Engineering Mathematics, Volume II, Second Edition

 

 

Hence the complete solution is

z = C.F. + P.I.

 

EXAMPLE 5.74

Solve (2D⁴ – 3D²D′ + D′²)z = 0.

Solution. We have

 

2D⁴ – 3D²D′ + D′² = (2D² – D′) (D² – D′)

     = ɸ₁(D, D′) ɸ₂(D, D′).

 

Since 2D² – D′ is irreducible, we have

 

ɸ₁(a, b) = 2a² – b = 0

 

or

b = 2a².

 

Therefore contribution to C.F. due to ɸ₁(D, D′)z = 0 is

 

c₁e^ax+2a2y.

 

Again, since D² – D′ is irreducible, we have

 

ɸ²(a′, b′) = a′² – b′ = 0 or b′ = a′².

 

Therefore contribution to C.F. due to ɸ₂(D, D′)z = 0 is

 

 

Hence the complete solution is

 

EXAMPLE 5.75

(D