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}

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}