What is the solution for y^{''}+3y^'=1-e^{-3x} ?

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

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y^{''}+3y^'=1-e^{-3x}

y^{^{''}} + 3 y^{^{'}} = 1 - e^{- 3 x}

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

Solution steps

y^{''} (x) + 3 y^{'} (x) = 1 - e^{- 3 x}

Solve linear ODE:

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

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

y^{^{''}} + 16 y = 6 x, y (0) = 1, y^{^{'}} (0) = 4

(D^{3} - 2 D^{2} - 3 D) y = 74 e^{- 3 x} cos (x)

y^{^{''}} - 2 y^{^{'}} - 3 y = - 11 t e^{- t}

y^{^{''}} - 4 y^{^{'}} + 8 y = x^{3} - x

y^{^{''}} - 4 y^{^{'}} + 4 y = \frac{1}{x} e^{2 x}

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