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

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

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

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

y = c_{1} e^{(- 2 + 6) x} + c_{2} e^{(- 2 - 6) x} + e^{- 2 x + 6 x} \cdot \int \frac{6}{12 e ^{6 x} x ^{2}} d x + e^{- 2 x - 6 x} \cdot \int - \frac{6 e ^{6 x}}{12 x ^{2}} d x

Solution steps

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

Solve linear ODE:

y = c_{1} e^{(- 2 + 6) x} + c_{2} e^{(- 2 - 6) x} + e^{- 2 x + 6 x} \cdot \int \frac{6}{12 e ^{6 x} x ^{2}} d x + e^{- 2 x - 6 x} \cdot \int - \frac{6 e ^{6 x}}{12 x ^{2}} d x

y = c_{1} e^{(- 2 + 6) x} + c_{2} e^{(- 2 - 6) x} + e^{- 2 x + 6 x} \cdot \int \frac{6}{12 e ^{6 x} x ^{2}} d x + e^{- 2 x - 6 x} \cdot \int - \frac{6 e ^{6 x}}{12 x ^{2}} d x

- y^{^{''}} + y = e^{x}

y^{^{''}} - y^{^{'}} + y = 4, y^{^{'}} (0) = - 1, y (0) = 4

y^{^{''}} + y - 2 y = 2 + 5 cos (x)

y^{^{''}} - 2 y^{^{'}} - 3 y = 4 e^{- t}

y^{^{''}} - 2 y - 4 x^{2} e^{2 x} = 0

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