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

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

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

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

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + e^{x} x^{2} + 4 x e^{x} + 6 e^{x}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + e^{x} x^{2} + 4 x e^{x} + 6 e^{x}

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + e^{x} x^{2} + 4 x e^{x} + 6 e^{x}

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

\frac{d}{d x} (\frac{1}{4} x^{4} - 2 x^{2})

d x + e^{8 x} d y = 0

\frac{\partial}{\partial v} (sin (v) cos (u))

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