1. r2 − 2r − 3 = (r − 3)(r + 1) = 0 ⇒ r = 3, −1. yc = c1 e3x + c2 e−x . yp = A cos(2x) +
B sin(2x). Then we have
(−4A − 4B − 3A) cos(2x) + (−4B + 4A − 3B) sin(2x) = cos(2x)
Hence A =
−7
,
65
B=
−4
65
and y = c1 e3x + c2 e−x −
√
7
65
cos(2x) −
4
65
sin(2x).
2
5. r2 − 4r + 5 = 0 ⇒ r = 4± 42 −4·5 = 2 ± i. yc = e2x (c1 cos x + c2 sin x). yp = Ae−x . Then
1
1 −x
we have (A + 4A + 5A)e−x = e−x . Hence A = 10
and y = e2x (c1 cos x + c2 sin x) + 10
e .
8. r2 − 4 = (r − 2)(r + 2) = 0 ⇒ r = ±2. yc = c1 e2x + c2 e−2x . yp = ex (A cos x + B sin x).
Then we have
ex [(2B − 4A) cos x − (2A + 4B) sin x] = ex cos x
Hence A = −0.2, B = 0.1 and y = c1 e2x + c2 e−2x + ex (−0.2 cos x + 0.1 sin x). As
y(0) = 1 = c1 + c2 − 0.2 and y 0 (0) = 2 = 2c1 − 2c2 − 0.2 + 0.1, we get c1 = 1.125,
c2 = 0.075 and y = 1.125e2x + 0.075e−2x + ex (−0.2 cos x + 0.1 sin x).
9. r2 − r = r(r − 1) = 0 ⇒ r = 0, 1. yc = c1 + c2 ex . yp = x(Ax + B)ex . Then we have
ex (2Ax + 2A + B) = xex . Hence A = 0.5, B = −1 and y = c1 + c2 ex + x(0.5x − 1)ex .
As y(0) = 2 = c1 + c2 and y 0 (0) = 1 = c2 − 1, we get c1 = 0, c2 = 2 and
y = 2ex + x(0.5x − 1)ex = ex (0.5x2 − x + 2)
13. r2 − r − 2 = (r − 2)(r + 1) = 0 ⇒ r = 2, −1. yc = c1 e2x + c2 e−x . yp = ex [(Ax +
B) cos x + (Cx + D) sin x].
16. r2 + 3r − 4 = (r + 4)(r − 1) = 0 ⇒ r = 1, −4. yc = c1 ex + c2 e−4x . yp = x(Ax3 + Bx2 +
Cx + D)ex .
18. r2 + 4 = 0 ⇒ r = ±2i. yc = c1 cos(2x) + c2 sin(2x). yp = Ae3x + x(Bx + C) cos(2x) +
x(Dx + E) sin(2x).
21. (a) r2 − 2r + 1 = (r − 1)2 = 0 ⇒ r = 1. yc = c1 ex + c2 xex . yp = Ae2x . Then
(4A − 4A + A)e2x = e2x ⇒ A = 1. Hence y = c1 ex + c2 xex + e2x .
(b) yp = u1 (x)ex +u2 (x)xex . yp0 = u01 ex +u1 ex +u02 xex +u2 (x+1)ex . Set u01 ex +u02 xex = 0.
Then
yp00 − 2yp0 + yp = u01 ex + u02 (x + 1)ex = e2x
R
Hence u02 ex = e2x ⇒ u02 = ex ⇒ u2 = ex and u01 = −xex ⇒ u1 = −xex dx = −xex +ex .
We obtain yp = (−xex + ex )ex + ex · xex = e2x .
23. r2 + 1 = 0 ⇒ y = ±i. yc = c1 cos x + c2 sin x. yp = u1 cos x + u2 sin x. yp0 =
u01 cos x − u1 sin x + u02 sin x + u2 cos x. Set u01 cos x + u02 sin x = 0. Then
yp00 + yp = −u01 sin x + u02 cos x = sec2 x
1
Hence u02 (cos x + sin x tan x) = sec2 x. Notice that u02 (cos x + sin x tan x) = u02 sec x.
Hence u02 = sec x ⇒ u2 = ln | sec x + tan x| and u01 = − tan x sec x ⇒ u1 = − sec x. We
obtain yp = (− sec x) cos x + ln | sec x + tan x| · sin x = −1 + sin x ln | sec x + tan x|.
y = c1 cos x + c2 sin x − 1 + sin x ln(sec x + tan x)
25. r2 − 3r + 2 = (r − 2)(r − 1) = 0 ⇒ r = 1, 2. yc = c1 ex + c2 e2x . yp = u1 ex + u2 e2x . Set
u01 ex + u02 e2x = 0. Then u01 ex + 2u02 e2x = 1+e1−x . Hence
Z Z
Z
1
e−2x
−v
1
=
dv =
− 1 dv = ln(1+e−x )−e−x
⇒ u2 =
dx =
−x
−x
1+e
1+e
1+v
1+v
R
R 1
e−x
and u1 = − 1+e
dv = ln(1 + v) = ln(1 + e−x ). We obtain yp = ln(1 +
−x dx =
1+v
e−x )ex + (ln(1 + e−x ) − e−x )e2x .
u02 e2x
y = c1 ex + c2 e2x + ln(1 + e−x )ex + (ln(1 + e−x ) − e−x )e2x
2