What is the solution for 4y^{''}-4y^'-8y=8e^{-t} ?

The solution for 4y^{''}-4y^'-8y=8e^{-t} is y=c_{1}e^{2t}+c_{2}e^{-t}-(2e^{-t}t)/3

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

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

y = c_{1} e^{2 t} + c_{2} e^{- t} - \frac{2 e ^{- t} t}{3}

Solution steps

4 y^{''} (t) - 4 y^{'} (t) - 8 y = 8 e^{- t}

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} e^{- t} - \frac{2 e ^{- t} t}{3}

y = c_{1} e^{2 t} + c_{2} e^{- t} - \frac{2 e ^{- t} t}{3}

y^{^{'''}} - 4 y^{^{'}} = t e^{- 2 t} + 3 cos (2 t)

y^{^{''}} + 49 y = 4 cos (7 x) + 6 sin (7 x)

y^{^{''}} + y^{^{'}} = - 8 e^{- x} - 4

x^{^{''}} + 2 x^{^{'}} + 5 x = 12 cos (2 t) + 3 sin (2 t)

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

What is the solution for 4y^{''}-4y^'-8y=8e^{-t} ?
The solution for 4y^{''}-4y^'-8y=8e^{-t} is y=c_{1}e^{2t}+c_{2}e^{-t}-(2e^{-t}t)/3