Question #223867

Solve d^{2}y/dx^{2}+2dy/dx+y=xsinx


1
Expert's answer
2021-08-24T06:18:07-0400
y(x)+2y(x)+y(x)=xsinxy''(x)+2y'(x)+y(x)=x\sin x

The characteristic equation of the homogeneous DE

k2+2k+1=0k^2+2k+1=0

Roots

k1=k2=1k_1=k_2=-1

Hence, the general solution of homogeneous DE

yh(x)=C1ex+C2xexy_h(x)=C_1e^{-x}+C_2xe^{-x}

The partial solution of inhomogeneous DE

yi(x)=(A+Bx)sinx+(C+Dx)cosxy_i(x)=(A+Bx)\sin x+(C+Dx)\cos x

We get

yi(x)=(A+Bx)cosx(C+Dx)sinx+Bsinx+Dcosxy'_i(x)=(A+Bx)\cos x-(C+Dx)\sin x\\+B\sin x+D\cos x

yi(x)=(A+Bx)sinx(C+Dx)cosx+2Bcosx2Dsinxy''_i(x)=-(A+Bx)\sin x-(C+Dx)\cos x\\+2B\cos x-2D\sin x

(A+Bx)sinx(C+Dx)cosx+2Bcosx2Dsinx+2((A+Bx)cosx(C+Dx)sinx+Bsinx+Dcosx)+(A+Bx)sinx+(C+Dx)cosx=xsinx-(A+Bx)\sin x-(C+Dx)\cos x\\+2B\cos x-2D\sin x+2((A+Bx)\cos x\\-(C+Dx)\sin x+B\sin x+D\cos x)\\+(A+Bx)\sin x+(C+Dx)\cos x\\=x \sin x

So, the partial solution of inhomogeneous DE

yi(x)=12(cosxxcosx+sinx)y_i(x)=\frac{1}{2}(\cos x-x\cos x+\sin x)

Finally,

y(x)=C1ex+C2xex+12(cosxxcosx+sinx)y(x)=C_1e^{-x}+C_2xe^{-x}\\+\frac{1}{2}(\cos x-x\cos x+\sin x)


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