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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\int \frac{1}{2} y^{2} d y

\frac{s + 2}{s ^{2} + 4 s + 5}

y^{^{''}} + 25 y = t sin (4 t)

\frac{\partial}{\partial w} (\frac{w}{u})

x \to 0 lim (\frac{e ^{x} - x - 1}{x})

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