What is the solution for y^{''}+2y^'+y=(1+x)e^{-x} ?

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

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y^{''}+2y^'+y=(1+x)e^{-x}

y^{^{''}} + 2 y^{^{'}} + y = (1 + x) e^{- x}

y = c_{1} e^{- x} + c_{2} x e^{- x} + \frac{e ^{- x} x ^{3}}{6} + \frac{e ^{- x} x ^{2}}{2}

Solution steps

y^{''} (x) + 2 y^{'} (x) + y = (1 + x) e^{- x}

Solve linear ODE:

y = c_{1} e^{- x} + c_{2} x e^{- x} + \frac{e ^{- x} x ^{3}}{6} + \frac{e ^{- x} x ^{2}}{2}

y = c_{1} e^{- x} + c_{2} x e^{- x} + \frac{e ^{- x} x ^{3}}{6} + \frac{e ^{- x} x ^{2}}{2}

(D^{2} - 2 D + 1) y = x sin (x)

y^{^{''}} + 4 y = sin (5 t), y (0) = 0, y^{^{'}} (0) = 0

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

y^{^{''}} + 3 y^{^{'}} - 10 y = 3 e^{- 5 x}

y^{^{''}} - 16 y = 80

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