Question #224865

Solve d²y/dx²+2 dy/dx+y=x sinx.


1
Expert's answer
2021-08-16T02:20:38-0400

d2ydx2+2dydx+y=xsinx\dfrac{d²y}{dx²}+2 \dfrac{dy}{dx}+y=x \sin x


(D2+2D+1)y=xsinx(D²+2D+1)y=x\sin x


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=-1


P.I.=1D2+2D+1xsinxP.I. =\dfrac1{D²+2D+1}x \sin x


=1(D+1)2xsinx;D    D1=\dfrac1{(D+1)²}x \sin x;\quad D\implies D-1


=1D2xsinx=1Dxsinx dx=\dfrac1{D²} x\sin x= \dfrac1D \int{x \sin x}\ dx


=1D(xcosxsinx)=\dfrac1D(-x\cos x-\sin x)


Hence P.I. = 2cosx2xsinx-2\cos x-2x\sin x


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