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

The solution for y^{''}-4y^'+4y=2xe^x is y=c_{1}e^{2x}+c_{2}xe^{2x}+2xe^x+4e^x

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

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

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + 2 x e^{x} + 4 e^{x}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + 2 x e^{x} + 4 e^{x}

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + 2 x e^{x} + 4 e^{x}

y^{^{''}} - y^{^{'}} + 12 y = e^{2 x}

y^{^{''}} - y = \frac{1}{cosh ( x )}

x^{^{''}} + 2 x^{^{'}} + x = e^{- 2 t}, x (0) = 2

y^{^{''}} + 4 y^{^{'}} + 13 y = cos (3 x)

4 y^{^{''}} - y = sin (x)

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