What is the solution for y^{''}+2y^'+2y=0,y(0)=2,y^'(0)=-3 ?

The solution for y^{''}+2y^'+2y=0,y(0)=2,y^'(0)=-3 is y=e^{-t}(2cos(t)-sin(t))

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y^{''}+2y^'+2y=0,y(0)=2,y^'(0)=-3

y^{^{''}} + 2 y^{^{'}} + 2 y = 0, y (0) = 2, y^{^{'}} (0) = - 3

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 8 y^{^{'}} + 25 y = 0, y (0) = 3

y^{^{'''''}} + y^{^{''''}} + 9 y^{^{'''}} + 9 y^{^{''}} = 0

15 y^{^{''''}} + 92 y^{^{''}} + 64 y = 0

y^{^{''}} + 11 y^{^{'}} + 24 y = 0, y (0) = 1, y^{^{'}} (0) = - 11

10 y^{^{''}} + 9 y^{^{'}} + 2 y = 0, y (0) = 0, y^{^{'}} (0) = 5

What is the solution for y^{''}+2y^'+2y=0,y(0)=2,y^'(0)=-3 ?
The solution for y^{''}+2y^'+2y=0,y(0)=2,y^'(0)=-3 is y=e^{-t}(2cos(t)-sin(t))