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

The solution for y^{''}+4y^'+5y=e^{-t} is y=e^{-2t}(c_{1}cos(t)+c_{2}sin(t))+1/2 e^{-t}

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 3 y^{^{'}} = e^{x}

x^{^{''}} + 4 x = 4 t

y^{^{''}} + 16 y = g (t), y (0) = 3, y^{^{'}} (0) = 0

\frac{d ^{2} y}{d x ^{2}} - 3 \frac{d y}{d x} + 2 y = x^{2} + 8 x + 9

y^{^{''}} + 4 y^{^{'}} + 4 y = 1

What is the solution for y^{''}+4y^'+5y=e^{-t} ?
The solution for y^{''}+4y^'+5y=e^{-t} is y=e^{-2t}(c_{1}cos(t)+c_{2}sin(t))+1/2 e^{-t}