What is the solution for y^{''}+5y^'+4y=-2e^{3t} ?

The solution for y^{''}+5y^'+4y=-2e^{3t} is y=c_{1}e^{-t}+c_{2}e^{-4t}-1/14 e^{3t}

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y^{''}+5y^'+4y=-2e^{3t}

y^{^{''}} + 5 y^{^{'}} + 4 y = - 2 e^{3 t}

y = c_{1} e^{- t} + c_{2} e^{- 4 t} - \frac{1}{14} e^{3 t}

Solution steps

y^{''} (t) + 5 y^{'} (t) + 4 y = - 2 e^{3 t}

Solve linear ODE:

y = c_{1} e^{- t} + c_{2} e^{- 4 t} - \frac{1}{14} e^{3 t}

y = c_{1} e^{- t} + c_{2} e^{- 4 t} - \frac{1}{14} e^{3 t}

y^{^{''}} + 4 y = 8 e^{2 t}

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

y^{^{''}} + 3 y^{^{'}} - 4 y = 7 e^{x}

y^{^{''}} + 25 y = 2 x, y (0) = 1, y^{^{'}} (0) = 5

y^{^{''}} - 10 y^{^{'}} + 25 y = 2 e^{- 3 x}

What is the solution for y^{''}+5y^'+4y=-2e^{3t} ?
The solution for y^{''}+5y^'+4y=-2e^{3t} is y=c_{1}e^{-t}+c_{2}e^{-4t}-1/14 e^{3t}