What is the solution for y^{''}-2y^'+2y=e^{-t},y(0)=0,y^'(0)=1 ?

The solution for y^{''}-2y^'+2y=e^{-t},y(0)=0,y^'(0)=1 is y=e^t(-1/5 cos(t)+7/5 sin(t))+1/5 e^{-t}

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y^{''}-2y^'+2y=e^{-t},y(0)=0,y^'(0)=1

y^{^{''}} - 2 y^{^{'}} + 2 y = e^{- t}, y (0) = 0, y^{^{'}} (0) = 1

y = e^{t} (- \frac{1}{5} cos (t) + \frac{7}{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{7}{5} sin (t)) + \frac{1}{5} e^{- t}

y = e^{t} (- \frac{1}{5} cos (t) + \frac{7}{5} sin (t)) + \frac{1}{5} e^{- t}

x^{2 e^{- 42}}

\frac{9}{x}, 2 x, \frac{1}{2} x

\int 4 \frac{cos ( x ) + sin ( 2 x )}{sin ( x )} d x

y = cos^{2} (1 - 3 x^{2})

x \to 0 - lim (e^{\frac{6}{x}})

What is the solution for y^{''}-2y^'+2y=e^{-t},y(0)=0,y^'(0)=1 ?
The solution for y^{''}-2y^'+2y=e^{-t},y(0)=0,y^'(0)=1 is y=e^t(-1/5 cos(t)+7/5 sin(t))+1/5 e^{-t}