What is the solution for y^{''}+y^'-20y=4e^2t ?

The solution for y^{''}+y^'-20y=4e^2t is y=c_{1}e^{4t}+c_{2}e^{-5t}-(e^2t)/5-(e^2)/(100)

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y^{''}+y^'-20y=4e^2t

y^{^{''}} + y^{^{'}} - 20 y = 4 e^{2} t

y = c_{1} e^{4 t} + c_{2} e^{- 5 t} - \frac{e ^{2} t}{5} - \frac{e ^{2}}{100}

Solution steps

y^{''} (t) + y^{'} (t) - 20 y = 4 e^{2} t

Solve linear ODE:

y = c_{1} e^{4 t} + c_{2} e^{- 5 t} - \frac{e ^{2} t}{5} - \frac{e ^{2}}{100}

y = c_{1} e^{4 t} + c_{2} e^{- 5 t} - \frac{e ^{2} t}{5} - \frac{e ^{2}}{100}

y^{^{''}} + 4 y^{^{'}} = 2 sin (2 x)

y^{^{''}} - y^{^{'}} - 6 y = 6 t^{2}

y^{^{''}} - 3 y^{^{'}} = 4 sin (x)

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

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

What is the solution for y^{''}+y^'-20y=4e^2t ?
The solution for y^{''}+y^'-20y=4e^2t is y=c_{1}e^{4t}+c_{2}e^{-5t}-(e^2t)/5-(e^2)/(100)