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

The solution for y^{''}-2y^'-3y=e^{-3t} is y=c_{1}e^{3t}+c_{2}e^{-t}+1/12 e^{-3t}

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

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

y = c_{1} e^{3 t} + c_{2} e^{- t} + \frac{1}{12} e^{- 3 t}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{3 t} + c_{2} e^{- t} + \frac{1}{12} e^{- 3 t}

y = c_{1} e^{3 t} + c_{2} e^{- t} + \frac{1}{12} e^{- 3 t}

y^{^{''}} + 4 y^{^{'}} + 4 y = 16 t + 13

y^{^{''}} - 14 y^{^{'}} + 49 y = 2 t e^{7 t}

v^{^{''}} - 1 v = 2 e^{4 x}

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

y^{^{''}} + 2 y^{^{'}} + 2 y = 5

What is the solution for y^{''}-2y^'-3y=e^{-3t} ?
The solution for y^{''}-2y^'-3y=e^{-3t} is y=c_{1}e^{3t}+c_{2}e^{-t}+1/12 e^{-3t}