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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

- 2 (x^{2} + 3 x)

2 t^{2} - 2 t + 1

\frac{d y}{d t} = \frac{2 y}{t}

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

f (x) = e^{- x + 1} x - 2 e^{- x + 1}

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