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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 4 y^{^{'}} + 4 y = t^{3} e^{2 t}

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

y^{^{''}} - 3 y^{^{'}} + 2 y = cos (x), y (0) = 0, y^{^{'}} (0) = 2

y^{^{''}} - 3 y^{^{'}} + 2 y = e^{t} sin (t)

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

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