What is the solution for y^{''}+2y^'+y=3x^4e^{-x} ?

The solution for y^{''}+2y^'+y=3x^4e^{-x} is y=c_{1}e^{-x}+c_{2}xe^{-x}+(e^{-x}x^6)/(10)

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

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

y = c_{1} e^{- x} + c_{2} x e^{- x} + \frac{e ^{- x} x ^{6}}{10}

Solution steps

y^{''} (x) + 2 y^{'} (x) + y = 3 x^{4} e^{- x}

Solve linear ODE:

y = c_{1} e^{- x} + c_{2} x e^{- x} + \frac{e ^{- x} x ^{6}}{10}

y = c_{1} e^{- x} + c_{2} x e^{- x} + \frac{e ^{- x} x ^{6}}{10}

9 y^{^{''}} - 6 y^{^{'}} + y = 9 - x^{2}

y^{^{''}} - y^{^{'}} - 2 y = sin (3 t), y (0) = 1, y^{^{'}} (0) = - 1

y^{^{''}} - 2 y^{^{'}} + 5 y = 1 + t, y (0) = 0, y^{^{'}} (0) = 1

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

y^{^{''}} - 2 y^{^{'}} + 5 = 0

What is the solution for y^{''}+2y^'+y=3x^4e^{-x} ?
The solution for y^{''}+2y^'+y=3x^4e^{-x} is y=c_{1}e^{-x}+c_{2}xe^{-x}+(e^{-x}x^6)/(10)