What is the solution for y^{''}+4y^'+3y=5e^{-3t} ?

The solution for y^{''}+4y^'+3y=5e^{-3t} is y=c_{1}e^{-t}+c_{2}e^{-3t}-(5e^{-3t}t)/2

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y^{''}+4y^'+3y=5e^{-3t}

y^{^{''}} + 4 y^{^{'}} + 3 y = 5 e^{- 3 t}

y = c_{1} e^{- t} + c_{2} e^{- 3 t} - \frac{5 e ^{- 3 t} t}{2}

Solution steps

y^{''} (t) + 4 y^{'} (t) + 3 y = 5 e^{- 3 t}

Solve linear ODE:

y = c_{1} e^{- t} + c_{2} e^{- 3 t} - \frac{5 e ^{- 3 t} t}{2}

y = c_{1} e^{- t} + c_{2} e^{- 3 t} - \frac{5 e ^{- 3 t} t}{2}

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

(D^{2} - 3 D + 2) (D^{2} + 2 D + 1) y = x

y^{^{''}} - 5 y^{^{'}} + 4 y = 2 t^{2}

x^{^{''}} - x^{^{'}} - 12 x = e^{4 t}

y^{^{''}} - 60 y^{^{'}} - 900 y = 5 e^{10 x}

What is the solution for y^{''}+4y^'+3y=5e^{-3t} ?
The solution for y^{''}+4y^'+3y=5e^{-3t} is y=c_{1}e^{-t}+c_{2}e^{-3t}-(5e^{-3t}t)/2