What is the solution for y^{''}+2y^'+y=12xe^{-x} ?

The solution for y^{''}+2y^'+y=12xe^{-x} is y=c_{1}e^{-x}+c_{2}xe^{-x}+2e^{-x}x^3

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

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

y = c_{1} e^{- x} + c_{2} x e^{- x} + 2 e^{- x} x^{3}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{- x} + c_{2} x e^{- x} + 2 e^{- x} x^{3}

y = c_{1} e^{- x} + c_{2} x e^{- x} + 2 e^{- x} x^{3}

2 y^{^{''}} + 5 y^{^{'}} + 4 x = 0

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

y^{^{''}} - 2 y^{^{'}} + y = 4 x^{2} - 3 + \frac{e ^{x}}{x}

(d^{3} + 4 d) y = e^{x} + sin (x)

2 y^{^{''}} + 18 y^{^{'}} = 6 tan (3 x)

What is the solution for y^{''}+2y^'+y=12xe^{-x} ?
The solution for y^{''}+2y^'+y=12xe^{-x} is y=c_{1}e^{-x}+c_{2}xe^{-x}+2e^{-x}x^3