1.3 Solutions
65
d
4 y
dx 4 +
4d
2 y
dx 2 = −8 cos(2x)
( 2 )
Multiply (1) by 4 and add to (2),
d
4 y
dx 4 +
8d
2 y
dx 2 + 16y = 0
D
4
+ 8D
2
+ 16 = 0
(D
2
+ 4)
2
= 0
D = ±2i
y = C 1 sin 2x + C 3 x sin 2x + C 2 cos 2x + C 4 x cos 2x
= U + C 3 x sin 2x + C 4 x cos 2x
= U + V
Y = V = C 3 x sin 2x + C 4 x cos 2x
(3)
Use (3) in (1) and compare the coefficients of sin 2x and cos 2x to find
C 3 = 1/2 and C 4 = 0. Thus the complete solution is
y = C 1 sin 2x + C 2 cos 2x +
1
2
x sin 2x
1.69
dy
dx
+
3y
x + 2
= x + 2
This equation is of the form
dy
dx
+ yp(x) = Q(x)
with P =
3
x+2
and Q = x + 2
The solution is obtained from
y exp
p(x)dx =
Q(x) exp
p(x)dx
dx + C
Now
p(x)dx = 3
dx
x + 2
= 3 ln(x + 2) = ln(x + 2)
3
∴ y exp (ln(x + 2)
3 ) =
(x + 2) exp (ln (x + 2)
3 )
dx + C
y(x + 2)
3
=
(x + 2)
4 dx + C
=
(x + 2)
5
5
+ C
or y =
(x + 2)
2
5
+ C
y = 2 when x = −1
Therefore C =
9
5
The complete solution is
y =
(x + 2)
2
5
+
9
5
=
x
2
+ 4x + 13
5