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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + y^{^{'}} + \frac{5}{4} y = 3 cos (x)

y^{^{''}} + y^{^{'}} - 6 y = 8 + 4 e^{t}

u^{^{''}} - 3 H (t) = 7 cos (t), H (0) = - \frac{3}{4}, u^{^{'}} (0) = 0

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

y^{^{''}} - 8 y^{^{'}} + 7 y = 14

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