What is the solution for y^{''}-8y^'-20y=5e^{-2x} ?

The solution for y^{''}-8y^'-20y=5e^{-2x} is y=c_{1}e^{10x}+c_{2}e^{-2x}-(5e^{-2x}x)/(12)

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y^{''}-8y^'-20y=5e^{-2x}

y^{^{''}} - 8 y^{^{'}} - 20 y = 5 e^{- 2 x}

y = c_{1} e^{10 x} + c_{2} e^{- 2 x} - \frac{5 e ^{- 2 x} x}{12}

Solution steps

y^{''} (x) - 8 y^{'} (x) - 20 y = 5 e^{- 2 x}

Solve linear ODE:

y = c_{1} e^{10 x} + c_{2} e^{- 2 x} - \frac{5 e ^{- 2 x} x}{12}

y = c_{1} e^{10 x} + c_{2} e^{- 2 x} - \frac{5 e ^{- 2 x} x}{12}

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

y^{^{''}} + 9 y^{^{'}} = sin (t), y (0) = 1, y^{^{'}} (0) = 1

y^{^{''}} + 9 y = 5 sin (3 x), y (0) = 6, y^{^{'}} (0) = 0

y^{^{''}} + 10 y^{^{'}} + 25 y = \frac{- 14 e ^{- 5 t}}{t ^{2} + 1}

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

What is the solution for y^{''}-8y^'-20y=5e^{-2x} ?
The solution for y^{''}-8y^'-20y=5e^{-2x} is y=c_{1}e^{10x}+c_{2}e^{-2x}-(5e^{-2x}x)/(12)