What is the solution for y^{''}-9y^'=(9x)/(e^{3x)} ?

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 4 y^{^{'}} = 13 e^{12 x}

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

y^{^{'''}} + 324 y^{^{'}} = tan (18 x)

2 y^{^{''}} + 3 y^{^{'}} + 4 y = - 5 cos (x) + sin (x)

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

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