What is the solution for y^{''}-2y^'-3y=8e^{-t} ?

The solution for y^{''}-2y^'-3y=8e^{-t} is y=c_{1}e^{3t}+c_{2}e^{-t}-2te^{-t}

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y^{''}-2y^'-3y=8e^{-t}

y^{^{''}} - 2 y^{^{'}} - 3 y = 8 e^{- t}

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

Solution steps

y^{''} (t) - 2 y^{'} (t) - 3 y = 8 e^{- t}

Solve linear ODE:

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

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

y^{^{''}} - y^{^{'}} + 4 y = x^{3} e^{2 x} + x e^{2 x}

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

9 y^{^{''}} + 6 y^{^{'}} + y = x^{2}

y^{^{''}} + 10 y^{^{'}} + 26 y = 2 e^{- 5 x} cos (x)

(D^{2} + 3 D + 2) y = 40 e^{3 x}

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