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

The solution for y^{''}+2y^'+y=x^2e^x is y=c_{1}e^{-x}+c_{2}xe^{-x}+(e^xx^2)/4-(e^xx)/2+3/8 e^x

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

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

y = c_{1} e^{- x} + c_{2} x e^{- x} + \frac{e ^{x} x ^{2}}{4} - \frac{e ^{x} x}{2} + \frac{3}{8} e^{x}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{- x} + c_{2} x e^{- x} + \frac{e ^{x} x ^{2}}{4} - \frac{e ^{x} x}{2} + \frac{3}{8} e^{x}

y = c_{1} e^{- x} + c_{2} x e^{- x} + \frac{e ^{x} x ^{2}}{4} - \frac{e ^{x} x}{2} + \frac{3}{8} e^{x}

y^{^{''}} - 6 y^{^{'}} + 5 y = - 1 + 5 x

y^{^{''}} - 6 y^{^{'}} + 9 y = 8 e^{3 x}

y^{^{''}} + 4 y = 8 tan^{2} (2 x)

y^{^{''}} + 4 y^{^{'}} = 8

y^{^{'''}} - 4 y^{^{''}} + 5 y^{^{'}} = 40 t - 7 - 6 e^{2 t}

What is the solution for y^{''}+2y^'+y=x^2e^x ?
The solution for y^{''}+2y^'+y=x^2e^x is y=c_{1}e^{-x}+c_{2}xe^{-x}+(e^xx^2)/4-(e^xx)/2+3/8 e^x