What is the solution for y^{''}-2y^'+2y=e^{-t},y(0)=0,y^'(0)=3 ?

The solution for y^{''}-2y^'+2y=e^{-t},y(0)=0,y^'(0)=3 is y=e^t(-1/5 cos(t)+17/5 sin(t))+1/5 e^{-t}

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y^{''}-2y^'+2y=e^{-t},y(0)=0,y^'(0)=3

y^{^{''}} - 2 y^{^{'}} + 2 y = e^{- t}, y (0) = 0, y^{^{'}} (0) = 3

y = e^{t} (- \frac{1}{5} cos (t) + \frac{17}{5} sin (t)) + \frac{1}{5} e^{- t}

Solution steps

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

Solve linear ODE:

y = e^{t} (- \frac{1}{5} cos (t) + \frac{17}{5} sin (t)) + \frac{1}{5} e^{- t}

y = e^{t} (- \frac{1}{5} cos (t) + \frac{17}{5} sin (t)) + \frac{1}{5} e^{- t}

What is the solution for y^{''}-2y^'+2y=e^{-t},y(0)=0,y^'(0)=3 ?
The solution for y^{''}-2y^'+2y=e^{-t},y(0)=0,y^'(0)=3 is y=e^t(-1/5 cos(t)+17/5 sin(t))+1/5 e^{-t}