Solve d²y/dx²+2 dy/dx+y=x sinx.
d2ydx2+2dydx+y=xsinx\dfrac{d²y}{dx²}+2 \dfrac{dy}{dx}+y=x \sin xdx2d2y+2dxdy+y=xsinx
(D2+2D+1)y=xsinx(D²+2D+1)y=x\sin x(D2+2D+1)y=xsinx
A.E is m2+2m+1=0(m+1)2=0m=−1; m=−1A.E \ is\ m²+2m+1 =0\\ (m+1)²= 0\\ m=-1;\ m=-1A.E is m2+2m+1=0(m+1)2=0m=−1; m=−1
P.I.=1D2+2D+1xsinxP.I. =\dfrac1{D²+2D+1}x \sin xP.I.=D2+2D+11xsinx
=1(D+1)2xsinx;D ⟹ D−1=\dfrac1{(D+1)²}x \sin x;\quad D\implies D-1=(D+1)21xsinx;D⟹D−1
=1D2xsinx=1D∫xsinx dx=\dfrac1{D²} x\sin x= \dfrac1D \int{x \sin x}\ dx=D21xsinx=D1∫xsinx dx
=1D(−xcosx−sinx)=\dfrac1D(-x\cos x-\sin x)=D1(−xcosx−sinx)
Hence P.I. = −2cosx−2xsinx-2\cos x-2x\sin x−2cosx−2xsinx
