What is the solution for y^{''}+2y^'+5y=8e^{-t} ?

The solution for y^{''}+2y^'+5y=8e^{-t} is y=e^{-t}(c_{1}cos(2t)+c_{2}sin(2t))+2e^{-t}

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 4 y^{^{'}} + 5 y = 35 e - 4 x, y (0) = - 5, y^{^{'}} (0) = 1

y^{^{''}} - 2 y^{^{'}} = 5 t^{4}, y (0) = 1, y^{^{'}} (0) = 0

y^{^{'''}} - 4 y^{^{''}} + 5 y^{^{'}} - 2 y = 5 e^{- x} + sin (x)

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

y^{^{''}} - 4 y^{^{'}} + 4 y = (x - 1) e^{2 x}

What is the solution for y^{''}+2y^'+5y=8e^{-t} ?
The solution for y^{''}+2y^'+5y=8e^{-t} is y=e^{-t}(c_{1}cos(2t)+c_{2}sin(2t))+2e^{-t}