1. SECOND-ORDER DIFFERENTIAL EQUATIONS18
SECOND-ORDER DIFFERENTIAL EQUATIONS

2. Nonhomogeneous Linear Equations
SECOND-ORDER DIFFERENTIAL EQUATIONS
18.2
Nonhomogeneous Linear Equations
In this section, we will learn how to solve:
Second-order nonhomogeneous linear
differential equations with constant coefficients.

3. NONHOMOGENEOUS LNR. EQNS.
Equation 1
Second-order nonhomogeneous linear differential equations with constant coefficients are equations of the form ay’’ + by’ + cy = G(x)
where:
a, b, and c are constants.
G is a continuous function.

4. COMPLEMENTARY EQUATION
The related homogeneous equation ay’’ + by’ + cy = 0 is called the complementary equation.
It plays an important role in the solution of the original nonhomogeneous equation 1.

5. NONHOMOGENEOUS LNR. EQNS.
Theorem 3
The general solution of the nonhomogeneous differential equation 1 can be written as y(x) = yp(x) + yc(x)
where:
yp is a particular solution of Equation 1.
yc is the general solution of Equation 2.

6. NONHOMOGENEOUS LNR. EQNS.
Proof
All we have to do is verify that, if y is any solution of Equation 1, then y – yp is a solution of the complementary Equation 2.
Indeed, a(y – yp)’’ + b(y – yp)’ + c(y – yp) = ay’’ – ayp’’ + by’ – byp’ + cy – cyp = (ay’’ + by’ + cy) – (ayp’’ + byp’ + cyp) = g(x) – g(x) = 0

7. NONHOMOGENEOUS LNR. EQNS.
We know from Section 17.1 how to solve the complementary equation.
Recall that the solution is: yc = c1y1 + c2y2 where y1 and y2 are linearly independent solutions of Equation 2.

8. NONHOMOGENEOUS LNR. EQNS.
Thus, Theorem 3 says that:
We know the general solution of the nonhomogeneous equation as soon as we know a particular solution yp.

9. METHODS TO FIND PARTICULAR SOLUTION
There are two methods for finding a particular solution:
The method of undetermined coefficients is straightforward, but works only for a restricted class of functions G.
The method of variation of parameters works for every function G, but is usually more difficult to apply in practice.

10. UNDETERMINED COEFFICIENTS
We first illustrate the method of undetermined coefficients for the equation
ay’’ + by’ + cy = G(x)
where G(x) is a polynomial.

11. UNDETERMINED COEFFICIENTS
It is reasonable to guess that there is a particular solution yp that is a polynomial of the same degree as G:
If y is a polynomial, then ay’’ + by’ + cy is also a polynomial.

12. UNDETERMINED COEFFICIENTS
Thus, we substitute yp(x) = a polynomial (of the same degree as G) into the differential equation and determine the coefficients.

13. UNDETERMINED COEFFICIENTS
Example 1
Solve the equation y’’ + y’ – 2y = x2
The auxiliary equation of y’’ + y’ – 2y = 0 is: r2 + r – 2 = (r – 1)(r + 2) = with roots r = 1, –2.
So, the solution of the complementary equation is: yc = c1ex + c2e–2x

14. UNDETERMINED COEFFICIENTS
Example 1
Since G(x) = x2 is a polynomial of degree 2, we seek a particular solution of the form
yp(x) = Ax2 + Bx + C
Then,
yp’ = 2Ax + B
yp’’ = 2A

15. UNDETERMINED COEFFICIENTS
Example 1
So, substituting into the given differential equation, we have:
(2A) + (2Ax + B) – 2(Ax2 + Bx + C) = x2 or
–2Ax2 + (2A – 2B)x + (2A + B – 2C) = x2

16. UNDETERMINED COEFFICIENTS
Example 1
Polynomials are equal when their coefficients are equal.
Thus, –2A = A – 2B = A + B – 2C = 0
The solution of this system of equations is: A = –½ B = –½ C = –¾

17. UNDETERMINED COEFFICIENTS
Example 1
A particular solution, therefore, is: yp(x) = –½x2 –½x – ¾
By Theorem 3, the general solution is: y = yc + yp = c1ex + c2e-2x – ½x2 – ½x – ¾

18. UNDETERMINED COEFFICIENTS
Suppose G(x) (right side of Equation 1) is of the form Cekx, where C and k are constants.
Then, we take as a trial solution a function of the same form, yp(x) = Aekx.
This is because the derivatives of ekx are constant multiples of ekx.

19. UNDETERMINED COEFFICIENTS
The figure shows four solutions of the differential equation in Example 1 in terms of:
The particular solution yp
The functions f(x) = ex and g(x) = e–2x

20. UNDETERMINED COEFFICIENTS
Example 2
Solve y’’ + 4y = e3x
The auxiliary equation is:
r2 + 4 = 0
with roots ±2i.
So, the solution of the complementary equation is: yc(x) = c1 cos 2x + c2 sin 2x

21. UNDETERMINED COEFFICIENTS
Example 2
For a particular solution, we try: yp(x) = Ae3x
Then,
yp’ = 3Ae3x
yp’’ = 9Ae3x

22. UNDETERMINED COEFFICIENTS
Example 2
Substituting into the differential equation, we have:
9Ae3x + 4(Ae3x) = e3x
So, 13Ae3x = e3x and A = 1/13

23. UNDETERMINED COEFFICIENTS
Example 2
Thus, a particular solution is:
yp(x) = 1/13 e3x
The general solution is:
y(x) = c1 cos 2x + c2 sin 2x + 1/13 e3x

24. UNDETERMINED COEFFICIENTS
Suppose G(x) is either C cos kx or C sin kx.
Then, because of the rules for differentiating the sine and cosine functions, we take as a trial particular solution a function of the form yp(x) = A cos kx + B sin kx

25. UNDETERMINED COEFFICIENTS
The figure shows solutions of the differential equation in Example 2 in terms of yp and the functions f(x) = cos 2x and g(x) = sin 2x.

26. UNDETERMINED COEFFICIENTS
Notice that:
All solutions approach ∞ as x → ∞.
All solutions (except yp) resemble sine functions when x is negative.

27. UNDETERMINED COEFFICIENTS
Example 3
Solve y’’ + y’ – 2y = sin x
We try a particular solution yp(x) = A cos x + B sin x
Then, yp’ = –A sin x + B cos x yp’’ = –A cos x – B sin x

28. UNDETERMINED COEFFICIENTS
Example 3
So, substitution in the differential equation gives: (–A cos x – B sin x) + (–A sin x + B cos x) – 2(A cos x + B sin x) = sin x or
(–3A + B) cos x + (–A – 3B) sin x = sin x

29. UNDETERMINED COEFFICIENTS
Example 3
This is true if:
–3A + B = 0 and –A – 3B = 1
The solution of this system is: A = –1/ B = –3/10
So, a particular solution is:
yp(x) = –1/10 cos x – 3/10 sin x

30. UNDETERMINED COEFFICIENTS
Example 3
In Example 1, we determined that the solution of the complementary equation is:
yc = c1ex + c2e–2x
So, the general solution of the given equation is: y(x) = c1ex + c2e–2x – 1/10 (cos x – 3 sin x)

31. UNDETERMINED COEFFICIENTS
If G(x) is a product of functions of the preceding types, we take the trial solution to be a product of functions of the same type.
For instance, in solving the differential equation y’’ + 2y’ + 4y = x cos 3x we could try yp(x) = (Ax + B) cos 3x + (Cx + D) sin 3x

32. UNDETERMINED COEFFICIENTS
If G(x) is a sum of functions of these types, we use the principle of superposition, which says that:
If yp1 and yp2 are solutions of ay’’ + by’ + cy = G1(x) ay’’ + by’ + cy = G2(x) respectively, then yp1 + yp2 is a solution of ay’’ + by’ + cy = G1(x) + G2(x)

33. UNDETERMINED COEFFICIENTS
Example 4
Solve y’’ – 4y = xex + cos 2x
The auxiliary equation is: r2 – 4 = 0 with roots ±2.
So, the solution of the complementary equation is: yc(x) = c1e2x + c2e–2x

34. UNDETERMINED COEFFICIENTS
Example 4
For the equation y’’ – 4y = xex, we try:
yp1(x) = (Ax + B)ex
Then,
y’p1= (Ax + A + B)ex
y’’p1= (Ax + 2A + B)ex

35. UNDETERMINED COEFFICIENTS
Example 4
So, substitution in the equation gives:
(Ax + 2A + B)ex – 4(Ax + B)ex = xex or
(–3Ax + 2A – 3B)ex = xex

36. UNDETERMINED COEFFICIENTS
Example 4
Thus, –3A = 1 and 2A – 3B = 0
So, A = –⅓, B = –2/9, and yp1(x) = (–⅓x – 2/9)ex

37. UNDETERMINED COEFFICIENTS
Example 4
For the equation y’’ – 4y = cos 2x, we try: yp2(x) = C cos 2x + D sin 2x
Substitution gives: –4C cos 2x – 4D sin 2x – 4(C cos 2x + D sin 2x) = cos 2x or – 8C cos 2x – 8D sin 2x = cos 2x

38. UNDETERMINED COEFFICIENTS
Example 4
Thus, –8C = 1, –8D = 0, and yp2(x) = –1/8 cos 2x
By the superposition principle, the general solution is: y = yc + yp1 + yp2
= c1e2x + c2e-2x – (1/3 x + 2/9)ex – 1/8 cos 2x

39. UNDETERMINED COEFFICIENTS
Here, we show the particular solution yp = yp1 + yp2 of the differential equation in Example 4.
The other solutions are given in terms of f(x) = e2x and g(x) = e–2x.

40. UNDETERMINED COEFFICIENTS
Finally, we note that the recommended trial solution yp sometimes turns out to be a solution of the complementary equation.
So, it can’t be a solution of the nonhomogeneous equation.
In such cases, we multiply the recommended trial solution by x (or by x2 if necessary) so that no term in yp(x) is a solution of the complementary equation.

41. UNDETERMINED COEFFICIENTS
Example 5
Solve y’’ + y = sin x
The auxiliary equation is: r2 + 1 = 0 with roots ±i.
So, the solution of the complementary equation is: yc(x) = c1 cos x + c2 sin x

42. UNDETERMINED COEFFICIENTS
Example 5
Ordinarily, we would use the trial solution
yp(x) = A cos x + B sin x
However, we observe that it is a solution of the complementary equation.
So, instead, we try: yp(x) = Ax cos x + Bx sin x

43. UNDETERMINED COEFFICIENTS
Example 5
Then,
yp’(x) = A cos x – Ax sin x + B sin x + Bx cos x
yp’’(x) = –2A sin x – Ax cos x + 2B cos x – Bx sin x

44. UNDETERMINED COEFFICIENTS
Example 5
Substitution in the differential equation gives:
yp’’ + yp = –2A sin x + 2B cos x = sin x
So, A = –½ , B = 0, and yp(x) = –½x cos x
The general solution is: y(x) = c1 cos x + c2 sin x – ½ x cos x

45. UNDETERMINED COEFFICIENTS
The graphs of four solutions of the differential equation in Example 5 are shown here.

46. UNDETERMINED COEFFICIENTS
We summarize the method of undetermined coefficients as follows.

47. If G(x) = ekxP(x), where P is a polynomial of degree n, then try:
SUMMARY—PART 1
If G(x) = ekxP(x), where P is a polynomial of degree n, then try:
yp(x) = ekxQ(x)
where Q(x) is an nth-degree polynomial (whose coefficients are determined by substituting in the differential equation).

48. yp(x) = ekxQ(x) cos mx + ekxR(x) sin mx
SUMMARY—PART 2
If G(x) = ekxP(x)cos mx or G(x) = ekxP(x) sin mx
where P is an nth-degree polynomial, then try:
yp(x) = ekxQ(x) cos mx + ekxR(x) sin mx
where Q and R are nth-degree polynomials.

49. SUMMARY—MODIFICATION
If any term of yp is a solution of the complementary equation, multiply yp by x (or by x2 if necessary).

50. UNDETERMINED COEFFICIENTS
Example 6
Determine the form of the trial solution for the differential equation
y’’ – 4y’ + 13y = e2x cos 3x

51. UNDETERMINED COEFFICIENTS
Example 6
G(x) has the form of part 2 of the summary, where k = 2, m = 3, and P(x) = 1.
So, at first glance, the form of the trial solution would be: yp(x) = e2x(A cos 3x + B sin 3x)

52. UNDETERMINED COEFFICIENTS
Example 6
However, the auxiliary equation is: r2 – 4r + 13 = 0 with roots r = 2 ± 3i.
So, the solution of the complementary equation is: yc(x) = e2x(c1 cos 3x + c2 sin 3x)

53. UNDETERMINED COEFFICIENTS
Example 6
This means that we have to multiply the suggested trial solution by x.
So, instead, we use: yp(x) = xe2x(A cos 3x + B sin 3x)

54. VARIATION OF PARAMETERS
Equation 4
Suppose we have already solved the homogeneous equation ay’’ + by’ + cy = 0 and written the solution as:
y(x) = c1y1(x) + c2y2(x)
where y1 and y2 are linearly independent solutions.

55. VARIATION OF PARAMETERS
Let’s replace the constants (or parameters) c1 and c2 in Equation 4 by arbitrary functions u1(x) and u2(x).

56. VARIATION OF PARAMETERS
Equation 5
We look for a particular solution of the nonhomogeneous equation ay’’ + by’ + cy = G(x) of the form
yp(x) = u1(x) y1(x) + u2(x) y2(x)

57. VARIATION OF PARAMETERS
This method is called variation of parameters because we have varied the parameters c1 and c2 to make them functions.

58. VARIATION OF PARAMETERS
Equation 6
Differentiating Equation 5, we get:
yp’ = (u1’y1 + u2’y2) + (u1y1’ + u2y2’)

59. VARIATION OF PARAMETERS
Since u1 and u2 are arbitrary functions, we can impose two conditions on them.
One condition is that yp is a solution of the differential equation.
We can choose the other condition so as to simplify our calculations.

60. VARIATION OF PARAMETERS
Equation 7
In view of the expression in Equation 6, let’s impose the condition that:
u1’y1 + u2’y2 = 0
Then, yp’’ = u1’y1’ + u2’y2’ + u1y1’’ + u2y2’’

61. VARIATION OF PARAMETERS
Equation 8
Substituting in the differential equation, we get: a(u1’y1’ + u2’y2’ + u1y1’’ + u2y2’’) + b(u1y1’ + u2y2’) + c(u1y1 + u2y2) = G or u1(ay1” + by1’ + cy1) + u2(ay2” + by2” + cy2) + a(u1’y1’ + u2’y2’) = G

62. VARIATION OF PARAMETERS
However, y1 and y2 are solutions of the complementary equation.
So, ay1’’ + by1’ + cy1 = 0 and ay2’’ + by2’ + cy2 = 0

63. VARIATION OF PARAMETERS
Equation 9
Thus, Equation 8 simplifies to: a(u1’y1’ + u2’y2’) = G

64. VARIATION OF PARAMETERS
Equations 7 and 9 form a system of two equations in the unknown functions u1’ and u2’.
After solving this system, we may be able to integrate to find u1 and u2 .
Then, the particular solution is given by Equation 5.

65. VARIATION OF PARAMETERS
Example 7
Solve the equation y’’ + y = tan x, 0 < x < π/2
The auxiliary equation is: r2 + 1 = 0 with roots ±i.
So, the solution of y’’ + y = 0 is: c1 sin x + c2 cos x

66. VARIATION OF PARAMETERS
Example 7
Using variation of parameters, we seek a solution of the form
yp(x) = u1(x) sin x + u2(x) cos x
Then, yp’ = (u1’ sin x + u2’ cos x) (u1 cos x – u2 sin x)

67. VARIATION OF PARAMETERS
E. g. 7—Equation 10
Set
u1’ sin x + u2’ cos x = 0
Then, yp’’ = u1’ cos x – u2’ sin x – u1 sin x – u2 cos x

68. VARIATION OF PARAMETERS
E. g. 7—Equation 11
For yp to be a solution, we must have: yp’’ + yp = u1’ cos x – u2’ sin x = tan x

69. VARIATION OF PARAMETERS
Example 7
Solving Equations 10 and 11, we get:
u1’(sin2x + cos2x) = cos x tan x
u1’ = sin x u1(x) = –cos x
We seek a particular solution.
So, we don’t need a constant of integration here.

70. VARIATION OF PARAMETERS
Example 7
Then, from Equation 10, we obtain:

71. VARIATION OF PARAMETERS
Example 7
So, u2(x) = sin x – ln(sec x + tan x)
Note that: sec x + tan x > 0 for 0 < x < π/2

72. VARIATION OF PARAMETERS
Example 7
Therefore,
yp(x) = –cos x sin x [sin x – ln(sec x + tan x)] cos x
= –cos x ln(sec x + tan x)
The general solution is: y(x) = c1 sin x + c2 cos x – cos x ln(sec x + tan x)

73. VARIATION OF PARAMETERS
The figure shows four solutions of the differential equation in Example 7.