r
+ l)(r - 3)
=
O.
It has the roots r = - 1 and r = 3 giving the complementary solution
Now G(x) = 1 - x 2 is a polynomial of degree 2. It would be reasonable to assume that a
particular solution to the given nonhomogeneous equation is also a polynomial of degree 2
because if y is a polynomial of degree 2, then y" - 2y' - 3y is also a polynomial of degree 2. So we seek a particular solution of the form
yp=Ax2+Bx+C.
We need to determine the unknown coefficients A, B, and C. When we substitote the polynomial YP and its derivatives into the given nonhomogeneous equation, we obtain
2A - 2(2Ax + B) - 3(Ax 2 + Bx + C) = 1 - x 2
or, collecting terms with like powers of x,
-3Ax 2 + (-4A - 3B)x
+ (2A -
2B - 3C) = 1 - x 2.
This last equation holds for all values of x if its two sides are identical polynomials of
degree 2. Thus, we equate corresponding powers of x to get
-3A = -1,
-4A - 3B = 0,
and
2A - 2B - 3C = 1.
These equations imply in turn that A = 1/3, B = -4/9, and C = 5/27. Substituting these
values into the quadratic expression for our particular solution gives
By Theorem 7, the general solution to the nonhomogeneous equation is
_
y-Yc
+ yp-cle
_
-. +
C2e
3.
+ :31 x 2 -94 x + 27'
5
•
17 -1 0
Chapter 17: Second-Order Differential Equations
EXAMPLE 2
Find a particular solution of y" - y' = 2 sin x.
Solution If we try to find a particular solution of the fann
yp=Asinx
and substitute the derivatives of yp in the given equation, we f"md that A must satisfy the
equation
-A sin x
+ A cos x
= 2 sin x
for all values of x. Since this requires A to equal both -2 and 0 at the same time, we conclude that the nonhomogeneous differential equation has no solution of the form A sin x.
It turns out that the required fann is the sum
yp = A sin x
+ B cosx.
The result of substitoting the derivatives of this new trial solution into the differential
equation is
-A sin x - B cos x - (A cos x - B sin x) = 2 sin x
or
(B - A) sin x - (A
+ B) cos x
= 2 sinx.
This last equation must be an identity. Equating the coefficients for like terms on each side
then gives
B-A=2
A
and
+B
=
O.
Simultaneous solution of these two equations gives A = - 1 and B = 1. Our particular
solution is
YP
EXAMPLE 3
=
•
cosx - sinx.
Find a particular solution of y' - 3y'
+ 2y
= 5e x •
Solution If we substitute
YP = Ae
x
and its derivatives in the differential equation, we f"md that
Aex
-
3Aex
+
2Ae x = 5e x
or
0= 5e x •
However, the exponential function is never zero. The trouble can be traced to the fact that
y = eX is already a solution of the related homogeneous equation
y" - 3y'
+ 2y
=
O.
The auxiliary equation is
r2 - 3r
+2
= (r -
1)(r - 2) = 0,
which has r = 1 as a root. So we would expect Ae x to become zero when substitoted into
the left-hand side of the differential equation.
The appropriate way to modify the trial solution in this case is to multiply Aex by x.
Thus, our new trial solution is
x
YP = Axe .
17.2
Nonhomogeneous Linear Equations
17-11
The result of substituting the derivatives of this new candidate into the differential equation is
(Axe X + ZAe") - 3(Axe X + Ae")
+ ZAxe x
= 5e x
or
Thus, A
=
-5 gives our sought-after particular solution
YP
=
•
-5xe x.
Find a particular solution of y" - 6y' + 9y
EXAMPLE 4
= e 3x •
Solutton The auxiliary equation for the complementary equation
+9
r2 - 6r
= (r - 3)2 = 0
as a repeated root. The appropriate cboice for yp in this case is neither Ae 3x nor
Axe because the complementary solution contains both of those terms already. Thus, we
choose a term containing the next higher power of x as a factor. When we substitute
has r
= 3
3x
YP =
Ax2e 3x
and its derivatives in the given differential equation, we get
(9Ax 2e 3X
- 6(3Ax 2e 3x
+ lZAxe 3x + ZAe~
+ ZAxe 3") + 9Ax 2e 3x
= e 3x
or
Thus, A
=
1/2, and the particular solution is
YP =
•
!x 2e"':
When we wish to fmd a particular solution of Equation (1) and the function G(x) is the
sum of two or more terms, we choose a tria1 function for each term in G(x) and add them.
EXAMPLE 5
Find the general solution to y" - y'
= 5e x -
sin 2x.
Solutton We fITSt check the auxiliary equation
r2 - r = O.
Its roots are r = I and r
mogeneous equation is
=
O. Therefore, the complementary solution to the associated hoYc
= cle
x
+
C2·
We now seek a particular solution Yp. That is, we seek a function that will produce
5e x - sin 2x when substituted into the left-hand side of the given differential equation.
One part of YP is to produce 5e', the other -sin 2x.
Since any function of the form CteX is a solution of the associated homogeneous equation, we choose our trial solution YP to be the sum
Yp = Axe'
+ B cos 2x + C sin 2x,
x
including xe where we might otherwise have included only eX. When the derivatives of YP
are substituted into the differential equation, the resulting equation is
(Axe'
+ ZAe'
- 4B cos 2x - 4C sin 2x)
- (Axe'
+ Ae'
- 2B sin 2x
+ 2C cos 2x)
= 5e' -
sin 2x
17 -12 Chapter 17: Second-Order Differential Equations
or
Ae' - (4B + 2C) cos 2x + (2B - 4C) sin 2x = 5e' - sin 2x.
TIris equation will hold if
A = 5,
4B
+ 2C
=
2B-4C= -I,
0,
or A = 5, B = -1/10, and C = 1/5. Our particular solution is
I .2x
I
2x+ Ssm.
Yp - 5
xe' - TOcOS
The general solution to the dllferential equation is
y =Yc + Yp =
cle'
+
C2
+ 5xe' -
l~cos2x
+ tsin2x.
•
You may find the following table helpful in solving the problems at the end of this
section.
TABLE 17.1
The method of undetermined coefficients for seLected equations
of the form
oy"
IT G(x) has a term
that is a constant
multiple of ...
e'"
+ by' + cy
=
And if
G(x).
Then include this
expression in the
trial function for Y,.
r is not a root of
the auxiliary equation
Ae'"
r is a single root of the
auxiliary equation
Axe'"
r is a double root of the
auxiliary equation
sin kx, cos kx
ki is not a root of
the auxiliary equation
Bcoskx + Csinkx
px 2 +qx+m
ois not a root of the
Dx2+Ex+F
auxiliary equation
ois a single root of the
Dx 3 + Ex 2 + Fx
auxiliary equation
ois a double root of the
Dx'
+ Ex 3 + Fx 2
auxiliary equation
The Method of Variation of Parameters
TIris is a general method for finding a particular solution of the nonhomogeneous equation
(I) once the general solution of the associated homogeneous equation is known. The
method consists of replacing the constants Cl and C2 in the complementary solution by
functions VI = VI(X) and V2 = V2(X) and requiring (in a way to be explained) that the
17.2
Nonhomogeneous Linear Equations
17-13
resulting expression satisfy the nonhomogeneous equation (I). There are two functions to
be detennined, and requiring that Equation (I) be satisfied is only one condition. As a second condition, we also require that
(4)
Then we have
Y = VIYI + V2Y2,
y' =
y" =
+ V2Y2',
VIY(" + V2Y2" +
VIYl'
Vl'Yl'
+ V2'Y2'.
lfwe substitote these expressions into the left-hand side of Equation (I), we obtain
vI(aYI"
+ bYI' +
CYI)
+ v2(aY2" + by2' + eY2) + a(vI'YI' + ViY2')
=
G(x).
The first two parenthetical terms are zero since YI and Y2 are solutions of the associated
homogeneous equation (2). So the nonhomogeneous equation (I) is satisfied if, in addition
to Equation (4), we require that
+ v2'yi)
a(VI'YI'
G(x).
=
(5)
Equations (4) and (5) can be solved together as a pair
Vt'Yt
+
V2'Y2
= 0,
for the unknown functions VI' and V2'. The usual procedure for solving this simple system
is to use the method ofdeterminants (also known as Cramer ~ Rule), which will be demonstrated in the examples to follow. Once the derivative functions VI' and V2' are known, the
two functions VI = VI(X) and V2 = V2(X) can be found by integration. Here is a summary
of the method.
Variation of Parameters Procedure
To use the method of variation of parameters to find a particular solution to the
nonhomogeneous equation
ay"
+ by' + cy
=
G(x),
we can work directly with Equations (4) and (5). It is not necessary to rederive
them. The steps are as follows.
1.
Solve the associated homogeneous equation
ay"+by'+cy=O
2.
to tmd the functions YI and Y2.
Solve the equations
+
Vt'Yt
Vt'Yt'
3.
4.
V2'Y2 =
+ 112'y2'
=
0,
G(x)
-a-
simultaneously for the derivative functions VI' and V2'.
Integrate VI' and V2' to find the functions VI = VI(X) and V2 = V2(X).
Write down the particular solution to nonhomogeneous equation (I) as
YP =
VJ)'1
+ V2Y2·
17 -14 Chapter 17: Second-Order Differential Equations
EXAMPLE 6
Find the general solution to the equation
y" + Y
= tanx.
Solution The solution of the homogeneous equation
y" + Y
=
0
= Cl COS X
+
Cz
is given by
Yc
sinx.
Since Yl(X) = cos x and Y2(X) = sin x, the conditions to be satisfied in Equations (4) and
(5) are
Vt'
-Vt'
+
V2'
sinx = 0,
sinx +
V2'
cosx = tanx.
cos x
a
= 1
Solution of this system gives
Vl' =
0
sin x I
tan x
cosx
I
-c"--co-,,-x--S-in--"XC1
-smx
-tan x sin x
cos2 x + 8in2 x
cosx·
cos X 1
Likewise,
I c~sx
V2' =
-smx
I-smx
c~sx
~xl
sinx
cos X
=
I
sinx.
After integmting VI' and V2', we have
Vl(X) =
j
= =
and
V2(X) =
-Sin2 x
cos x dx
j
(sec x - cos x) dx
-In Isecx
j
+ tanxl + sinx,
sin x dx = -cos x.
Note that we have omitted the constants of integration in detennining Vl and V2. They
would merely be absorbed into the arbitrary constants in the complementary solution.
Substitutiog Vl and V2 into the expression for YP in Step 4 gives
+ tanxl + sin x] cos x + (-cos x) sin x
(-cos x) In lsecx + tanxl.
YP = [-In Isecx
=
The general solution is
Y =
Cl
cos x
+ C2 sin x
- (cos x) In Isecx
+ tanxl.
•
17.2
EXAMPLE 7
Nonhomogeneous Linear Equations
17-15
Solve the nonhomogeneous equation
y" + y' - 2y
=
xe x.
Solution The auxiliary equation is
r2
+r
- 2 = (r
+ 2)(r
- I) = 0
giving the complementary solution
Yc
c}e- 2x
=
+ cze x .
The conditions to be satisfied in Equations (4) and (5) are
Vt'e- 2x + v2'e x = 0,
+ v2'e x =
-2vt'e-2x
xe x.
a
= 1
Solving the above system for VI' and vi gives
Ix~x
Vt' =
I-2ee
eXI
eX
2x
2x
eXI
eX
_ - xe 2x _
- 3e-x -
I
-3 xe
'-'
.
Likewise,
Integrating to obtain the parameter functions, we have
VI(X) =
J-t
=
xe 3x dx
_t(X~3X
- Je;x dx )
_ l( 1 3xe
)'-'
-27
,
and
Therefore,
The general solution to the differential equation is
Y=
lx
cle-
+
C2ex -
~xex +
kx
2 X
e ,
where the term (1/27)e Xin YP has been absorbed into the term C2ex in the complementary
solution.
•
17 -16 Chapter 17: Second-Order Differential Equations
EXERCISES 17.2
Solve the equations in Exercises 1-16 by the method of undetermined
coefficients.
L y" - 3y' - lOy = -3
2_ y" - 3y' - lOy = 2x - 3
4. y" + 2y' + Y
3. y" - y' = sinx
5. y"
+Y
7. y" - y' - 2y = 20 cos x
+x
= e2x
+ Y = a+ 3ex
+ 2y' + Y = 6 sin 2x
6. y" + Y
= cos3x
9. y" - y = eX
= x2
2
8. y"
10. y"
11. y" - y' - 6y = e-x - 7 cosX'
12. y"
+ 3y' + 2y
d'y
= e-x
+ e-2.t" -
dy,
+ 5 dx
13_ dx'
x
d'y
d'y
dy]x
15_ dx' - 3 dx = e - 12x
d'y
dy _
16_ dx'
,
+ 7 dx - 42x + 5x + I
Solve the equations in Exercises 17-28 by variation of parameters.
17. y"
+ y'
18. y"
+Y
19. y"
+ Y = sinx
+ 2y' + Y =
21. y"
= x
=
20. y"
23. y" - y = eX
25. y"
+ 4y' + Sy
d'y
27. dx'
+Y
d'y
28. dx 2
-
=
= secx,
dy
dx
=
+ 2y' + Y
= eX
22.y"-y=x
24. y" - y = sinx
e-x
10
26. y" - y'
1f
=
2x
eXcosx, x> 0
31. y" + Y = 2 cosx
32. y"
+ y'
+ Bxe~
+ sinx, Yp = A cosX' + B sinx
+ sinx, YP = Ax cosX' + Bx sinx
- 2y = xe x , YP = Ax 2e x
+ Bxe x
coefficients.
dy
33 - - = eX
• dx'
dx
+ e-r
d'y
dy
35. dx,-4
-5y =ex +4
dx
=
cscx,
39. y" - 8y' =
o<x<w
43. y"
+
45. y"
+Y
42. y"
2y' = x 2
=
46. y" - 3y'
e~
-
44. y"
secxtanx,
+
+ 4y = sinx
+ 4y' + 5y = x + 2
+ 9y = 9x - cosX'
40. y"
e!b:
41. y" - y' = x 3
_'!!
2
< x < '!!
2
2y = eX - e'h
The method of undetermined coefficients can sometimes be used to
solve flISt-order ordinary differential equations. Use the method to
solve the equations in Exercises 47-50.
47. y' - 3y = eX
48. y'
+ 4y = x
49. y' - 3y = 5e'"
SO. y'
+Y
d'y
51. dx'
+ Y = sec x, -2 < x < 2;
d'y
52. dx'
+ Y = e2x;
,1f
y(O)
= 0,
1f
= sinx
y(O)
= y'(O) = I
2
=5
y'(O)
d'y
53. y"
x' - x, y(0) = 0,
+ y' = x, YP = T
54. y"
+ Y = x, YP = 2 sin x + x,
55.
In Exercises 33-36, solve the given differential equations (a) by
variation of parameters and (h) by the method of undetermined
d'y
cotx, O<X<'7T
isfYing the equation and the given initial conditions.
In each of Exercises 29-32, the given differential equation has a particular solution YP of the form given. Determine the coefficients in Yp.
30. y" - y' = cosX'
=
In Exercises 53-58, verify that the given function is a particular solution to the specified nonhomogeneous equation. Find the general solution and evaluate its arbitrary constants to f'md the unique solution sat-
1f
-2 < x < 2
Then solve the differential equation.
29. y" - 5y' = xeil:, YP = Ax 2e 5.l:
38. y"
+Y
+Y
Solve the differential equations in Exercises 51 and 52 sui!ject to the
given initial conditions.
_'!! < x < '!!
2
2
tanx,
37. y"
dy
14_ dx' - dx = -8x + 3
= 15x
Solve the differential equations in Exercises 37-46. Some of the equations can be solved by the method of undetermined coefficients, but
others cannot.
dy
34. - 4dx'
dx
+
4. = 2e 2x
'
d'y
dy
9x
36. dx' - 9dx = ge
ty" + y' + Y
57. y" - 2y'
58. y" - 2y'
YP
+Y
+Y
= xexJnx,
= 0,
=0
y'(O)
=0
4el{cosx - sinx),
=
YP = 2e x cos x, y(0)
56. y" - y' - 2y
y(O)
y'(O)
=I
= 0,
y' (0)
- 2x, YP
=I
=x
- I, y(O)
= 0,
y'(O)
=
2e x, YP = x 2e x, y(O) = 1, y'(O) = 0
=
x -leX, x
y(1)
=I
> 0,
= e, y'(I) = a
In Exercises 59 and 60, two linearly independent solutions YI and Y2
are given to the associated homogeneous equation of the variable-
coefficient nonhomogeneous equation. Use the method of variation of
parameters to fmd a particular solution to the nonhomogeneous equation. Assmne x > 0 in each exercise.
59. x7" + 2xy' - 2y = x 2, Yl = x- 2 , Y2 = x
60.
x7" + xy'
- Y=
X,
YI = x-I, Y2 = x
17.3
I ,,,
Applications
17-17
AppLications
In this section we apply second-order differential equations to the study of vibrating
springs and electric circuits.
Vibrations
A spring has its upper end fastened to a rigid support, as shown in Figure 17.2. An object
of mass m is suspended from the spring and stretches it a length s when the spring comes
to rest in an equilibrium position. According to Hooke's Law (Section 6.5), the tension
force in the spring is ks, where k is the spring constant. The force due to gravity pulling
down on the spring is mg, and equilibrium requires that
massm
ks = mg.
at equilibrium
y
FIGURE 17.2
Massm
stretches a spring by
length s to the equilibrium
position at y = O.
(1)
Suppose that the object is pulled down an additionsl amount Yo beyond the equilibrium position and then released We want to study the object's motion, that is, the vertical position
of its center of mass at any future time.
Let y, with positive direction downward, denote the displacement position of the object away from the equilibrium position y = 0 at any time t after the motion has started.
Then the forces acting on the object are (see Figure 17.3)
Fp
=
mg,
F, = k(s
F,=ll
the propulsion force due to gravity,
+ y),
the restoring force of the spring's tension,
dy
,
dt
a frictional force assumed proportional to velocity.
The frictional force tends to retard the motion of the object. The resultant of these forces is
F = Fp - F. - F" and by Newton's second law F = ma, we must then have
a"ly
m dt 2
=
dy
mg - ks - ky - Il dt'
By Equation (1), mg - ks = 0, so this last equation becomes
dly
m dt 2
y
Yo
start
position
y
FIGURE 17.3 Thepropulsion
force ("";ght) Fp pulls iIle mass
downward, but !he spring
restoring force F, aod fric1iooa1
force F, pullille mass upward.
The motion starts at Y = Yo with
iIle mass vihratiog up aod down.
+
dy
Il dt
+ ky
=
(2)
0,
subject to the initial conditions y(O) = Yo and y' (0) = O. (Here we use the prime notation
to denote differentiation with respect to time t.)
You might expect that the motion predicted by Equation (2) will be oscillatory about
the equilibrium position y = 0 and eventually dsmp to zero because of the retarding frictional force. This is indeed the case, and we will show how the constants m, Il, and k determine the nature of the dsmping. You will also see that if there is no friction (so Il = 0),
then the object will simply oscillate indefinitely.
Simple Harmonic Motion
Suppose first that there is no retarding frictional force. Then Il = 0 and there is no dsmping. lfwe substitote w = Vkj;. to simplify our calculations, then the second-order equation (2) becomes
with
y(O) = Yo
and
y'(O) = O.
17 -18 Chapter 17: Second-Order Differential Equations
The auxiliary equation is
r2
+ if
0,
=
having the imaginary roots r = ±wi. The general solution to the differential equation in
(2) is
Y =
Cl
cos rut
+
sin wt.
C2
(3)
To fit the initial conditions, we compute
y'
-Cl Cd sin
=
rut +
C2W cos
wt
and then substitute the conditions. This yields c, = Yo and C2 =
o. The particular solution
Y=Yocoswl
(4)
describes the motion of the object. Equation (4) represents simple harmonic motion of
amplitude Yo and period T = 27T/W.
The general solution given by Equation (3) can be combined into a single term by
using the trigonometric identity
sin(WI
+ cf»
+ sin wi cos cf>.
wi sin cf>
= cos
To apply the identity, we take (see Figure 17.4)
c,
=
C sin cf>
and
C2
=
C cos cf>,
where
C= Vc,2
FIGURE 17_4
= Ceosc/>.
C2
c, =
C sin c/> and
+
co'
_
and
-1 Cl
cf> - tan
C2·
Then the general solution in Equation (3) can be written in the alternative form
Y = C sin (WI
+ cf».
(5)
Here C and cf> may he taken as two new arbitrary constants, replacing the two constants c,
and C2. Equation (5) represents simple harmonic motion of amplitude C and period
T = 27T/W. The angle wi + cf> is called the phase angie, and cf> may be interpreted as its
initial value. A graph of the simple harmonic motion represented by Equation (5) is given
in Figure 17.5.
y
Period
~ T=~ ~
C
Csin."
I
I
I
I
o
-C
y=Csin(w,+.,,)
FIGURE 17.5 Simple hannonie motion of amplitude C
and period T wi1h initial phase angle c/> (Equation 5).
17.3 Applications
17 -19
Damped Motion
Assume now that there is friction in the spring system, so [; "" O. If we substitute
w = ~ and 2b = [;/m, then the differential equation (2) is
y"
+ 2by' + .Jy
=
O.
(6)
The auxiliary equation is
,2+2b,+w2 =O,
with roots , = -b ± yCb"2-_-w'''. Three cases now present themselves, depending upon
the relative sizes of b and w.
Case 1: b = Cd. The double root of the auxiliary equation is real and equals, = w. The
general solution to Equation (6) is
y = (Cl
+ c2t)e-"".
This situation of motion is called critical damping and is not oscillatory. Figure 17.6a
shows an example of this kind of damped motion.
Case 2:
b>
-b +
is given by
'1 =
The roots of the auxiliary equation are real and unequal, given by
2
Vb - w' and '2 = -b - Vb 2 - w'. The general solution to Equation (6)
Cd.
y = Cl e (-b+v'b'-w'p
+ C2e (-b-v'b'-w'Jt.
Here again the motion is not oscillatory and both '1 and '2 are negative. Thus y approaches
zero as time goes on. This motion is referred to as overdamping (see Figure l7.6b).
Case 3: b
, =
<
Cd.
-b ± iVw2
The roots to the auxiliary equation are complex and given by
b 2• The general solution to Equation (6) is given by
-
y = e- bl(cl cosVw' - b 2 t
+ C2SinVw' - b 2 t).
This situation, called underdamping, represents damved oscillatory motion. It is analogous to simple harmonic motion of period T = 21T/V w' - b 2except that the amplitude
is not constant but damped by the factor e -bl. Therefore, the motion tends to zero as t
increases, so the vibrations tend to die out as time goes on. Notice that the period
T = 21T/Vw2 - b 2 is larger than the period To = 21T/W in the friction-free system.
Moreover, the larger the value of b = [;/2m in the exponential damping factor, the more
quickly the vihrations tend to become unnoticeable. A curve illustrating underdamped motion is shown in Figure l7.6c.
y
y
y
t
o
y = e-t sin(St + 1114)
y=(l +t)e-t
(a) Critical damping
(b) Overdamping
(c) Underdamping
FIGURE 17.6 Three examples of damped vibra1xny motion for a spring system with
friction, so B oF O.
17-20
Chapter 17: Second-Order Differential Equations
An external force F(t) can also be added to lbe spring system modeled by Equation
(2). The forcing function may represent an external disturbance on lbe system. For instance, if lbe equation models an automobile suspension system, lbe forcing function
might represent periodic bumps or polboles in lbe road affecting lbe performance of lbe
suspension system; or it might represent lbe effects of winds when modeling lbe vertical
motion of a suspension bridge. Inclusion of a forcing function results in lbe second-order
nonhomogeneous equation
d 2y
m dt 2
dy
+ lJ dt + ky
= F(t).
(7)
We leave lbe study of such spring systems tu a more advanced course.
Electric Circuits
The basic quantity in electricity is lbe charge q (analogous to lbe idea of mass). In an electric field we use lbe flow of charge, or current I = dql dt, as we might use velocity in a
gravitational field. There are many similarities between motion in a gravitational field and
lbe flow of electrons (lbe carriers of charge) in an electric field.
Consider lbe electric circuit shown in Figure 17.7. It consists of four components:
voltage source, resistor, inductor, and capacitor. Think of electrical flow as being like a
fluid flow, where lbe voltage source is lbe pump and lbe resistor, inductor, and capacitor
tend to block lbe flow. A battery or generator is an example of a source, producing a voltage that causes lbe current to flow through lbe circuit when lbe switch is closed. An electric light bulb or appliance would provide resistance. The inductance is due to a magnetic
field that opposes any change in lbe current as it flows through a coil. The capacitance is
normally created by two metal plates that alternate charges and lbus reverse lbe current
flow. The following symbols specify lbe quantities relevant to lbe circuit:
q: charge at a cross section ofa conductor measured in coulombs (abbreviated c);
I: current or rate of change of charge dq/dt (flow of electrons) at a cross section of a
conductor measured in amperes (abbreviated A);
E: electric (potential) source measured in volt. (abbreviated V);
V: difference in potential between two points along lbe conductor measured in volts (V).
R, Resistor
Voltage E
soun:e
L.lnd~
C, Capacitor
FIGURE 17.7 An elec1ric circuit.
Ohm observed lbat lbe current I flowing through a resistor, caused by a potential difference across it, is (approximately) proportional to lbe potential difference (voltage drop).
He named his constant of proportionality IIR and called R lbe resistance. So Ohm ~ law is
1
I=JiV,
17.3
Applications
17 -21
Similarly, it is known from physics that the voltage drops across an inductor and a capacitor are
and
q
C'
where L is the inductance and C is the capacitance (with q the charge on the capacitor).
The German physicist Gustav R. Kirchhoff (1824-1887) formulated the law that the
sum of the voltage drops in a closed circuit is equal to the supplied voltage E(t). Symbolically, this says that
RI
+L
dI
dt
q
+C
= E(t).
Since I = dqldt, Kirchhoff's law becomes
d 2q
dq
I
L dt 2 +R dt +Cq=E(t).
(8)
The second-order differential equation (8), which models an electric circuit, has exactly
the same form as Equation (7) modeling vibratory motion. Both models can be solved
using the methods developed in Section 17.2.
Summary
The following chart summarizes our analogies for the physics of motion of an object in a
spring system versus the flow of charged particles in an electrical circuit.
Linear Second-Order Constant-Coefficient Models
Mecbllllical System
my"
y:
y/:
y":
m:
8:
k:
F(t):
+ 8y' + ky
Electrical System
= F(t)
displacement
velocity
acceleration
mass
damping constant
spring constant
forcing function
Lq"
q:
q/:
q":
L:
R:
I/C:
E(t):
+ Rq' + lq
C
= E(t)
charge
current
change in current
inductance
resistance
where C is the capacitance
voltage source
EXERCISES 17.3
1. A 16·lb weight is attached to the lower eod of a coil spring sus·
peoded from the ceiling aod having a spring coostaot of I Ib/ft.
The resistance in the spring-mass system is numerically equal to
the instaotaoeous velocity. At t = 0 the weight is set in motion
from a position 2 ft below its eqnilibrium position by giving it a
downward velocity of2 ft/sec. Write ao initial value problem that
models the giveo situation.
2. An g·lb weight stretches a spring 4 ft. The spring-mass system resides in a medium offering a resistance to the motion that is numerically equal to 1.5 times the instantaoeous velocity. If the
weight is released at a position 2 ft above its equilibrium position
with a downward velocity of3 ft/sec. write ao initial value problem modeling the giveo situation.
17-22
Chapter 17: Second-Order Differential Equations
3. A 2()'lb weight is bung on an IS-in. spring and stretches it 6 in.
The weight is pulled dawn 5 in. and 5 lb are added to the weight !f
the weight is now released with a downward ",Iocity of". in./sec,
write an initial value problem modeling the vertical disp1acement
4. A I ()'Ib weight is suspended by a spring that is stretched 2 in. by
the weight. Assume a resistance whose magnitude is 20/ Vg Ib
times the instantaneous velocity v in feet per second. !fthe weight
is pulled down 3 in. below its equilibrium position and released,
formulate an initial value problem modeling the behavior of the
spring-mass system.
S. Ao (opea) electrical circuit coosists of an inductor, a resistor, and
a capacitor. There is an initial cbarge of 2 coulombs on the capac-
itor. At the instant the circuit is closed. a current of 3 amperes is
present and a voltage of E(t) = 20 cos 1 is applied. In this circuit
the voltage drop across the resistor is 4 times the instantaneous
cbange in the charge, the voltage drop acroas the capacitor is
10 times the charge, and the voltage drop across the inductor is
2 times the instantaneous change in the current Write an initial
value problem to model the circuit
6. Ao inductor of 2 henrys is connected in series with a resistor
of 12 obms, a capacitor of 1/ 16 farad, and a 300 volt battery.
Initially, the charge on the capacitor is zero and the current is
zero. Formulate an initial value problem modeling this electrical
circuit.
Mechanical units in the British and metric systems may be helpful
in doing the following problems.
Unit
Brlli.h System
MKSSystem
Distance
Feet (ft)
Slugs
Seconds (sec)
Pounds (lb)
32 ftjsec"
Meters (m)
Kilugrams (kg)
Seconds (sec)
Msss
Time
Force
g(earth)
Newtoos(N)
9.81 m/sec"
7. A 16-lb weight is attached to the lower end of a coil spring suspeoded from the ceiling and having a spriog coostant of Ilb/ ft.
The resistance in the spring-maas system is numerically equal to
the instantaneous velocity. At t = a the weight is set in motion
from a position 2 ft below its equilibrium position by giving it a
downward ",Iocity of 2 ft/sec. At the end of 'IT sec, determine
whether the mass is above or below the equilibrium position and
by what distance.
8. Ao 8-lb weight stretches a spring 4 ft. The spring-msss system
resides in a medium offering a resistance to the motion equal to
1.5 times the instantaneous velocity. If the weight is released at a
position 2 It shove its equilibrium position with a downward
velocity of 3 ft/ sec, fmd its position relative to the equilibrium
position 2 sec later.
9. A 2()'lb weight is hung on an 18-in. spring stretching it 6 in. The
weight is pulled down 5 in. and 5 lb are added to the weight. (fthe
weight is now released with a downward velocity of". in./sec,
frod the position of mass relative to the equilibrium in terms of".
and valid for any time 1 '" O.
10. A mass of 1 slug is attached to a spring whose constant is 25/4
Ib/ ft. Initially the mass is released I ft above the equilibrium position with a downward ",Iocity of 3 ftjsec, and the subsequent
motion takes place in a medium that offers a dsmping force numerically equal to 3 times the instantaneous velocity. Ao extema1
force /(1) is driving the system, but assume that initially /(1) ~ O.
Formulate and solve an initial value problem that models the
given system. Ioterpret your results.
11. A IO-lb weight is suspended by a spring that is stretched 2 in. by
the weight. Assume a resistance whose magnitude is 40/Vg Ib
times the instantaneous velocity in feet per second. If the weight
is pulled down 3 in. below its equilibrium position and released,
fmd the time required to reach the equilibrium position for the
fll'St time.
12- A ...,ight stretches a spring 6 in. It is set in motion at a point 2 in. below its equilibrium position with a downward ",Iocity of2 in./sec.
a. When docs the weight retum to its starting position?
b. When docs it reach its highest point1
c. Show that the maximum velocity is 2V2g + 1 in./sec.
13. A weight of 10 Ib stretches a spring lOin. The weight is drawn
down 2 in. below its equilibriwn position and given an initial velocity of 4 in./sec. Ao identical spring has a different weight attached to it. This second weight is drawn down from its equilibrium position a distance equal to the amplitude of the frrst motion
and then given an initial velocity of 2 ft/sec. If the amplitude of
the second motion is twice that of the frrst, what weight is attached to the second spriog?
14. A weight stretches one spring 3 in. and a second weight stretches
another spring 9 in. If both weights are simultaneously pulled
down I in. below their respective equilibrium positions and then
released, fmd the frrst time after 1 = 0 whco their velocities are
equal.
IS. A weight of 16 Ib stretches a spriog 4 ft. The weight is pulled
dawn 5 ft below the equilibrium position and then released. What
initial velocity ". given to the weight would have the effeet of
doubling the amplitude of the vibration?
16. A mass weighing 8 Ib stretches a spring 3 in. The spring-mass system resides in a medium with a dampins constant of 2 Ib-sec/ft. If
the mass is released from its equilibrium position with a velocity
of 4 in./sec in the downward direction, flud the time required for
the mass to return to its equilibrium position for the frrst time.
17. A weight suspended from a spring execute. damped vibrations with
a period of2 sec. !fthe damping factor decreases by 90% in 10 sec,
fmd the acceleration of the weight when it is 3 in. below its eqoilibrium position and is moving upward with a speed of2 ft/sec.
18. A 1()'lb weight stretches a spring 2 ft. If the weight is pulled down
6 in. below its equilibrium position and released, fmd the highest
point reached by the weight. Assume the spring-maas system residea in a medium offering a resistance of IO/ Vg lb times the instantaneous ",Iocity in feet per second.
17.4 Euler Equations
19. An LRC circuit is set up wi1h an inductance of liS henry, a resist·
ance of! ohm, aod a capacitance of516 farad. Assuming 1he initial
charge is 2 coulombs and 1he initial current is 4 amperes, fmd 1he
solutioo fimction descnbing 1he charge 00 1he capacitor at any time.
What is 1he charge 00 1he capacitor after a loog period of time?
20. An (opoo) electrical circuit coosists of an inductor, a resistor, aod
a capacitor. There is an initial charge of 2 coulombs on the capacitor. At the instant the circuit is closed, a current of 3 amperes is
preseot but no external voltage is being applied. In 1his circuit 1he
voltage drops at 1hree points are numerically related as follows:
across the capacitor. 10 times the charge; across the resistor, 4
times the instantaneous change in the charge; and across the inductor, 2 times 1he instantaneous chaoge in 1he current. Find 1he
charge on the capacitor as a function of time.
21. A 16-lb weight stretches a spring 4 It This spring-mass system is
in a medium wi1h a damping constant of 4.5Ib-sec/ft, and ao external force giveo by /(t) = 4 + e- 2t (in pouods) is being applied. What is 1he solution fimction describing 1he position of 1he
mass at any time if 1he mass is released from 2 ft below 1he equilibrium position wi1h ao initial velocity of 4 ft/ sec downward?
22. A IO-kg mass is a_hed to a spring hsving a spring coostant of
140 N/m. The mass is started in mntioo from 1he equilibrium position wi1h ao initial velocity of 1 mf sec in 1he upward direction
and wi1h an applied extemal force giveo by /(t) = 5 sin t (in newtons). The mass is in a viscous medium. with a coefficient of resistance eqoa1 to 90 N-sec/m. Formulate ao initial value problem
iliat models 1he given system; solve 1he model aod interpret 1he
results.
23. A 2-kg mass is a_hed to 1he lower end of a coil spring suspended ftom 1he ceiling. The mass comes to rest in its equilibrium
17-23
positioo 1hereby stretcbing 1he spring 1.96 m. The mass is in a
viscous medium. that offers a resistance in newtons numerically
equal to 4 times the instantaneous velocity measmed in meters
per secood. The mass is 1hoo pulled down 2 m below its equilibrium positioo and released wi1h a downward velocity of3 mfsec.
At this same instant an external force givoo by /(t) = 20 cos t (in
newtons) is applied to 1he system. At 1he ood of 1r sec detennine
if1he mass is above or below its equilibrium position aod by how
much.
24. An 8-lb weight stretches a spring 4 ft. The spring-mass system r<>sides in a medium offering a resistance to the motion equal to 1.5
times the instantaneous velocity, and an extemal force given by
/(t) = 6 + e -, (in pounds) is being applied. If 1he weight is released at a position 2 ft above its equilibrium positioo wi1h downward velocity of 3 ft/sec, fmd its position relative to 1he equilibrium after 2 sec hsve elapsed.
25. Suppose L = 10 henrys, R = 10 ohms, C = 1/500 farads,
E = 100 volts, q(O) = 10 coulombs, and q'(O) = i(O) = O. Formulate aod solve an initial value problem iliat models 1he given
LRC circuit. Interpret your results.
26. A series circuit consisting of an inductor, a resistor, and a capacitor is open. There is an initial charge of 2 coulombs on 1he capacitor, and 3 amperes of current is present in the circuit at the instant
1he circuit is closed. A voltage giveo by E(t) = 20 cos t is applied. In this circuit 1he voltage drops are numerically eqoa1 to 1he
following: across the resistor to 4 times the instantaneous change
in the charge, across the capacitor to 10 times the charge, and
across the inductor to 2 times the instantaneous change in the current. Find 1he chsrge 00 1he capacitor as a fimctioo of time. Determine 1he chsrge 00 1he capacitor and 1he current at time t = 10.
uations
In Section 17.1 we introduced the aecond-order linear homogeneous differential equation
P(x)y"(x)
+ Q(x)y'(x) + R(x)y(x)
= 0
and showed how to solve this equation when the coefficients P, Q, and R are constants. If
the coefficients are not constant, we cannot generally solve this differential equation in
tenns of elementary functions we have studied in calculus. In this section you will learn
how to solve the equation when the coefficients have the special forms
P(x)
= ax 2,
Q(x)
= bx,
and
R(x)
= c,
where a, b, and c are constants. These special types of equations are called Euler equations, in honor of Leonhard Euler who studied them and showed how to solve them. Such
equations arise in the study of mechanical vibrations.
The General Solution of Euler Equations
Consider the Euler equation
ax"y"
+ bxy' + cy
=
0, x
> O.
(l)
17-24
Chapter 17: Second-Order Differential Equations
To solve Equation (I), we rlISt make the change of variables
z = Inx
y(x) = Y(z).
and
We next use the chain rule to find the derivatives y'(x) and y"(x):
y' (x) =
fx Y(z)
=
~Y(z):
= Y' (z){
and
-~2 Y'(z) + ! y"(z)dz
y"(x) = i£y'(x) = i£y'(z)!
x =
dx
dx
x
x
dx
=
-~2 Y'(z) + ~2 Y"(z).
x
x
Substituting these two derivatives into the left-hand side of Equation (I), we rmd
ax"y"
+ bxy' +
:2
cy = ax 2 ( = aY"(z)
Y'(z)
+ (b
:2
+
+ bX({ Y'(Z)) +
Y"(Z))
- a)Y'(z)
cY(z)
+ cY(z).
Therefore, the substitutions give us the second-order linear clifferential equation with constant coefficients
aY"(z)
+ (b
- a)Y'(z)
+ cY(z)
=
o.
(2)
We can solve Equation (2) using the method of Section 17.1. That is, we find the roots to
the associated auxiliary equation
ar2
+ (b
- a)r
+C
= 0
(3)
to find the general solution for Y(z). After rmding Y(z), we can detennine y(x) from the
substitutionz = Inx.
EXAMPLE 1
Find the general solution of the equation x 2y"
+
2xy' - 2y =
o.
Solution This is an Euler equation with a = I, b = 2, and C = -2. The auxiliary equation (3) for Y(z) is
r2
+ (2
+ 2)
- I)r - 2 = (r - I)(r
=
0,
with roots r = -2 and r = 1. The solution for Y(z) is given by
Y(z) = Ct e -2z
+ C2ez.
Substituting z = In x gives the general solution for y(x):
y(x) = cte-2lnx
EXAMPLE 2
+ c2e lnx
= CtX- 2
Solve the Euler equation x"y" - 5xy'
+
+ C2X
•
9y = O.
Solution Since a = I, b = -5, and C = 9, the auxiliary equation (3) for Y(z) is
r2
+ (-5
- I)r
+9
= (r -
3)2 = O.
The auxiliary equation has the double root r = 3 giving
Y(z) = cte 3z
+
c2ze3~
Substituting z = In x into this expression gives the general solution
y(x) = Cte3lnx
+ c2lnxe3lnx
= CtX3
+ C2x3lnx
•
17.4 Euler Equations
EXAMPLE 3
Find the particular solution to x"y" - 3xy'
initial conditions y(1) = 0 andy'(I) = 1.
Solution Here a = I, b = -3, and
gives
C
+ 68y
17-25
= 0 that satisfies the
= 68 substitoted into the auxiliary equation (3)
r 2 -4r+68=0.
The roots are r = 2
+
8i and r = 2 - 8i giving the solution
Y(z) = e2z(cl cos 8z
+ C2 sin 8z).
Substitoting z = In x into this expression gives
y(x) =
e
21nx
( Cl
cos (8 In x)
+ C2 sin (8 In x»).
From the initial condition y(l) = 0, we see that Cl = 0 and
y(x) = C2x2 sin (8 In x).
y
To fit the second initial condition, we need the derivative
10
y'(x) = c2(8x cos (8 In x)
5
Since y'(I) = I, we immediately obtain C2 = 1/8. Therefore, the particular solution satisfying both initial conditions is
o
2
10
y(x) =
2
-5
+ 2xsin(8lnx»).
t
x 2 sin (8 In x).
y= "ssin(8lnx)
Since -I :5 sin (8 In x) :5 I, the solution satisfies
-10
x2
-8 :5 y(x)
FIGURE 17.8
Example 3.
:5
x2
8'
Graph of the solution to
A graph of the solution is shown in Figure 17.8.
EXERCISES 17.4
In Exercises 1-24, rmd the general solution to the given Euler
equation. Assume x > 0 throughout
1. x"y" + 2xy' - 2y
=
0
=
0
3.x"y"-6y=0
5. xV - Sxy'
+ 8y
=
0
4.x"y"+xy'-y=0
7. 3xV + 4xy' = 0
9. x"y" - xy' + Y = 0
11. x"y" - xy' + Sy = 0
+ 3xy' + lOy = 0
15. 4xV + 8xy' + Sy = 0
17. x"y" + 3xy' + Y = 0
19. xV + xy' = 0
13. x"y"
2. x"y" + xy' - 4y
+ 7xy' + 2y = 0
8. xV + 6xy' + 4y = 0
10. x"y" - xy' + 2y = 0
12. x"y" + 7xy' + l3y = 0
14. x"y" - Sxy' + lOy = 0
16. 4x"y" - 4xy' + Sy = 0
18. x"y" - 3xy' + 9y = 0
20. 4xV + Y = 0
6. 2xV
21. 9x"y' + ISxy'
22. 16xV - 8xy'
+Y = 0
+ 9y = 0
23. 16xV + S6xy' + 2Sy = 0
24. 4x"y' - 16xy' + 2Sy = 0
In Exercises 25-30, solve the given initial value problem.
25. xV
26. 6x"y'
27. xV
+ 3xy' - 3y = 0, y(l) = I, y'(I) = -I
+ 7xy' - 2y = 0, y(1) = 0, y'(I) = I
- xy' + Y = 0, y(l) = I, y'(I) = I
28. xV + 7xy' + 9y = 0, y(l) = I, y'(I) = 0
29. xV - xy' + 2y = 0, y(1) = -I, y'(I) = I
30. xV + 3xy' + Sy = 0, y(l) = I, y'(I) = 0
•
17 -26
Chapter 17: Second-Order Differential Equations
Power-Series SoLutions
In this section we exteod our stody of second-order linear homogeoeous equations with
variable coefficieots. With the Euler equations in Section 17.4, the power of the variable x
in the nonconstant coefficieot had to match the order of the derivative with which it was
paired: x 2 with y', Xl with y', and x O(=I) with y. Here we drop that requiremeot so we
ean solve more geoeral equations.
Method of Solution
The poweMeries method for solving a second-order homogeneous differential equation
consists of finding the coefficieots of a power series
00
y(x) = ~ c.x' = Co
.-0
+ ClX + C2x2 + ...
(1)
which solves the equation. To apply the method we substitote the series and its derivatives
into the differential equation to determine the coefficieots Co, Ct, C2, .... The technique for
finding the coefficieots is similar to that used in the method of uodetermined coefficieots
preseoted in Section 17.2.
In our first example we demonstrate the method in the setting of a simple equation
whose general solution we already know. This is to help you become more comfortable
with solutions expressed in series form.
EXAMPLE 1
Solve the equation y'
+Y
= 0
by the power-series method.
Solution We assume the series solution takes the form of
and calculate the derivatives
00
y'
=
~
00
ncll x n -
1
y"
and
n(n - 1)c.x·- 2.
= ~
11=1
11=2
Substitotion of these forms into the second-order equation gives us
00
00
11 - 2
11 - 0
~ n(n - l)c.x·- 2 + ~ c.x· = O.
Next, we equate the coefficients of each power of x to zero as summarized in the followiog
table.
Powerofx
Coefficient Equation
1
xo
2(I)C2
+ Co
=
0
or
C2 = -ZCo
xl
3(2)C3
+ Cl
=
0
or
C3 = -3-2 Cl
x2
4(3)C4
+ C2
=
0
or
C4 = -4'3 C2
x3
5(4)cs
+ C3
=
0
or
Cs = -5'4 C3
x4
6(5)C6
+ C4
=
0
or
C6 = -6'5 C4
+ C.-2
=
0
or
en =
X,,-2
n(n - l)c.
1
1
1
1
1
n(n - 1) C.-2
17.5 Power-Series Solutions
17-27
From the table we notice that the coefficients with even indices (n = 2k, k = 1,2,3, ... )
are related to each other and the coefficients with odd indices (n = 2k + 1) are also interrelated. We treat each group in turn.
Even indices: Here n = 2k, so the power is x 2k--2. From tlie last line of the table, we have
2k(2k - l)c2k
+ C2k-2
=
0
or
C2k =
I
2k(2k _ I) C2k-2·
From this recursive relation we find
c2k = [ 2k(2;- 1)][ (2k - 2;(2k - 3)]-" [-
=
4(~)][-!]co
(-1)1
(2k)! co·
Odd indices: Here n = 2k
line of the table yields
(2k
+ 1, so the power is X2k-I. Substituting this into the last
+ 1)(2k)c2k+ I + C2k-1
=
0
or
c2k+ I =
(2k
I
+ I )(2k) C2k-I·
Thus,
C2k+1 = [ (2k+ \)(2k)] [ (2k -
=
1)~2k -
2J . ht4)] ht2)]cl
(_1)1
(2k + I)! CI·
Writing the power series by grouping its even and odd powers together and substituting for the coefficients yields
00
=
~
<-0
00
C2kx2k
+ ~ C2k+l X2H1
1- 0
00 (-1)1
'" (2k)! X 2k
-_ CotEl,
00
(_1)1
+ CI tEl,
'" (2k + I)! x 2.<+1.
From Table 9.1 in Section 9.10, we see that the trrst series on the right-hand side of the last
equation represents the cosine function and the second series represents the sine. Thus, the
general solution to y" + Y = 0 is
y = Co cos x
+ Cl sinx.
•
17-28 Chapter 17: Second-Order Differential Equations
EXAMPLE 2
+ ~' + Y
Find the general solution to y"
= O.
Solution We assume the series solution form
00
y = ~CnXll
n=O
and calculate the derivatives
00
y'
L ncn x
=
.-1
ll
00
t
-
y"
and
=
Ln(n - l)c.x·- 2•
.-2
Substitution of these forms into the second-order equation yields
00
00
~ n(n - 1)cnx"- 2
+
n=2
L
ncnX
n+
11=1
00
L
n=
CnX
o.
11=0
We equate the coefficients of each power of x to zero as summarized in the following table.
Powerofx
Coefficient Equation
+ Co
3(2)C3 + Cl + CI
4(3)C4 + 2C2 + C2
5(4)cs + 3C3 + C3
6(5)C6 + 4C4 + C4
=
+ 2)(n + l)c.+2 + (n + l)c.
2(I)c2
x'
(n
=
=
=
=
=
or
or
or
C3
or
Cs
=
0
0
0
0
0
or
C6
=
0
or
Cn+2 =
=
=
=
C2
C4
-!co
-let
-lc2
-lC3
-~C4
-
I
n + 2 en
From the table notice that the coefficients with even indices are interrelated and the coefficients with odd indices are also interrelated.
Even indices: Here n = 2k - 2, so the power is x2k-2. From the last line in the table,
we have
C2k
= -
I
2k C2k-2·
From this recurrence relation we obtain
=
(2)(4)(6) ... (2k) Co·
Odd indices: Here n
we have
=
2k - 1, so the power is X2k-I. From the last line in the table,
C2k+1 = -
1
2k + 1 C2k-I·
From this recurrence relation we obtain
C2k+l =
=
(-2k
~ 1)(-2k ~
(-I)'
(3)(5)'" (2k
+ 1) CI.
J . (-t) (
-t)Cl
17-29
17.5 Power-Series Solutions
Writing the power series by grouping its even and odd powers and substituting for the
coefficients yields
00
Y
=
00
L C2kx2k + L C21-+ 1X2k+ 1
1=0
1=0
(_I)k
00
~
- 0&'0 (2)(4) .. ·(2k)
-
EXAMPLE 3
C
X2k
+C
(_I)k
00
~
2k+ l
1&'0 (3)(5)"'(2k + I) X
•
.
Find the general solution to
(I -
X
2
Ixl <
)y" - &y' - 4y = 0,
1.
Solution Notice that the leading coefficient is zero when x = ± 1. Thus, we assume the
solution interval I: -I < x < I. Substitution of the series form
and its derivatives gives us
(I - X2)
00
L
n(n - I )C.X·- 2 - 6
11=2
00
L
n=2
L
00
nc.x' - 4
n=1
00
n(n - I )C.X·- 2 -
00
L
n=2
L
c.x· = 0,
n=O
00
n(n - I )C.X' - 6
L
00
nc.x' - 4
n=1
L
c.x· = O.
n=O
Next, we equate the coefficients of each power of x to zero as summarized in the following
table.
Powerofx
Coefficient Equation
xO
Xl
x2
x3
x'
(n
_4
- 4co=0
or
Cz - leO
- 6(I)Cl - 4Cl = 0
3(2)C3
4(3)C4 - 2(I)c2 - 6(2)C2 - 4C2 = 0
5(4)c, - 3(2)C3 - 6(3)C3 - 4C3 = 0
or
C3 = iCl
or
C4 -
2(I)c2
+ 2)(n + I)C.+2 - [n(n - I) + 6n + 4]c.
(n + 2)(n + l)c.+2 - (n + 4)(n + I)c.
or
_6
4C2
_7
Cs -
SC3
= 0
= 0
or
n+4
Cn+2
= n + 2 en
Again we notice that the coefficients with even indices are interrelated and those with odd
indices are interrelated
Even indices: Here n = 2k - 2, so the power is x2k. From the right-hand column and
last line of the table, we get
=
(2\~ 2)(2/~ 2)(;::= ~)- .. ~(~)CO
= (k
+
I)co.
17 -30
Chapter 17: Second-Order Differential Equations
Odd indices: Here n
=
2k - I, so the power is x 2k+l. The right-hand column and last
line of the table gives us
=
=
(2k + 3)(~)(~)
2k+1
2k-1
2,\;-3
... l(i)
53
Cl
2k + 3
--3- C1 •
The general solution is
00
Y
=
L cnx"
.-0
=
L C2kx2k + L C2k+ 1x 2k+ 1
00
00
k~O
k~O
00
= CO~
(k
+
00
~O
EXAMPLE 4
2k+3
•
+ Cl~ -3-X2k+1.
l)x2k
~O
Find the general solution to y" - 2>y'
+Y
=
o.
Solution Assuming that
substitution into the differential equation gives us
00
00
~ n(n - l)c.x·- 2
-
n=2
00
2 ~ nc.x·
n=1
+
~ c.x· =
o.
11=0
We next determine the coefficients, listiog them in the following table.
Powerofx
Coefficient Equation
2(I)c2
3(2)C3 -
2Cl
Co =
0
or
C2 = -Zco
+
Cl =
0
or
c3
3
c4=4_3 c2
4(3)C4 - 4c2
+ C2
=
0
or
5(4)c, - 6C3
+ C3
=
0
or
or
x·
(n
+ 2)(n + l)c.+2 -
I
+
(2n - I)c. = 0
I
=3_2
c1
5
c'=5_4 c3
7
c6=6_5 c4
2n - I
or
C.+2 = (n
+ 2)(n + I) c.
17.5 Power-Series Solutions
17-31
From the recursive relation
2n - I
+ 2)(n +
e.+2 = (n
I) e.,
we write out the first few terms of each series for the general solution:
Y =
eO(1 - lx
2
2
-
lx'
41
n61 x• -
... )
•
EXERCISES 17.5
In Exercises 1-18, use power series to f'md 1he general solution of1he
differential equation.
1. y"
2. y"
+ 2y' = 0
+ 2y' + Y =
0
+ 2y = 0
2xy' + 2y = 0
11. (x 2
-
-
+ 2xy'
x"y
l)y" - 6y
- 2y = 0
0
=
0
=
15. y" -2xy' +3y=O
6.y"-.>;Y'+y=O
7. (I + x)y" - Y = 0
8. (1 - x 2)y" - 4.>;Y' + 6y
l)y"
+ y'
+ 2)y' + 2y = 0
2
13. (x - l)y" + 4'>;y' + 2y = 0
14. y" - 2xy' + 4y = 0
4. y" - 3y'
xV -
-
10. y"
12. .>;y" - (x
3.y"+4y=O
5.
9. (x 2
16. (I - x 2)y" - .>;y'
=0
=
0
+ 6y
=
17. y" - xy' + 3y
18.
xV -
4.>;Y'
+ 4y
0
= 0