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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 4 y^{^{'}} + 3 y = 2 e^{5} x

\frac{d ^{2} x}{d t ^{2}} + 5 \frac{d x}{d t} + 4 x = e^{6 t}

y^{^{''}} - 6 y^{^{'}} + 9 y = 27 e^{6 x}

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

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

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