What is the solution for y^{''}+2y^'+2y=e^{-1} ?

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

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y^{''}+2y^'+2y=e^{-1}

y^{^{''}} + 2 y^{^{'}} + 2 y = e^{- 1}

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

Solution steps

y^{''} (t) + 2 y^{'} (t) + 2 y = e^{- 1}

Solve linear ODE:

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

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

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