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

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

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

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

y = e^{- t} (c_{1} cos (2 t) + c_{2} sin (2 t)) + \frac{1}{4} e^{- t}

Solution steps

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

Solve linear ODE:

y = e^{- t} (c_{1} cos (2 t) + c_{2} sin (2 t)) + \frac{1}{4} e^{- t}

y = e^{- t} (c_{1} cos (2 t) + c_{2} sin (2 t)) + \frac{1}{4} e^{- t}

x^{^{''}} + 5 x^{^{'}} + 4 x = 5 e^{10 t}

\frac{d ^{2} y}{d x ^{2}} + y = 1

3 y^{^{''}} + 8 y^{^{'}} + 4 y = e^{- 2 x}

y^{^{''}} - 3 y^{^{'}} - 4 y = 3 e^{2 x} + 2 sin (x)

y^{^{''}} + 2 y^{^{'}} - 2 y = x^{2} - e^{x} cos (2 x)

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