What is the solution for y^{''}+2y^'+2y=0,y(0)=1,y^'(0)=1 ?

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

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y^{''}+2y^'+2y=0,y(0)=1,y^'(0)=1

y^{^{''}} + 2 y^{^{'}} + 2 y = 0, y (0) = 1, y^{^{'}} (0) = 1

y = e^{- t} (cos (t) + 2 sin (t))

Solution steps

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

Solve linear ODE:

y = e^{- t} (cos (t) + 2 sin (t))

y = e^{- t} (cos (t) + 2 sin (t))

\frac{d ^{3} y}{d x ^{3}} = \frac{d y}{d x}

16 y^{^{''''}} = 625 y

y^{^{''}} + 64 y = 0, y (0) = \frac{2}{3}, y^{^{'}} (0) = - \frac{4}{3}

(4 D^{3} + 4 D^{2} + D) y = 0

y^{^{''}} + 2 y - 3 y = 0

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