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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

(D^{2} - 4 D + 13) y = 24 e^{2 x} cos (3 x)

y^{^{''}} - 2 y^{^{'}} + 10 y = 3 e^{x} tan (3 x)

y^{^{''}} + 4 y^{^{'}} + 4 y = 7 e^{- 2 t}

y^{^{''}} - 2 y^{^{'}} + 5 y = - 8 e^{- t}

y^{^{''}} - 5 y = x^{2} - 2 x

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