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}

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}