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

The solution for y^{''}-2y^'+2y=e^{-t},y(0)=0,y^'(0)=2 is y=e^t(-1/5 cos(t)+12/5 sin(t))+1/5 e^{-t}

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

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

y = e^{t} (- \frac{1}{5} cos (t) + \frac{12}{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{12}{5} sin (t)) + \frac{1}{5} e^{- t}

y = e^{t} (- \frac{1}{5} cos (t) + \frac{12}{5} sin (t)) + \frac{1}{5} e^{- t}

y^{^{''}} - 5 y^{^{'}} - 6 y = 10, y (0) = 2, y^{^{'}} (0) = 1

y^{^{''}} + 8 y^{^{'}} + 16 y = 3 e^{- 4 x}

y^{^{'''}} + 3 y^{^{''}} + 2 y^{^{'}} = x^{2}

y^{^{''}} + y^{^{'}} = sec (t)

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

What is the solution for y^{''}-2y^'+2y=e^{-t},y(0)=0,y^'(0)=2 ?
The solution for y^{''}-2y^'+2y=e^{-t},y(0)=0,y^'(0)=2 is y=e^t(-1/5 cos(t)+12/5 sin(t))+1/5 e^{-t}