Section 17.2 Nonhomogeneous Linear Equations
EX.1
The auxiliary equation is r2 − 2r − 3 = 0 ⇒ (r − 3)(r + 1) = 0 ⇒ r = 3 or r = −1. Then the
complementary solution is
yc = C1 e3x + C2 e−x , C1 , C2 ∈ R.
Take the particular solution yp = A sin 2x + B cos 2x. Then
yp = A sin 2x + B cos 2x,
yp0 = 2A cos 2x − 2B sin 2x,
yp00 = −4A sin 2x − 4B cos 2x,
yp00 − 2yp0 − 3yp = (−7A + 4B) sin 2x + (−4A − 7B) cos 2x = cos 2x.
So we have −7A + 4B = 0 and −4A − 7B = 1 and A =
y = yc + yp = C1 e3x + C2 e−x −
−4
65
and B =
−7
.
65
So the general solution is
4
7
sin 2x −
cos 2x, C1 , C2 ∈ R.
65
65
EX.3
The auxiliary equation is r2 +9 = 0 ⇒ (r+3i)(r−3i) = 0 ⇒ r = 3i or r = −3i. Then the complementary
solution is
yc = C1 cos 3x + C2 sin 3x, C1 , C2 ∈ R.
Take the particular solution yp = Ae−2x . Then
yp = Ae−2x ,
yp0 = −2Ae−2x ,
yp00 = 4Ae−2x ,
yp00 + 9yp = 13Ae−2x = e−2x .
So we have 13A = 1 and A =
1
.
13
So the general solution is
y = yc + yp = C1 cos 3x + C2 sin 3x +
1
1 −2x
e , C1 , C2 ∈ R.
13
EX.5
The auxiliary equation is r2 − 4r + 5 = 0 ⇒ r = 2 ± i. Then the complementary solution is
yc = e2x (C1 cos x + C2 sin x), C1 , C2 ∈ R.
Take the particular solution yp = Ae−x . Then
yp = Ae−x ,
yp0 = −Ae−x ,
yp00 = Ae−x ,
yp00 − 4yp + 5yp = 10Ae−x = e−x .
So we have 10A = 1 and A =
1
.
10
So the general solution is
y = yc + yp = e2x (C1 cos x + C2 sin x) +
1 −x
e , C1 , C2 ∈ R.
10
EX.13
The auxiliary equation is r2 − r − 2 = 0 ⇒ r = 2 or − 1. Then the complementary solution is
yc = C1 e2x + C2 e−x , C1 , C2 ∈ R.
So that we can take the trial solution as
yp = (Ax + B) ex cos x + (Cx + D) ex sin x.
EX.15
The auxiliary equation is r2 − 3r + 2 = 0 ⇒ (r − 2)(r − 1) = 0 ⇒ r = 2 or 1. Then the complementary
solution is
yc = C1 e2x + C2 ex , C1 , C2 ∈ R.
So that we can take the trial solution as
yp = Axex + B cos x + C sin x.
2
EX.19
(a) The auxiliary equation is 4r2 + 1 = 0 ⇒ (2r + i)(2r − i) = 0 ⇒ r = ± 2i . Then the complementary
solution is
1
1
yc = C1 cos x + C2 sin x, C1 , C2 ∈ R.
2
2
Take the particular solution yp = A cos x + B sin x. Then
yp = A cos x + B sin x,
yp0 = −A sin x + B cos x,
yp00 = −A cos x − B sin x,
4yp00 + yp = −3A cos x − 3B sin x = cos x.
So we have −3A = 1 and −3B = 0 and A = − 31 and B = 0. So the general solution is
y = yc + yp = C1 cos
x
x 1
+ C2 sin − cos x, C1 , C2 ∈ R.
2
2 3
(b) From (a) we know the complementary solution is yc = C1 cos x2 + C2 sin x2 , C1 , C2 ∈ R. Set y1 = cos x2
and y2 = sin x2 . We have y1 y20 − y2 y10 = 21 cos2
+ 12 sin2 x2 = 12 . Thus, by the product-to-sum identity,
cos x · sin x2
1
1
3
0
u1 = −
= − sin
x − sin
x ,
1
4
2
2
4· 2
cos x · cos x2
1
3
1
0
u2 =
=
cos
x + cos
x ,
1
4
2
2
4· 2
x
2
and
3
1
1
1
x − cos
x ,
u1 =
= cos
6
2
2
2
Z
1
3
1
1
0
u2 = u2 dx = sin
x + sin
x .
6
2
2
2
Z
u01 dx
Thus
x
x
yp = u1 (x) cos + u2 (x) sin
2
2
1
3
1
1
1
1
3
1
1 2 1
2
= cos
x cos
x − cos
x + sin
x sin
x + sin
x
6
2
2
2
2
6
2
2
2
2
1
3
1
3
1
1
1
1
=
cos
x cos
x + sin
x sin
x −
cos2
x − sin2
x
6
2
2
2
2
2
2
2
1
1
1
= cos x − cos x = − cos x.
6
2
3
Then the general solution is
y = yc + yp = C1 cos
x
x 1
+ C2 sin − cos x, C1 , C2 ∈ R.
2
2 3
3
EX.21
(a) The auxiliary equation is r2 − 2r + 1 = 0 ⇒ r = 1. Then the complementary solution is
yc = C1 ex + C2 xex , C1 , C2 ∈ R.
Take the particular solution yp = Ae2x . Then
yp = Ae2x ,
yp0 = 2Ae2x ,
yp00 = 4Ae2x ,
yp00 − 2yp0 + yp = Ae2x = e2x .
So we have A = 1. And the general solution is
y = yc + yp = C1 ex + C2 xex + e2x , C1 , C2 ∈ R.
(b) From (a) we know the complementary solution is yc = C1 ex + C2 xex , C1 , C2 ∈ R. Set y1 = ex and
y2 = xex . Then y1 y20 − y2 y10 = e2x and so
u01
x
Z
= −xe ⇒ u1 (x) = −xex dx = −(x − 1)ex ,
Z
0
x
u2 = e ⇒ u2 (x) = ex dx = ex .
Thus,
yp (x) = (1 − x)e2x + xe2x = e2x
and the general solution is
y = yc + yp = C1 ex + C2 xex + e2x , C1 , C2 ∈ R.
EX.23
The auxiliary equation is r2 + 1 = 0 ⇒ r = ±i. Then the complementary solution is
yc = C1 cos x + C2 sin x, C1 , C2 ∈ R.
Set y1 = cos x and y2 = sin x. Then y1 y20 − y2 y10 = 1 and so
Z
sec2 x sin x
0
u1 = −
= − sec x tan x ⇒ u1 (x) = − sec x tan xdx = − sec x,
1
Z
sec2 x cos x
π
0
u2 =
= sec x ⇒ u2 (x) = sec xdx = ln (sec x + tan x) f or 0 < x < .
1
2
4
Thus,
yp (x) = − sec x · cos x + ln |sec x + tan x| · sin x = ln (sec x + tan x) · sin x − 1
and the general solution is
y = yc + yp = C1 cos x + C2 sin x + ln (sec x + tan x) · sin x − 1, C1 , C2 ∈ R.
EX.25
The auxiliary equation is r2 −3r+2 = 0 ⇒ (r−2)(r−1) = 0 ⇒ r = 1 or r = 2. Then the complementary
solution is
yc = C1 ex + C2 e2x , C1 , C2 ∈ R.
Set y1 = ex and y2 = e2x . Then y1 y20 − y2 y10 = e3x and so
Z
−e2x
−e−x
−e−x
0
u1 =
=
⇒
u
(x)
=
dx = ln(1 + e−x ),
1
−x
3x
−x
−x
(1 + e )e
1+e
1+e
Z
x
x
e
ex
ex + 1
e
−x
−x
u02 =
=
⇒
u
(x)
=
dx
=
ln(
)
−
e
=
ln
1
+
e
− e−x .
2
(1 + e−x )e3x
e3x + e2x
e3x + e2x
ex
Thus,
yp (x) = ex ln(1 + e−x ) + e2x ln 1 + e−x − ex ,
and the general solution is
y = yc + yp = ex C1 + ln(1 + e−x ) − 1 + e2x C2 + ln 1 + e−x , C1 , C2 ∈ R.
EX.27
The auxiliary equation is r2 − 2r + 1 = 0 ⇒ r = 1. Then the complementary solution is
yc = C1 ex + C2 xex , C1 , C2 ∈ R.
Set y1 = ex and y2 = xex . Then y1 y20 − y2 y10 = e2x and so
Z
ex
xex 1+x
x
x
1
2
0
=
−
⇒
u
(x)
=
−
dx
=
−
ln(1 + x2 )
u1 = −
1
2x
2
2
e
1+x
1+x
2
Z
x ex
e
1
1
2
u02 = 1+x
=
⇒ u2 (x) =
dx = tan−1 x.
2x
2
e
1+x
1 + x2
Thus,
1
yp (x) = − ex ln(1 + x2 ) + xex tan−1 x
2
and the general solution is
1
2
−1
y = yc + yp = e C1 + C2 x − ln(1 + x ) + x tan x , C1 , C2 ∈ R.
2
x
5