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

The solution for y^{''}-4y^'+4y=xe^{2x} is y=c_{1}e^{2x}+c_{2}xe^{2x}+(e^{2x}x^3)/6

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

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

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + \frac{e ^{2 x} x ^{3}}{6}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + \frac{e ^{2 x} x ^{3}}{6}

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + \frac{e ^{2 x} x ^{3}}{6}

\frac{d}{d x} ((1 + x + x^{2})^{99})

\frac{\partial}{\partial u} ((u^{2} + v^{2})^{\frac{3}{2}})

x \to 4 - lim (\frac{1}{x - 4})

\int (x^{2} - 1)^{- 1} d x

x^{2} + 2 y^{2} = 2

What is the solution for y^{''}-4y^'+4y=xe^{2x} ?
The solution for y^{''}-4y^'+4y=xe^{2x} is y=c_{1}e^{2x}+c_{2}xe^{2x}+(e^{2x}x^3)/6