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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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

y^{^{''}} + 6 y^{^{'}} + 5 y = 5 x^{3} + 13 x^{2} + 4 x

y^{^{''''}} + 8 y^{^{''}} + 16 y = cos (3 x) + 5

y^{^{''}} - y^{^{'}} - 2 y = cos (2 t)

y^{^{''}} + 4 y = cos (9 x)

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