What is the solution for y^{''}-9y^'=9xe^{-3x} ?

The solution for y^{''}-9y^'=9xe^{-3x} is y=c_{1}e^{9x}+c_{2}+(e^{-3x}x)/4+5/48 e^{-3x}

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y^{''}-9y^'=9xe^{-3x}

y^{^{''}} - 9 y^{^{'}} = 9 x e^{- 3 x}

y = c_{1} e^{9 x} + c_{2} + \frac{e ^{- 3 x} x}{4} + \frac{5}{48} e^{- 3 x}

Solution steps

y^{''} (x) - 9 y^{'} (x) = 9 x e^{- 3 x}

Solve linear ODE:

y = c_{1} e^{9 x} + c_{2} + \frac{e ^{- 3 x} x}{4} + \frac{5}{48} e^{- 3 x}

y = c_{1} e^{9 x} + c_{2} + \frac{e ^{- 3 x} x}{4} + \frac{5}{48} e^{- 3 x}

6 y^{^{''}} - 15 y^{^{'}} = 3 x^{2}

y^{^{''}} + 14 y^{^{'}} + 49 y = t - 2 e - 7 t

y^{^{''}} + 9 y = (x^{2} + 1) e^{3 x}

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

3 x^{^{''}} + 12 x = sin (2 t)

What is the solution for y^{''}-9y^'=9xe^{-3x} ?
The solution for y^{''}-9y^'=9xe^{-3x} is y=c_{1}e^{9x}+c_{2}+(e^{-3x}x)/4+5/48 e^{-3x}