second-order differential equations - minificciones · 17.1 second-order linear equations 17-3...

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 to the linear homogeneous equation (2), then for any constants Cl and C2, the 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)

= 0, YI is a solution = 0, Y2 is a solution

= CI(O) + C2(0) = O.

Therefore, Y = CIYI + 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. A constant multiple kYI of any solution YI to Equation (2) is also a solution. (Choose CI = k and C2 = 0.)

3. 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) = xex 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)

17.1 Second-Order Linear Equations 17-3

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' = r2e"', which are just constant multiples of the original exponential. Ifwe substitute y = e'" into Equation (3), we obtain

ar2en: + bre1"X + cerx = 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 If rl and r2 are two real and unequal roots to the auxiliary equation ar2 + br + c = 0, then

is the general solution to ay" + by' + cy = O.

EXAMPLE 1 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 = cle3x + 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'" + r2xe"') + 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 If r is the only (repeated) real root to the auxiliary equation ar2 + br + c = 0, then

is the general solution to ay" + by' + cy = o.

EXAMPLE 2 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• • Case 3: b2 - 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 i2 = -1. (These real

numbers are a = -bl2a and {3 = Y4ac - b2/2a.) 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

I +1 ax Q Y3 = ZYI ZY2 = e 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.


THEOREM 5 If rl = a + if3 and r2 = a - if3 are two complex roots to the auxiliary equation ar2 + br + c = 0, then

y = e""(cI cos f3x + C2Sin f3x)

is the general solution to ay" + by' + cy = o.

EXAMPLE 3 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 rl = 2 + i and r2 = 2 - i. Thus, a = 2 and f3 = 1 give the general solution

•

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.

THEOREM 6 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

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.


y

-4

-6

-8

FIGURE 17"1 Particular solution curve for Example 4.

EXAMPLE 4 Find the particular solution to the initial value problem

y"-2y'+y=0, y(O) = 1, y'(0) = -I.


r2 - 2r + I = (r - 1)2 = O.

The repeated real root is r = I, giving the general solution

Y = CteX + C2 xex.

Then,

y' = clex + C2(X + I)ex•

From the initial conditions we have

and

Thus, CI = I and C2 = -2. The unique solution satisfying the initial conditions is

y = eX - bex.

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. •

EXERCISES 17.1

In Exercises 1-30, fmd the general solution of the given equation.

1. y' - y' - 12y ~ a 3. y' + 3y' - 4y ~ a 5.y·-4y~0

7.2y·-y'-3y~0

9. 8y" - lOy' - 3y ~ 0

11.y·+9y~0

13.y·+25y~0

15. y' - 2y' + 5y ~ a 17. y' + 2y' + 4y ~ a 19. y' + 4y' + 9y ~ a 21. y" = 0

d'y dy 23. tb;2+4tb;+4y~0

d'y dy 25. tb;2+6tb;+9y~0

d'y dy 27.4tb;2+4tb;+y~0

d'y dy 29. 9tb;2+6tb;+y~0

2. 3y" - y' = 0

4.y"-9y=O

6.y"-64y~0

8. 9y" - y ~ 0

10. 3y' - 2Oy' + 12y ~ 0

12. y" + 4y' + 5y ~ 0

14. y" + Y = 0

16. y" + 16y ~ a 18. y" -2y' +3y~0

20. 4y" - 4y' + \3y ~ 0

22. y" + 8y' + 16y ~ 0

d'y dy 24. tb;2-6tb;+9y~0

d'y dy 26. 4tb;2 - 12tb; + 9y ~ 0

d'y dy 28.4tb;2-4tb;+y~0

d'y 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 ~ 0, y(O) ~ 0, y'(O) ~ I

37. y" - 4y' + 4y ~ 0, y(O) ~ I, y'(O) ~ a 38. 4y" - 4y' + y ~ 0, y(O) ~ 4, y'(O) ~ 4

d'y dy dy 39.4 tb;2 + 12tb; + 9y ~ 0, y(0) ~ 2, tb;(0) ~ I

d'y dy dy 40.9 tb;2 - 12tb; + 4y ~ 0, y(0) ~ -I, tb;(0) ~ I


In Exercises 41-55, fmd the general solution.

41. y" - 2y' - 3y ~ 0

43. 4y" + 4y' + y ~ a 45. 4y' + 20y ~ 0

47. 25y" + lOy' + y ~ 0

49. 4y' + 4y' + 5y ~ 0

51. 16y" - 24y' + 9y ~ a 53. 9y" + 24y' + 16y ~ a 55. 6y" - 5y' - 4y ~ a

42.6y·-y'-y~0

44. 9y' + 12y' + 4y ~ 0

46.y·+2y'+2y~0

48. 6y' + \3y' - 5y ~ 0

50.y·+4y'+6y~0

52. 6y'-5y'-6y~0

54. 4y' + 16y' + 52y ~ a

In Exercises 56-60, solve the initial value problem.

56. y" - 2y' + 2y ~ 0, y(0) ~ 0, y'(O) ~ 2

57. y" + 2y' + y ~ 0, y(0) ~ I, y'(O) ~ I

58. 4y" - 4y' + y ~ 0, y(O) ~ -I, y'(O) ~ 2

59. 3y" + y' - 14y ~ 0, y(O) ~ 2, y'(O) ~ -I

60. 4y" + 4y' + 5y ~ 0, y('IJ') ~ I, Y'(1T) ~ a 61. Prove that the two solution fimctions in Theorem 3 are linearly in

dependent.

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

y" + 4y ~ 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

approach zero as x ~ 00.


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 of parameters. 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 = G(x), (1)

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

also solves the noohomogeneous equation (I) because

a(ye + Yp)" + b(ye + Yp)' + c(ye + Yp)

= (aye" + bye' + eye) + (ayp' + byp' + eyp)

= 0 + G(x)

= G(x).

y, solves Eq. (2) and YP solves Eq. (\)

(3)

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 - Yp)' + c(y - Yp)

= (ay' + by' + ey) - (ayp" + byp' + eyp)

= G(x) - G(x) = 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 differen-tial 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)eUX cos {3x, P3(x)eax sin {3x.

For instance, 1 - x, e2x, xex, 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(Ax2 + Bx + C) = 1 - x 2

or, collecting terms with like powers of x,

-3Ax2 + (-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

_ + _ -. + 3. + 1 2 4 + 5 y-Yc yp-cle C2e :3 x -9x 27' •

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 and A + B = O.

Simultaneous solution of these two equations gives A = - 1 and B = 1. Our particular solution is

YP = cosx - sinx.

EXAMPLE 3 Find a particular solution of y' - 3y' + 2y = 5ex•

Solution If we substitute

YP = Aex

and its derivatives in the differential equation, we f"md that

Aex - 3Aex + 2Aex = 5ex

or

0= 5ex•

•

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 Aex 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

YP = Axex.


The result of substituting the derivatives of this new candidate into the differential equation is

(AxeX + ZAe") - 3(AxeX + Ae") + ZAxex = 5ex

or

Thus, A = -5 gives our sought-after particular solution

YP = -5xex.

EXAMPLE 4 Find a particular solution of y" - 6y' + 9y = e3x

•

Solutton The auxiliary equation for the complementary equation

r2 - 6r + 9 = (r - 3)2 = 0

•

has r = 3 as a repeated root. The appropriate cboice for yp in this case is neither Ae3x nor Axe3x 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

YP = Ax2e

3x

and its derivatives in the given differential equation, we get

(9Ax2e3X + lZAxe3x + ZAe~ - 6(3Ax2e 3x + ZAxe3") + 9Ax2e 3x = e3x

or

Thus, A = 1/2, and the particular solution is

YP = !x2e"': • 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' = 5ex - sin 2x.

Solutton We fITSt check the auxiliary equation

r2 - r = O.

Its roots are r = I and r = O. Therefore, the complementary solution to the associated homogeneous equation is

Yc = clex + C2·

We now seek a particular solution Yp. That is, we seek a function that will produce 5ex - 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,

including xex 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


or

Ae' - (4B + 2C) cos 2x + (2B - 4C) sin 2x = 5e' - sin 2x.

TIris equation will hold if

A = 5, 4B + 2C = 0, 2B-4C= -I,

or A = 5, B = -1/10, and C = 1/5. Our particular solution is

-5' I 2x+ I .2x Yp - xe - TOcOS Ssm.

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

IT G(x) has a term that is a constant multiple of ...

e'"

sin kx, cos kx

px2 +qx+m

oy" + by' + cy = G(x).

And if

r is not a root of the auxiliary equation

r is a single root of the auxiliary equation

r is a double root of the auxiliary equation

ki is not a root of the auxiliary equation

o is not a root of the auxiliary equation

o is a single root of the auxiliary equation

o is a double root of the auxiliary equation

The Method of Variation of Parameters

Then include this expression in the trial function for Y,.

Ae'"

Axe'"

Bcoskx + Csinkx

Dx2+Ex+F

Dx3 + Ex2 + Fx

Dx' + Ex3 + Fx2

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


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

Then we have

Y = VIYI + V2Y2,

y' = VIYl' + V2Y2',

y" = VIY(" + V2Y2" + 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).

(4)

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

a(VI'YI' + v2'yi) = G(x).

Equations (4) and (5) can be solved together as a pair

Vt'Yt + V2'Y2 = 0,

(5)

for the unknown functions VI' and V2'. The usual procedure for solving this simple system is to use the method of determinants (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

to tmd the functions YI and Y2. 2. Solve the equations

Vt'Yt + V2'Y2 = 0, G(x)

Vt'Yt' + 112'y2' = -a-

simultaneously for the derivative functions VI' and V2'. 3. Integrate VI' and V2' to find the functions VI = VI(X) and V2 = V2(X). 4. Write down the particular solution to nonhomogeneous equation (I) as

YP = VJ)'1 + V2Y2·


EXAMPLE 6 Find the general solution to the equation

y" + Y = tanx.

Solution The solution of the homogeneous equation

y" + Y = 0

is given by

Yc = Cl COS X + Cz sinx.

Since Yl(X) = cos x and Y2(X) = sin x, the conditions to be satisfied in Equations (4) and (5) are

Vt' cos x + V2' sinx = 0,

-Vt' sinx + V2' cosx = tanx. a = 1

Solution of this system gives

Likewise,

I 0 sin x I

tan x cosx Vl' = -c

1

"--co-,,-x--S-in--"XC-

1 -smx cos X

I c~sx V2' =

-smx

I c~sx -smx

-tan x sin x cos2 x + 8in2 x

~xl = sinx.

sinx I cos X

After integmting VI' and V2', we have

j -Sin2 x Vl(X) = cos x dx

= - j (sec x - cos x) dx

= -In Isecx + tanxl + sinx,

and

V2(X) = j sin x dx = -cos x.

cosx·

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

YP = [-In Isecx + tanxl + sin x] cos x + (-cos x) sin x

= (-cos x) In lsecx + tanxl.

The general solution is

Y = Cl cos x + C2 sin x - (cos x) In Isecx + tanxl. •

EXAMPLE 7


Solve the nonhomogeneous equation

y" + y' - 2y = xex.


r2 + r - 2 = (r + 2)(r - I) = 0

giving the complementary solution

Yc = c}e-2x + czex.

The conditions to be satisfied in Equations (4) and (5) are

Vt'e-2x + v2'ex = 0,

-2vt'e-2x + v2'ex = xex. a = 1

Solving the above system for VI' and vi gives

Ix~x eXI

Vt' = eX _ - xe2x _ I '-'

I e 2x eXI - 3e-x - -3xe .

-2e-2x eX

Likewise,

Integrating to obtain the parameter functions, we have

and

Therefore,

VI(X) = J -t xe3x dx

= _t(X~3X -J e;x dx )

_ l( )'-' -27 1 - 3xe ,

The general solution to the differential equation is

Y = cle-lx + C2ex - ~xex + kx2e X,

where the term (1/27)eX in YP has been absorbed into the term C2ex in the complementary solution. •


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

3. y" - y' = sinx 4. y" + 2y' + Y = x 2

5. y" + Y = cos3x 6. y" + Y = e2x

7. y" - y' - 2y = 20 cos x 8. y" + Y = a+ 3ex

9. y" - y = eX + x 2 10. y" + 2y' + Y = 6 sin 2x

11. y" - y' - 6y = e-x - 7 cosX'

12. y" + 3y' + 2y = e-x + e-2.t" - x

d'y dy, d'y dy 13_ dx' + 5 dx = 15x 14_ dx' - dx = -8x + 3

d'y dy]x d'y dy _ , 15_ dx' - 3 dx = e - 12x 16_ dx' + 7 dx - 42x + 5x + I

Solve the equations in Exercises 17-28 by variation of parameters.

17. y" + y' = x

18. y" + Y = tanx, _'!! < x < '!! 2 2

19. y" + Y = sinx

21. y" + 2y' + Y = e-x

20. y" + 2y' + Y = eX

22.y"-y=x

23. y" - y = eX 24. y" - y = sinx

25. y" + 4y' + Sy = 10 26. y" - y' = 2x

d'y 1f 1f 27. dx' + Y = secx, -2 < x < 2

d'y dy 28. dx2 - dx = eXcosx, x> 0

In each of Exercises 29-32, the given differential equation has a particular solution YP of the form given. Determine the coefficients in Yp. Then solve the differential equation.

29. y" - 5y' = xeil:, YP = Ax2e5.l: + Bxe~ 30. y" - y' = cosX' + sinx, Yp = A cosX' + B sinx

31. y" + Y = 2 cosx + sinx, YP = Ax cosX' + Bx sinx

32. y" + y' - 2y = xex, YP = Ax2ex + Bxex

In Exercises 33-36, solve the given differential equations (a) by variation of parameters and (h) by the method of undetermined coefficients.

d'y dy 33 - - - = eX + e-r

• dx' dx d'y dy

34. - - 4- + 4. = 2e2x dx' dx '

d'y dy 35. dx,-4 dx -5y =ex +4

d'y dy 9x 36. dx' - 9dx = ge

Solve the differential equations in Exercises 37-46. Some of the equations can be solved by the method of undetermined coefficients, but others cannot.

37. y" + Y = cotx, O<X<'7T 38. y" + Y = cscx, o<x<w 39. y" - 8y' = e!b:

41. y" - y' = x3

43. y" + 2y' = x2 - e~

45. y" + Y = secxtanx,

40. y" + 4y = sinx

42. y" + 4y' + 5y = x + 2

44. y" + 9y = 9x - cosX'

_'!! < x < '!! 2 2

46. y" - 3y' + 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 = sinx

Solve the differential equations in Exercises 51 and 52 sui!ject to the given initial conditions.

d'y ,1f 1f 51. dx' + Y = sec x, -2 < x < 2; y(O) = y'(O) = I

d'y 2 52. dx' + Y = e2x; y(O) = 0, y'(O) = 5

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 satisfYing the equation and the given initial conditions.

x' 53. y" + y' = x, YP = T - x, y(0) = 0, y'(O) = 0

54. y" + Y = x, YP = 2 sin x + x, y(O) = 0, y'(O) = 0

55. ty" + y' + Y = 4el{cosx - sinx),

YP = 2ex cos x, y(0) = 0, y' (0) = I

56. y" - y' - 2y = I - 2x, YP = x - I, y(O) = 0, y'(O) = I 57. y" - 2y' + Y = 2ex, YP = x 2ex, y(O) = 1, y'(O) = 0

58. y" - 2y' + Y = x -leX, x > 0,

YP = xexJnx, y(1) = 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 variablecoefficient 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

I ,,, AppLications

massm at equilibrium

y

FIGURE 17.2 Massm stretches a spring by length s to the equilibrium position at y = O.

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.

17.3 Applications 17-17

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

ks = mg. (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 + y),

dy F,=ll

dt,

the propulsion force due to gravity,

the restoring force of the spring's tension,

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 dy m dt2 = mg - ks - ky - Il dt'

By Equation (1), mg - ks = 0, so this last equation becomes

dly dy m dt2 + Il dt + ky = 0, (2)

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.


FIGURE 17_4 c, = C sin c/> and C2 = Ceosc/>.


r2 + if = 0,

having the imaginary roots r = ±wi. The general solution to the differential equation in (2) is

Y = Cl cos rut + C2 sin wt. (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» = cos wi sin cf> + sin wi cos cf>.

To apply the identity, we take (see Figure 17.4)

c, = C sin cf> and C2 = C cos cf>,

where

C= Vc,2 + 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=~~ I I I I C

Csin."

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)


,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 > Cd. The roots of the auxiliary equation are real and unequal, given by

'1 = -b + Vb 2 - w' and '2 = -b - Vb 2 - w'. The general solution to Equation (6) is given by

y = Cle(-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. The roots to the auxiliary equation are complex and given by , = -b ± iV w2 - b2• The general solution to Equation (6) is given by

y = e-bl(cl cosVw' - b2t + C2SinVw' - b2t). This situation, called underdamping, represents damved oscillatory motion. It is analogous to simple harmonic motion of period T = 21T/V w' - b2 except 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/V w2 - 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

o y=(l +t)e-t

(a) Critical damping

y t

y

y = e-t sin(St + 1114)

(b) Overdamping (c) Underdamping

FIGURE 17.6 Three examples of damped vibra1xny motion for a spring system with friction, so B oF O.


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

d2y dy m dt2 + 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).

Voltage E soun:e

R, Resistor

C, Capacitor

FIGURE 17.7 An elec1ric circuit.

L.lnd~

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 II R and called R lbe resistance. So Ohm ~ law is

1 I=JiV,

EXERCISES 17.3

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

dI q RI + L dt + C = E(t).

Since I = dqldt, Kirchhoff's law becomes

d2q dq I L dt2 +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 Electrical System

my" + 8y' + ky = F(t) Lq" + Rq' + lq = E(t) C

y: displacement q: charge y/: velocity q/: current y": acceleration q": change in current m: mass L: inductance 8: damping constant R: resistance k: spring constant I/C: where C is the capacitance F(t): forcing function E(t): voltage source

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.


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 capacitor. 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) Meters (m) Msss Slugs Kilugrams (kg) Time Seconds (sec) Seconds (sec) Force Pounds (lb) Newtoos(N) g(earth) 32 ftjsec" 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.

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

uations

17.4 Euler Equations 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.

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) = ax2, 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)


To solve Equation (I), we rlISt make the change of variables

z = Inx and y(x) = Y(z).

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

y"(x) = i£y'(x) = i£y'(z)! = -~ Y'(z) + ! y"(z)dz = -~ Y'(z) + ~ Y"(z). dx dx x x2 x dx x 2 x 2

Substituting these two derivatives into the left-hand side of Equation (I), we rmd

ax"y" + bxy' + cy = ax2( - :2 Y'(z) + :2 Y"(Z)) + bX({ Y'(Z)) + cY(z)

= aY"(z) + (b - a)Y'(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 x2y" + 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 - I)r - 2 = (r - I)(r + 2) = 0,

with roots r = -2 and r = 1. The solution for Y(z) is given by

Y(z) = Cte -2z + C2ez.

Substituting z = In x gives the general solution for y(x):

y(x) = cte-2lnx + c2elnx = CtX-2 + C2X

EXAMPLE 2 Solve the Euler equation x"y" - 5xy' + 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) = cte3z + c2ze3~

Substituting z = In x into this expression gives the general solution

y(x) = Cte3lnx + c2lnxe3lnx = CtX3 + C2x3lnx

•

•

17.4 Euler Equations 17-25

EXAMPLE 3 Find the particular solution to x"y" - 3xy' + 68y = 0 that satisfies the initial conditions y(1) = 0 andy'(I) = 1.

Solution Here a = I, b = -3, and C = 68 substitoted into the auxiliary equation (3) gives

y

10

r2 -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) = e21nx

( 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).

To fit the second initial condition, we need the derivative

y'(x) = c2(8x cos (8 In x) + 2xsin(8lnx»).

5 Since y'(I) = I, we immediately obtain C2 = 1/8. Therefore, the particular solution satisfying both initial conditions is

10 o 2

2 -5 y= "ssin(8lnx)

y(x) = t x2 sin (8 In x).

Since -I :5 sin (8 In x) :5 I, the solution satisfies

-10 x2 x2

-8 :5 y(x) :5 8' FIGURE 17.8 Graph of the solution to

Example 3. 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 2. x"y" + xy' - 4y = 0

3.x"y"-6y=0 4.x"y"+xy'-y=0

5. xV - Sxy' + 8y = 0 6. 2xV + 7xy' + 2y = 0

7. 3xV + 4xy' = 0 8. xV + 6xy' + 4y = 0

9. x"y" - xy' + Y = 0 10. x"y" - xy' + 2y = 0

11. x"y" - xy' + Sy = 0 12. x"y" + 7xy' + l3y = 0

13. x"y" + 3xy' + lOy = 0 14. x"y" - Sxy' + lOy = 0

15. 4xV + 8xy' + Sy = 0 16. 4x"y" - 4xy' + Sy = 0

17. x"y" + 3xy' + Y = 0 18. x"y" - 3xy' + 9y = 0

19. xV + xy' = 0 20. 4xV + Y = 0

21. 9x"y' + ISxy' + Y = 0

22. 16xV - 8xy' + 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 + 3xy' - 3y = 0, y(l) = I, y'(I) = -I

26. 6x"y' + 7xy' - 2y = 0, y(1) = 0, y'(I) = I

27. xV - 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

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17 -2 6 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: x2 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 + ClX + C2x2 + ... (1) .-0

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' = ~ ncll xn- 1

11=1 and

00

y" = ~ n(n - 1)c.x·-2. 11=2

Substitotion of these forms into the second-order equation gives us 00 00

~ n(n - l)c.x·-2 + ~ c.x· = O. 11 - 2 11 - 0

Next, we equate the coefficients of each power of x to zero as summarized in the followiog table.

Powerofx Coefficient Equation

x o 2(I)C2 + Co = 0 1

or C2 = -ZCo

xl 3(2)C3 + Cl = 0 1

or C3 = -3-2 Cl

x2 4(3)C4 + C2 = 0 1 or C4 = -4'3 C2

x3 5(4)cs + C3 = 0 1

or Cs = -5'4 C3

x4 6(5)C6 + C4 = 0 1 or C6 = -6'5 C4

X,,-2 n(n - l)c. + C.-2 = 0 1

or en = 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

I C2k = 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 + 1, so the power is X2k-I. Substituting this into the last line of the table yields

(2k + 1 )(2k)c2k+ I + C2k-1 = 0

or

I c2k+ I = (2k + 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 00

= ~ C2kx2k + ~ C2k+lX2H1

<-0 1- 0

00 (-1)1 00 (_1)1 _ '" 2k + '" 2.<+1 - Co tEl, (2k)! X CI tEl, (2k + I)! x .

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. •


EXAMPLE 2 Find the general solution to y" + ~' + Y = O.

Solution We assume the series solution form

and calculate the derivatives 00

00

y = ~CnXll n=O

00

y' = L ncnxll-

t

.-1 and y" = Ln(n - l)c.x·-2• .-2

Substitution of these forms into the second-order equation yields 00 00 00

~ n(n - 1)cnx"-2 + L ncnXn + L CnX

n = o. n=2 11=1 11=0

We equate the coefficients of each power of x to zero as summarized in the following table.


x'

2(I)c2 + Co = 0

3(2)C3 + Cl + CI = 0

4(3)C4 + 2C2 + C2 = 0

5(4)cs + 3C3 + C3 = 0

6(5)C6 + 4C4 + C4 = 0

(n + 2)(n + l)c.+2 + (n + l)c. = 0

or C2 = -!co

or C3 = -let

or C4 = -lc2

or Cs = -lC3 or C6 = -~C4

I or Cn+2 = - 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

I C2k = - 2k C2k-2·

From this recurrence relation we obtain

= (2)(4)(6) ... (2k) Co·

Odd indices: Here n = 2k - 1, so the power is X2k-I. From the last line in the table, we have

1 C2k+1 = - 2k + 1 C2k-I·

From this recurrence relation we obtain

C2k+l = (-2k ~ 1)( -2k ~ J .. (-t) ( -t)Cl

(-I)' = (3)(5)'" (2k + 1) CI.


Writing the power series by grouping its even and odd powers and substituting for the coefficients yields

00 00

Y = L C2kx2k + L C21-+ 1 X2k+ 1

1=0 1=0

00 (_I)k 00 (_I)k - C ~ X2k + C ~ X2k+l • - 0&'0 (2)(4) .. ·(2k) 1&'0 (3)(5)"'(2k + I) .

EXAMPLE 3 Find the general solution to

(I - X 2)y" - &y' - 4y = 0, Ixl < 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 00 00 00

(I - X2) L n(n - I )C.X·-2 - 6 L nc.x' - 4 L c.x· = 0, 11=2 n=1 n=O

00 00 00 00

L n(n - I )C.X·-2 - L n(n - I )C.X' - 6 L nc.x' - 4 L c.x· = O. n=2 n=2 n=1 n=O

Next, we equate the coefficients of each power of x to zero as summarized in the following table.


xO 2(I)c2 - 4co=0 or _4

Cz - leO Xl 3(2)C3 - 6(I)Cl - 4Cl = 0 or C3 = iCl

x2 4(3)C4 - 2(I)c2 - 6(2)C2 - 4C2 = 0 or _6

C4 - 4C2

x3 5(4)c, - 3(2)C3 - 6(3)C3 - 4C3 = 0 or _7

Cs - SC3

x' (n + 2)(n + I)C.+2 - [n(n - I) + 6n + 4]c. = 0

(n + 2)(n + l)c.+2 - (n + 4)(n + I)c. = 0 n+4

or 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.


Odd indices: Here n = 2k - I, so the power is x2k+l. The right-hand column and last line of the table gives us

= (2k + 3)(~)(~) ... l(i) 2k+1 2k-1 2,\;-3 53 Cl

2k + 3 = --3-C1 •

The general solution is

00

Y = L cnx" .-0

00 00

= L C2kx2k + L C2k+ 1 x 2k+ 1 k~O k~O

00 00 2k+3 = CO~ (k + l)x2k + Cl~ -3-X2k+1.

~O ~O

EXAMPLE 4 Find the general solution to y" - 2>y' + Y = o.

Solution Assuming that

substitution into the differential equation gives us

00 00 00

~ n(n - l)c.x·-2 - 2 ~ nc.x· + ~ c.x· = o.

n=2 n=1 11=0

We next determine the coefficients, listiog them in the following table.


2(I)c2 + Co = 0 or

3(2)C3 - 2Cl + Cl = 0 or

4(3)C4 - 4c2 + C2 = 0 or

5(4)c, - 6C3 + C3 = 0 or

or

I C2 = -Zco

I c3 =3_2 c1

3 c4=4_3 c2

5 c'=5_4 c3

7 c6=6_5 c4

2n - I

•

x· (n + 2)(n + l)c.+2 - (2n - I)c. = 0 or C.+2 = (n + 2)(n + I) c.


From the recursive relation

2n - I e.+2 = (n + 2)(n + I) e.,

we write out the first few terms of each series for the general solution:

EXERCISES 17.5

In Exercises 1-18, use power series to f'md 1he general solution of1he differential equation.

1. y" + 2y' = 0

2. y" + 2y' + Y = 0

3.y"+4y=O

4. y" - 3y' + 2y = 0

5. xV - 2xy' + 2y = 0

6.y"-.>;Y'+y=O

7. (I + x)y" - Y = 0

8. (1 - x2)y" - 4.>;Y' + 6y = 0

Y = eO(1 - lx2 - lx' - n x• - ... ) 2 41 61

9. (x 2 - l)y" + 2xy' - 2y = 0

10. y" + y' - x"y = 0

11. (x 2 - l)y" - 6y = 0

12. .>;y" - (x + 2)y' + 2y = 0

13. (x2 - l)y" + 4'>;y' + 2y = 0

14. y" - 2xy' + 4y = 0

15. y" -2xy' +3y=O

16. (I - x 2)y" - .>;y' + 4y = 0

17. y" - xy' + 3y = 0

18. xV - 4.>;Y' + 6y = 0

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second-order differential equations - minificciones · 17.1 second-order linear equations 17-3...

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