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))-e^{-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)) - 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)) - e^{- t}

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

y^{^{''}} - 5 y = x^{2} - 2 x

y^{^{''}} - 3 y^{^{'}} - 4 y = 2 e^{- t}

y^{^{''}} + 8 y^{^{'}} + 2 y = cos (x), y (0) = 0, y^{^{'}} (0) = 1

y^{^{''}} - 2 y^{^{'}} + y = x

y^{^{''}} - 3 y^{^{'}} + 2 y = 4 x + 6

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))-e^{-t}