Find particular integral. D^2(y) +4y = sin^2 x

Question

D^2y +4y = sin^2 x

Answer 1

 Symbolic form of the given equation,

(D²+4)y = sin 2x

Corresponding auxiliary equation,

D²+4 = 0

i.e. D = ±2i

Thus, y(C.F.) = (c₁ cos 2x + c₂ sin 2x)

y(P.I.) \(=\frac1{D^2+4}\)sin 2x; Replace

D² = -a² = -4, but here f(-a²) = 0

[In case of failure, \(\frac1{f(D^2)}\)sin(ax+b) \(=x\frac1{f'(-a^2)}\)sin(ax+b)]

Implying y(P.I.)

\(=x\frac{1}{2.D}sin\,2x=\frac{x}{2}(\frac{-cos\,2x}{2})=-\frac{x\,cos\,2x}{4}.\)

Hence the complete solution,

y = (c₁ cos 2x + c₂ sin 2x) \(-\frac{x\,cos\,2x}4.\)