# Nonhomogeneous Linear Equations

```Nonhomogeneous Linear Equations
In this section we learn how to solve second-order nonhomogeneous linear differential equations with constant coefficients, that is, equations of the form
1
ay by cy Gx
where a, b, and c are constants and G is a continuous function. The related homogeneous
equation
2
ay by cy 0
is called the complementary equation and plays an important role in the solution of the
original nonhomogeneous equation (1).
3 Theorem The general solution of the nonhomogeneous differential equation (1)
can be written as
yx ypx ycx
where yp is a particular solution of Equation 1 and yc is the general solution of the
complementary Equation 2.
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
ay yp by yp cy yp ay ayp by byp cy cyp
ay by cy ayp byp cyp tx tx 0
We know from Additional Topics: Second-Order Linear Differential Equations how to
solve the complementary equation. (Recall that the solution is yc c1 y1 c2 y2 , where y1
and y2 are linearly independent solutions of Equation 2.) Therefore, Theorem 3 says that
we know the general solution of the nonhomogeneous equation as soon as we know a particular solution yp . 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 i0s usually
more difficult to apply in practice.
The Method of Undetermined Coefficients
We first illustrate the method of undetermined coefficients for the equation
ay by cy Gx
where Gx) is a polynomial. It is reasonable to guess that there is a particular solution
yp that is a polynomial of the same degree as G because if y is a polynomial, then
ay by cy is also a polynomial. We therefore substitute ypx a polynomial (of the
same degree as G ) into the differential equation and determine the coefficients.
EXAMPLE 1 Solve the equation y y 2y x 2.
SOLUTION The auxiliary equation of y y 2y 0 is
r 2 r 2 r 1r 2 0
1
2 ■ NONHOMOGENEOUS LINEAR EQUATIONS
with roots r 1, 2. So the solution of the complementary equation is
yc c1 e x c2 e2x
Since Gx x 2 is a polynomial of degree 2, we seek a particular solution of the form
ypx Ax 2 Bx C
Then yp 2Ax B and yp 2A so, substituting into the given differential equation, we
have
2A 2Ax B 2Ax 2 Bx C x 2
or
2Ax 2 2A 2Bx 2A B 2C x 2
Polynomials are equal when their coefficients are equal. Thus
Figure 1 shows four solutions of the differential equation in Example 1 in terms of the particular solution yp and the functions f x e x
and tx e2 x.
■ ■
8
2A 2B 0
2A B 2C 0
The solution of this system of equations is
A 12
yp+2f+3g
yp+3g
2A 1
B 12
C 34
A particular solution is therefore
yp+2f
_3
3
ypx 12 x 2 12 x 34
yp
and, by Theorem 3, the general solution is
_5
y yc yp c1 e x c2 e2x 12 x 2 12 x 34
FIGURE 1
If Gx (the right side of Equation 1) is of the form Ce k x, where C and k are constants,
then we take as a trial solution a function of the same form, ypx Ae k x, because the
derivatives of e k x are constant multiples of e k x.
EXAMPLE 2 Solve y 4y e 3x.
SOLUTION The auxiliary equation is r 2 4 0 with roots 2i, so the solution of the
Figure 2 shows solutions of the differential
equation in Example 2 in terms of yp and the
functions f x cos 2x and tx sin 2x.
Notice that all solutions approach as x l and all solutions resemble sine functions when
x is negative.
■ ■
4
complementary equation is
ycx c1 cos 2x c2 sin 2x
For a particular solution we try ypx Ae 3x. Then yp 3Ae 3x and yp 9Ae 3x. Substituting into the differential equation, we have
9Ae 3x 4Ae 3x e 3x
yp+f+g
yp+g
yp
_4
yp+f
_2
FIGURE 2
so 13Ae 3x e 3x and A 131 . Thus, a particular solution is
2
ypx 131 e 3x
and the general solution is
yx c1 cos 2x c2 sin 2x 131 e 3x
If Gx 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
ypx A cos kx B sin kx
NONHOMOGENEOUS LINEAR EQUATIONS ■ 3
EXAMPLE 3 Solve y y 2y sin x.
SOLUTION We try a particular solution
ypx A cos x B sin x
yp A sin x B cos x
Then
yp A cos x B sin x
so substitution in the differential equation gives
A cos x B sin x A sin x B cos x 2A cos x B sin x sin x
3A B cos x A 3B sin x sin x
or
This is true if
3A B 0
and
A 3B 1
The solution of this system is
A 101
B 103
so a particular solution is
ypx 101 cos x 103 sin x
In Example 1 we determined that the solution of the complementary equation is
yc c1 e x c2 e2x. Thus, the general solution of the given equation is
yx c1 e x c2 e2x 101 cos x 3 sin x
If Gx is a product of functions of the preceding types, then 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 would try
ypx Ax B cos 3x Cx D sin 3x
If Gx is a sum of functions of these types, we use the easily verified principle of superposition, which says that if yp1 and yp2 are solutions of
ay by cy G1x
ay by cy G2x
respectively, then yp1 yp2 is a solution of
ay by cy G1x G2x
EXAMPLE 4 Solve y 4y xe x cos 2x.
SOLUTION The auxiliary equation is r 2 4 0 with roots 2, so the solution of the com-
plementary equation is ycx c1 e 2x c2 e2x. For the equation y 4y xe x we try
yp1x Ax Be x
Then yp1 Ax A Be x, yp1 Ax 2A Be x, so substitution in the equation
gives
Ax 2A Be x 4Ax Be x xe x
or
3Ax 2A 3Be x xe x
4 ■ NONHOMOGENEOUS LINEAR EQUATIONS
Thus, 3A 1 and 2A 3B 0, so A 13 , B 29 , and
yp1x ( 13 x 29 )e x
For the equation y 4y cos 2x, we try
yp2x C cos 2x D sin 2x
In Figure 3 we show the particular solution
yp yp1 yp 2 of the differential equation in
Example 4. The other solutions are given in
terms of f x e 2 x and tx e2 x.
■ ■
5
Substitution gives
4C cos 2x 4D sin 2x 4C cos 2x D sin 2x cos 2x
8C cos 2x 8D sin 2x cos 2x
or
yp+2f+g
Therefore, 8C 1, 8D 0, and
yp+g
yp+f
_4
yp2x 18 cos 2x
1
yp
By the superposition principle, the general solution is
_2
y yc yp1 yp2 c1 e 2x c2 e2x ( 13 x 29 )e x 18 cos 2x
FIGURE 3
Finally we note that the recommended trial solution yp sometimes turns out to be a solution of the complementary equation and therefore can’t be a solution of the nonhomogeneous equation. In such cases we multiply the recommended trial solution by x (or by x 2
if necessary) so that no term in ypx is a solution of the complementary equation.
EXAMPLE 5 Solve y y sin x.
SOLUTION The auxiliary equation is r 2 1 0 with roots i, so the solution of the com-
plementary equation is
ycx c1 cos x c2 sin x
Ordinarily, we would use the trial solution
ypx A cos x B sin x
but we observe that it is a solution of the complementary equation, so instead we try
ypx Ax cos x Bx sin x
Then
ypx A cos x Ax sin x B sin x Bx cos x
ypx 2A sin x Ax cos x 2B cos x Bx sin x
The graphs of four solutions of the differential equation in Example 5 are shown in Figure 4.
■ ■
Substitution in the differential equation gives
4
yp yp 2A sin x 2B cos x sin x
so A 12 , B 0, and
_2π
2π
ypx 12 x cos x
yp
_4
FIGURE 4
The general solution is
yx c1 cos x c2 sin x 12 x cos x
NONHOMOGENEOUS LINEAR EQUATIONS ■ 5
We summarize the method of undetermined coefficients as follows:
1. If Gx e kxPx, where P is a polynomial of degree n, then try ypx e kxQx,
where Qx is an nth-degree polynomial (whose coefficients are determined by
substituting in the differential equation.)
2. If Gx e kxPx cos mx or Gx e kxPx sin mx, where P is an nth-degree
polynomial, then try
ypx e kxQx cos mx e kxRx sin mx
where Q and R are nth-degree polynomials.
Modification: If any term of yp is a solution of the complementary equation, multiply yp
by x (or by x 2 if necessary).
EXAMPLE 6 Determine the form of the trial solution for the differential equation
y 4y 13y e 2x cos 3x.
SOLUTION Here Gx has the form of part 2 of the summary, where k 2, m 3, and
Px 1. So, at first glance, the form of the trial solution would be
ypx e 2xA cos 3x B sin 3x
But the auxiliary equation is r 2 4r 13 0, with roots r 2 3i, so the solution
of the complementary equation is
ycx e 2xc1 cos 3x c2 sin 3x
This means that we have to multiply the suggested trial solution by x. So, instead, we
use
ypx xe 2xA cos 3x B sin 3x
The Method of Variation of Parameters
Suppose we have already solved the homogeneous equation ay by cy 0 and written the solution as
4
yx c1 y1x c2 y2x
where y1 and y2 are linearly independent solutions. Let’s replace the constants (or parameters) c1 and c2 in Equation 4 by arbitrary functions u1x and u2x. We look for a particular solution of the nonhomogeneous equation ay by cy Gx of the form
5
ypx u1xy1x u2xy2x
(This method is called variation of parameters because we have varied the parameters c1
and c2 to make them functions.) Differentiating Equation 5, we get
6
yp u1 y1 u2 y2 u1 y1 u2 y2 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. In view of the expression in Equation 6, let’s impose the
condition that
7
u1 y1 u2 y2 0
6 ■ NONHOMOGENEOUS LINEAR EQUATIONS
yp u1 y1 u2 y2 u1 y1 u2 y2
Then
Substituting in the differential equation, we get
au1 y1 u2 y2 u1 y1 u2 y2 bu1 y1 u2 y2 cu1 y1 u2 y2 G
or
8
u1ay1 by1 cy1 u2ay2 by2 cy2 au1 y1 u2 y2 G
But y1 and y2 are solutions of the complementary equation, so
ay1 by1 cy1 0
and
ay2 by2 cy2 0
and Equation 8 simplifies to
au1 y1 u2 y2 G
9
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 and then the particular solution is given by Equation 5.
EXAMPLE 7 Solve the equation y y tan x, 0 x 2.
SOLUTION The auxiliary equation is r 2 1 0 with roots i, so the solution of
y y 0 is c1 sin x c2 cos x. Using variation of parameters, we seek a solution
of the form
ypx u1x sin x u2x cos x
Then
yp u1 sin x u2 cos x u1 cos x u2 sin x
Set
u1 sin x u2 cos x 0
10
Then
yp u1 cos x u2 sin x u1 sin x u2 cos x
For yp to be a solution we must have
yp yp u1 cos x u2 sin x tan x
11
Solving Equations 10 and 11, we get
u1sin 2x cos 2x cos x tan x
■ ■ Figure 5 shows four solutions of the
differential equation in Example 7.
u1 sin x
2.5
u1x cos x
(We seek a particular solution, so we don’t need a constant of integration here.) Then,
from Equation 10, we obtain
0
yp
π
2
u2 sin x
sin 2x
cos 2x 1
u1 cos x sec x
cos x
cos x
cos x
_1
FIGURE 5
So
u2x sin x lnsec x tan x
NONHOMOGENEOUS LINEAR EQUATIONS ■ 7
(Note that sec x tan x 0 for 0 x 2.) Therefore
ypx cos x sin x sin x lnsec x tan x cos x
cos x lnsec x tan x
and the general solution is
yx c1 sin x c2 cos x cos x lnsec x tan x
Exercises
S
16. y 3y 4 y x 3 xe x
17. y 2 y 10 y x 2ex cos 3x
18. y 4y e 3x x sin 2x
1–10
Solve the differential equation or initial-value problem
using the method of undetermined coefficients.
1. y 3y 2y x
2. y 9y e
2
3. y 2y sin 4x
7. y y e x x 3,
y0 2,
8. y 4y e x cos x,
y0 2,
■
■
■
■
■
■
y0 0
■
■
■
■
■
; 11–12
Graph the particular solution and several other solutions.
What characteristics do these solutions have in common?
■
■
■
■
13 –18
■
13. y 9 y e
x sin x
2
x
14. y 9 y xe
cos x
15. y 9 y 1 xe 9x
■
■
■
■
■
■
■
■
■
■
■
■
■
■
■
■
23–28
Solve the differential equation using the method of variation of parameters.
1
1 ex
25. y 3y 2y ■
■
■
■
■
Write a trial solution for the method of undetermined
coefficients. Do not determine the coefficients.
2x
■
24. y y cot x, 0 x 2
12. 2y 3y y 1 cos 2x
■
■
23. y y sec x, 0 x 2
11. 4y 5y y e x
■
■
22. y y e x
y0 1,
■
■
21. y 2y y e 2x
y0 2
■
■
■
20. y 3y 2y sin x
y0 1
10. y y 2y x sin 2x,
■
19. y 4y x
y0 0
y0 1,
■
Solve the differential equation using (a) undetermined
coefficients and (b) variation of parameters.
6. y 2y y xex
5. y 4y 5y e
■
19–22
4. y 6y 9y 1 x
x
9. y y xe x,
■
3x
■
26. y 3y 2y sine x 27. y y 1
x
e2x
x3
28. y 4y 4y ■
■
■
■
■
■
■
■
■
■
■
■
■
8 ■ NONHOMOGENEOUS LINEAR EQUATIONS
S
1. y c1 e2x c2 ex 2 x 2 2 x 1
3. y c1 c2 e 2x 3
7
4
cos 4x 201 sin 4x
1
5. y e c1 cos x c2 sin x 10 ex
3
11
1 x
7. y 2 cos x 2 sin x 2 e x 3 6x
1
9. y e x ( 2 x 2 x 2)
5
11.
The solutions are all
asymptotic to yp e x10 as
x l . Except for yp ,
all solutions approach
_2
4
yp
either or as x l .
1
40
2x
_4
13.
15.
17.
19.
21.
23.
25.
27.
yp Ae 2x Bx 2 Cx D cos x Ex 2 Fx G sin x
yp Ax Bx C e 9x
yp xex Ax 2 Bx C cos 3x Dx 2 Ex F sin 3x
y c1 cos 2x c2 sin 2x 14 x
y c1e x c2 xe x e 2x
y c1 x sin x c2 ln cos x cos x
y c1 ln1 ex e x c2 ex ln1 ex e 2x
y [c1 12 x e xx dx]ex [c2 12 x exx dx]e x
NONHOMOGENEOUS LINEAR EQUATIONS ■ 9
Solutions: Nonhomogeneous Linear Equations
1. The auxiliary equation is r 2 + 3r + 2 = (r + 2)(r + 1) = 0, so the complementary solution is
yc (x) = c1 e−2x + c2 e−x . We try the particular solution yp (x) = Ax2 + Bx + C, so yp0 = 2Ax + B and
yp00 = 2A. Substituting into the differential equation, we have (2A) + 3(2Ax + B) + 2(Ax2 + Bx + C) = x2 or
2Ax2 + (6A + 2B)x + (2A + 3B + 2C) = x2 . Comparing coefficients gives 2A = 1, 6A + 2B = 0, and
2A + 3B + 2C = 0, so A = 12 , B = − 32 , and C =
−2x
y(x) = yc (x) + yp (x) = c1 e
+ c2 e
−x
+
1 2
x
2
−
7
.
4
3
x
2
Thus the general solution is
+ 74 .
3. The auxiliary equation is r 2 − 2r = r(r − 2) = 0, so the complementary solution is yc (x) = c1 + c2 e2x . Try the
particular solution yp (x) = A cos 4x + B sin 4x, so yp0 = −4A sin 4x + 4B cos 4x
and yp00 = −16A cos 4x − 16B sin 4x. Substitution into the differential equation
gives (−16A cos 4x − 16B sin 4x) − 2(−4A sin 4x + 4B cos 4x) = sin 4x
⇒
(−16A − 8B) cos 4x + (8A − 16B) sin 4x = sin 4x. Then −16A − 8B = 0 and 8A − 16B = 1
1
and B = − 20
. Thus the general solution is y(x) = yc (x) + yp (x) = c1 + c2 e2x +
1
40
cos 4x −
1
20
⇒ A=
1
40
sin 4x.
5. The auxiliary equation is r 2 − 4r + 5 = 0 with roots r = 2 ± i, so the complementary solution is
yc (x) = e2x (c1 cos x + c2 sin x). Try yp (x) = Ae−x , so yp0 = −Ae−x and yp00 = Ae−x . Substitution gives
Ae−x − 4(−Ae−x ) + 5(Ae−x ) = e−x
2x
y(x) = e (c1 cos x + c2 sin x) +
⇒
10Ae−x = e−x
⇒
A=
1
.
10
Thus the general solution is
1 −x
e .
10
7. The auxiliary equation is r 2 + 1 = 0 with roots r = ±i, so the complementary solution is
yc (x) = c1 cos x + c2 sin x. For y00 + y = ex try yp1 (x) = Aex . Then yp0 1 = yp001 = Aex and substitution gives
Aex + Aex = ex
A = 12 , so yp1 (x) = 12 ex . For y00 + y = x3 try yp2 (x) = Ax3 + Bx2 + Cx + D.
⇒
Then yp0 2 = 3Ax2 + 2Bx + C and yp002 = 6Ax + 2B. Substituting, we have
6Ax + 2B + Ax3 + Bx2 + Cx + D = x3 , so A = 1, B = 0, 6A + C = 0 ⇒ C = −6, and 2B + D = 0
⇒
D = 0. Thus yp2 (x) = x3 − 6x and the general solution is
y(x) = yc (x) + yp1 (x) + yp2 (x) = c1 cos x + c2 sin x + 12 ex + x3 − 6x. But 2 = y(0) = c1 +
and 0 = y0 (0) = c2 +
y(x) =
3
2
cos x +
11
2
1
2
−6 ⇒
c2 =
11
.
2
1
2
⇒ c1 =
3
2
Thus the solution to the initial-value problem is
sin x + 12 ex + x3 − 6x.
9. The auxiliary equation is r 2 − r = 0 with roots r = 0, r = 1 so the complementary solution is yc (x) = c1 + c2 ex .
Try yp (x) = x(Ax + B)ex so that no term in yp is a solution of the complementary equation. Then
yp0 = (Ax2 + (2A + B)x + B)ex and yp00 = (Ax2 + (4A + B)x + (2A + 2B))ex . Substitution into the
differential equation gives (Ax2 + (4A + B)x + (2A + 2B))ex − (Ax2 + (2A + B)x + B)ex = xex ⇒
¡
¢
(2Ax + (2A + B))ex = xex ⇒ A = 12 , B = −1. Thus yp (x) = 12 x2 − x ex and the general solution is
¡
¢
y(x) = c1 + c2 ex + 12 x2 − x ex . But 2 = y(0) = c1 + c2 and 1 = y0 (0) = c2 − 1, so c2 = 2 and c1 = 0. The
¡
¡
¢
¢
solution to the initial-value problem is y(x) = 2ex + 12 x2 − x ex = ex 12 x2 − x + 2 .
10 ■ NONHOMOGENEOUS LINEAR EQUATIONS
11. yc (x) = c1 e−x/4 + c2 e−x . Try yp (x) = Aex . Then
10Aex = ex , so A =
1
10
and the general solution is
y(x) = c1 e−x/4 + c2 e−x +
1 x
e .
10
The solutions are all composed
of exponential curves and with the exception of the particular
solution (which approaches 0 as x → −∞), they all approach
either ∞ or −∞ as x → −∞. As x → ∞, all solutions are
asymptotic to yp =
1 x
e .
10
13. Here yc (x) = c1 cos 3x + c2 sin 3x. For y00 + 9y = e2x try yp1 (x) = Ae2x and for y 00 + 9y = x2 sin x try
yp2 (x) = (Bx2 + Cx + D) cos x + (Ex2 + F x + G) sin x. Thus a trial solution is
yp (x) = yp1 (x) + yp2 (x) = Ae2x + (Bx2 + Cx + D) cos x + (Ex2 + F x + G) sin x.
15. Here yc (x) = c1 + c2 e−9x . For y 00 + 9y0 = 1 try yp1 (x) = Ax (since y = A is a solution to the complementary
equation) and for y 00 + 9y0 = xe9x try yp2 (x) = (Bx + C)e9x .
17. Since yc (x) = e−x (c1 cos 3x + c2 sin 3x) we try
yp (x) = x(Ax2 + Bx + C)e−x cos 3x + x(Dx2 + Ex + F )e−x sin 3x (so that no term of yp is a solution of the
complementary equation).
Note: Solving Equations (7) and (9) in The Method of Variation of Parameters gives
u01 = −
Gy2
a (y1 y20 − y2 y10 )
u02 =
and
Gy1
a (y1 y20 − y2 y10 )
We will use these equations rather than resolving the system in each of the remaining exercises in this section.
19. (a) The complementary solution is yc (x) = c1 cos 2x + c2 sin 2x. A particular solution is of the form
yp (x) = Ax + B. Thus, 4Ax + 4B = x
⇒ A=
1
4
solution is y = yc + yp = c1 cos 2x + c2 sin 2x + 14 x.
and B = 0 ⇒ yp (x) = 14 x. Thus, the general
(b) In (a), yc (x) = c1 cos 2x + c2 sin 2x, so set y1 = cos 2x, y2 = sin 2x. Then
y1 y20 − y2 y10 = 2 cos2 2x + 2 sin2 2x = 2 so u01 = − 12 x sin 2x ⇒
R
¡
¢
u1 (x) = − 12 x sin 2x dx = − 14 −x cos 2x + 12 sin 2x [by parts] and u02 = 12 x cos 2x
R
¡
¢
⇒ u2 (x) = 12 x cos 2xdx = 14 x sin 2x + 12 cos 2x [by parts]. Hence
¢
¢
¡
¡
yp (x) = − 14 −x cos 2x + 12 sin 2x cos 2x + 14 x sin 2x + 12 cos 2x sin 2x = 14 x. Thus
y(x) = yc (x) + yp (x) = c1 cos 2x + c2 sin 2x + 14 x.
21. (a) r 2 − r = r(r − 1) = 0 ⇒
r = 0, 1, so the complementary solution is yc (x) = c1 ex + c2 xex . A particular
solution is of the form yp (x) = Ae2x . Thus 4Ae2x − 4Ae2x + Ae2x = e2x
⇒
2x
x
⇒
Ae2x = e2x
x
yp (x) = e . So a general solution is y(x) = yc (x) + yp (x) = c1 e + c2 xe + e2x .
⇒ A=1
(b) From (a), yc (x) = c1 ex + c2 xex , so set y1 = ex , y2 = xex . Then, y1 y20 − y2 y10 = e2x (1 + x) − xe2x = e2x
R
and so u01 = −xex ⇒ u1 (x) = − xex dx = −(x − 1)ex [by parts] and u02 = ex ⇒
R
u2 (x) = ex dx = ex . Hence yp (x) = (1 − x)e2x + xe2x = e2x and the general solution is
y(x) = yc (x) + yp (x) = c1 ex + c2 xex + e2x .
NONHOMOGENEOUS LINEAR EQUATIONS ■ 11
23. As in Example 6, yc (x) = c1 sin x + c2 cos x, so set y1 = sin x, y2 = cos x. Then
sec x cos x
y1 y20 − y2 y10 = − sin2 x − cos2 x = −1, so u01 = −
= 1 ⇒ u1 (x) = x and
−1
R
sec x sin x
u02 =
= − tan x ⇒ u2 (x) = − tan xdx = ln |cos x| = ln(cos x) on 0 < x < π2 . Hence
−1
yp (x) = x sin x + cos x ln(cos x) and the general solution is y(x) = (c1 + x) sin x + [c2 + ln(cos x)] cos x.
25. y1 = ex , y2 = e2x and y1 y20 − y2 y10 = e3x . So u01 =
Z
−e2x
e−x
=−
and
−x
3x
(1 + e )e
1 + e−x
e−x
ex
ex
dx = ln(1 + e−x ). u02 =
= 3x
so
1 + e−x
(1 + e−x )e3x
e + e2x
¶
µ
Z
ex
ex + 1
u2 (x) =
− e−x = ln(1 + e−x ) − e−x . Hence
dx = ln
3x
2x
e +e
ex
u1 (x) =
−
yp (x) = ex ln(1 + e−x ) + e2x [ln(1 + e−x ) − e−x ] and the general solution is
y(x) = [c1 + ln(1 + e−x )]ex + [c2 − e−x + ln(1 + e−x )]e2x .
ex
e−x
and
27. y1 = e−x , y2 = ex and y1 y20 − y2 y10 = 2. So u01 = − , u02 =
2x
2x
Z x
Z −x
e
e
dx + ex
dx. Hence the general solution is
yp (x) = −e−x
2x
2x
µ
µ
¶
Z x
Z −x ¶
e
e
y(x) = c1 −
dx e−x + c2 +
dx ex .
2x
2x
```