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

The solution for y^{''}-y^'-20y=27e^{-4c} is y=c_{1}e^{5t}+c_{2}e^{-4t}-(27e^{-4c})/(20)

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

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

y = c_{1} e^{5 t} + c_{2} e^{- 4 t} - \frac{27 e ^{- 4 c}}{20}

Solution steps

y^{''} (t) - y^{'} (t) - 20 y = 27 e^{- 4 c}

Solve linear ODE:

y = c_{1} e^{5 t} + c_{2} e^{- 4 t} - \frac{27 e ^{- 4 c}}{20}

y = c_{1} e^{5 t} + c_{2} e^{- 4 t} - \frac{27 e ^{- 4 c}}{20}

(D - 2)^{2} y = 8 e^{2 x}

y^{^{''}} - 10 y^{^{'}} + 25 y = 102 t^{2} e^{5 t}

2 y^{^{''}} + 4 y^{^{'}} + 8 y = 8

y^{^{''}} - 10 y^{^{'}} + 25 y = 30 x^{2} + 3

3 y^{^{''}} + 12 y = - 4 sin (2 t)

What is the solution for y^{''}-y^'-20y=27e^{-4c} ?
The solution for y^{''}-y^'-20y=27e^{-4c} is y=c_{1}e^{5t}+c_{2}e^{-4t}-(27e^{-4c})/(20)