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}(-cos(t)+sin(t))+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} (- cos (t) + sin (t)) + e^{- t}

Solution steps

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

Solve linear ODE:

y = e^{- t} (- cos (t) + sin (t)) + e^{- t}

y = e^{- t} (- cos (t) + sin (t)) + e^{- t}

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\frac{d}{d x} (- \frac{11}{x ^{6}})

f (x) = 3 x - 4 x^{\frac{9}{10}}

\int_{2}^{\infty} y e^{- 6 y} d y

y = \frac{π}{3} (l^{2} h - h^{3})

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}(-cos(t)+sin(t))+e^{-t}