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

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

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

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

y = e^{t} (c_{1} cos (t) + c_{2} sin (t)) + \frac{1}{5} e^{- t}

Solution steps

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

Solve linear ODE:

y = e^{t} (c_{1} cos (t) + c_{2} sin (t)) + \frac{1}{5} e^{- t}

y = e^{t} (c_{1} cos (t) + c_{2} sin (t)) + \frac{1}{5} e^{- t}

y^{^{''}} - y^{^{'}} + 196 y = 14 sin (14 t)

y^{^{''}} + 2 y^{^{'}} + 5 y = 6 sin (2 x) + 7 cos (2 x)

(D^{2} - 3 D + 2) y = 2 x^{3} - 9 x^{2} + 6 x

y^{^{''}} + 6 y^{^{'}} + 13 y = 7 cos (3 t + \frac{π}{4})

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

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