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)

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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)

Solution steps

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

Solve linear ODE:

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

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

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)