58
1 Mathematical Physics
1.58
d
2 y
dx 2 + m
2 y = cos bx
(1)
Replacing the right-hand member by zero,
d
2 y
dx 2 + m
2 y = 0.
(2)
Solving, we get the complimentary function
y = C 1 sin mx + C 2 cos mx = U.
(3)
Differentiating (1) twice, we get
d
4 y
dx 4 + m
2 d
2 y
dx 2 = −b
2 cos bx.
(4)
Multiply (1) by b
2 and adding the result to (4) gives
d
4 y
dx 4 + (m
2
+ b
2 )
d
2 y
dx 2 + b
2 m
2 y = 0.
(5)
The complete solution of (5) is
y = C 1 sin mx + C 2 cos mx + C 3 sin bx + C 4 cos bx
or y = U + C 3 sin bx + C 4 cos bx = U + V
We shall now determine C 3 and C 4 so that C 3 sin bx + C 4 cos bx shall be a
particular solution V of (1)
Substituting
y = C 3 sin bx + C 4 cos bx,
dy
dx
= C 3 b cos bx − C 4 b sin bx,
d
2 y
dx 2 = −C 3 b
2 sin bx − C 4 b
2 cos bx in (1), we get
C 4 (m
2
− b
2 ) cos bx + C 3 (m
2
− b
2 ) sin bx = cos bx
Equating the coefficients of like terms in this identity we get
C 4 (m
2
− b
2 ) = 1 → C 4 =
1
m 2 − b 2
C 3 (m
2
− b
2 ) = 0 → C 3 = 0
Hence a particular solution of (1) is
V =
cos bx
m 2 − b 2
and the complete solution is
y = 0 + V = C 1 sin mx + C 2 cos mx +
cos bx
m 2 − b 2
1.59
d
2 y
dx 2 − 5
dy
dx
+ 6y = x
(1)
Replace the right-hand member by zero to form the auxiliary equation
D
2
− 5D + 6 = 0
( 2 )
The roots are D = 2 and 3. The solution is
y = C 1 e
2x
+ C 2 e
3x
= 0
( 3 )