66
1 Mathematical Physics
1.70 (i) d
2 y
dx 2 −
4dy
dx
+ 4y = 8x
2
− 4x − 4
( 1 )
Replace the RHS member by zero to get the auxiliary solution.
D
2
− 4D + 4 = 0
The roots are D = 2 and 2. Therefore the auxiliary solution is
y = Ae
2x
+ Bxe
2x
(2)
Complete solution is
y = ( A + Bx)e
2x
+ C x
2
+ Dx + E
(3)
The derivatives are
dy
dx
= (2A + 2Bx + B)e
2x
+ 2C x + D
(4)
d
2 y/dx
2
= 4(A + B + Bx)e
2x
+ 2C
(5)
Use (3), (4) and (5) in (1) and compare the coefficients of like terms. We
get three equations. Two more equations are obtained from the conditions
y = −2 and
dy
dx
= 0 when x = 0.
Solving the five equations we get, A = −3, B = 3, C = 2, D = 3 and
E = 1. Hence the complete solution is y = 3(x − 1)e
2x
+ 2x
2
+ 3x + 1
(ii) d
2 y
dx 2 + 4y = sin x
(1)
Replace the RHS member by zero and write down the auxiliary equation
D
2
+ 4 = 0
The roots are ±2i. The auxiliary solution is
Y = A cos 2x + B sin 2x
The complete solution is
Y = A cos 2x + B sin 2x + C sin x
(2)
d
2 y
dx 2 = −4(A cos 2x + B sin 2x) − C sin x
(3)
Substitute (2) and (3) in (1) to find C = 1/3. Thus
y = A cos 2x + b sin 2x +
1
3
sin x
1.71 y
− y
+ y
− y = 0
Auxiliary equation is
D
3
− D
2
+ D − 1 = 0
(D − 1)(D
2
+ 1) = 0 The roots are D = 1, ±i
The solution is
y = A sin x + B cos x + ce
x