Answer to Question #224865 in Chemical Engineering for Lokika

Question

Solve d²y/dx²+2 dy/dx+y=x sinx.

Answer 1

$\dfrac{d²y}{dx²}+2 \dfrac{dy}{dx}+y=x \sin x$

$(D²+2D+1)y=x\sin x$

$A.E \ is\ m²+2m+1 =0\\ (m+1)²= 0\\ m=-1;\ m=-1$

$P.I. =\dfrac1{D²+2D+1}x \sin x$

$=\dfrac1{(D+1)²}x \sin x;\quad D\implies D-1$

$=\dfrac1{D²} x\sin x= \dfrac1D \int{x \sin x}\ dx$

$=\dfrac1D(-x\cos x-\sin x)$

Hence P.I. = $-2\cos x-2x\sin x$