What is the solution for y^{''}-4y=xe^{2x} ?

The solution for y^{''}-4y=xe^{2x} is y=c_{1}e^{2x}+c_{2}e^{-2x}+(e^{2x}x^2)/8-(e^{2x}x)/(16)

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y^{''}-4y=xe^{2x}

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

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} + \frac{e ^{2 x} x ^{2}}{8} - \frac{e ^{2 x} x}{16}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} + \frac{e ^{2 x} x ^{2}}{8} - \frac{e ^{2 x} x}{16}

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} + \frac{e ^{2 x} x ^{2}}{8} - \frac{e ^{2 x} x}{16}

(D^{2} - D - 2) y = 6 x + 6 e^{- x}

2 y^{^{''}} - 3 y^{^{'}} - 2 y = 14 t^{2} - 4 t - 11

y^{^{''}} - 3 y^{^{'}} + 2 y = e x

4 y^{^{''}} + y = 4 cos (x)

y^{^{''}} + 2 y^{^{'}} + 2 y = e^{x} sin (x)

What is the solution for y^{''}-4y=xe^{2x} ?
The solution for y^{''}-4y=xe^{2x} is y=c_{1}e^{2x}+c_{2}e^{-2x}+(e^{2x}x^2)/8-(e^{2x}x)/(16)