What is the solution for y^{''}+2y^'+4y= x/(e^{2x)(1+x^2)} ?

The solution for y^{''}+2y^'+4y= x/(e^{2x)(1+x^2)} is

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y^{''}+2y^'+4y= x/(e^{2x)(1+x^2)}

y^{^{''}} + 2 y^{^{'}} + 4 y = \frac{x}{e ^{2 x} ( 1 + x ^{2} )}

y = e^{- x} (c_{1} cos (3 x) + c_{2} sin (3 x)) + e^{- x} cos (3 x) \cdot \int - \frac{3 x sin ( 3 x )}{3 e ^{x} ( 1 + x ^{2} )} d x + e^{- x} sin (3 x) \cdot \int \frac{3 x cos ( 3 x )}{3 e ^{x} ( 1 + x ^{2} )} d x

Solution steps

y^{''} (x) + 2 y^{'} (x) + 4 y = \frac{x}{e ^{2 x} ( 1 + x ^{2} )}

Solve linear ODE:

y = e^{- x} (c_{1} cos (3 x) + c_{2} sin (3 x)) + e^{- x} cos (3 x) \cdot \int - \frac{3 x sin ( 3 x )}{3 e ^{x} ( 1 + x ^{2} )} d x + e^{- x} sin (3 x) \cdot \int \frac{3 x cos ( 3 x )}{3 e ^{x} ( 1 + x ^{2} )} d x

y = e^{- x} (c_{1} cos (3 x) + c_{2} sin (3 x)) + e^{- x} cos (3 x) \cdot \int - \frac{3 x sin ( 3 x )}{3 e ^{x} ( 1 + x ^{2} )} d x + e^{- x} sin (3 x) \cdot \int \frac{3 x cos ( 3 x )}{3 e ^{x} ( 1 + x ^{2} )} d x

What is the solution for y^{''}+2y^'+4y= x/(e^{2x)(1+x^2)} ?
The solution for y^{''}+2y^'+4y= x/(e^{2x)(1+x^2)} is