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# Answer to Question #223867 in Chemical Engineering for Lokika

Question #223867

Solve d^{2}y/dx^{2}+2dy/dx+y=xsinx

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

The characteristic equation of the homogeneous DE

"k^2+2k+1=0"

Roots

"k_1=k_2=-1"

Hence, the general solution of homogeneous DE

"y_h(x)=C_1e^{-x}+C_2xe^{-x}"

The partial solution of inhomogeneous DE

"y_i(x)=(A+Bx)\\sin x+(C+Dx)\\cos x"

We get

"y'_i(x)=(A+Bx)\\cos x-(C+Dx)\\sin x\\\\+B\\sin x+D\\cos x"

"y''_i(x)=-(A+Bx)\\sin x-(C+Dx)\\cos x\\\\+2B\\cos x-2D\\sin x"

"-(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

"y_i(x)=\\frac{1}{2}(\\cos x-x\\cos x+\\sin x)"

Finally,

"y(x)=C_1e^{-x}+C_2xe^{-x}\\\\+\\frac{1}{2}(\\cos x-x\\cos x+\\sin x)"

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