What is the solution for y^{''}-y^'-20y=27e^{-4x} ?

The solution for y^{''}-y^'-20y=27e^{-4x} is y=c_{1}e^{5x}+c_{2}e^{-4x}-3xe^{-4x}

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y^{''}-y^'-20y=27e^{-4x}

y^{^{''}} - y^{^{'}} - 20 y = 27 e^{- 4 x}

y = c_{1} e^{5 x} + c_{2} e^{- 4 x} - 3 x e^{- 4 x}

Solution steps

y^{''} (x) - y^{'} (x) - 20 y = 27 e^{- 4 x}

Solve linear ODE:

y = c_{1} e^{5 x} + c_{2} e^{- 4 x} - 3 x e^{- 4 x}

y = c_{1} e^{5 x} + c_{2} e^{- 4 x} - 3 x e^{- 4 x}

(D^{3} - D^{2} + D - 1) y = 4 sin (x)

y^{^{''}} - 2 y^{^{'}} - 8 y = 18 t e^{4 t}

y^{^{''}} - 4 y = 8 x^{2}

x^{^{''}} - 4 x^{^{'}} + 3 x = \frac{e ^{t}}{1 + e ^{t}}

y^{^{''}} - 9 y^{^{'}} + 18 y = 9 + 18 x

What is the solution for y^{''}-y^'-20y=27e^{-4x} ?
The solution for y^{''}-y^'-20y=27e^{-4x} is y=c_{1}e^{5x}+c_{2}e^{-4x}-3xe^{-4x}