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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d ^{2} y}{d x ^{2}} - 25 \cdot y = 0

x^{^{''}} = - x^{^{'}} + 3 x

y^{^{'''}} - 7 y^{^{''}} + 14 y^{^{'}} - 8 y = 0

y^{^{''}} + 2 y^{^{'}} + y = - y^{^{''}} + 5 y + 2 y^{^{'}} - y

\frac{d ^{2} y}{d x ^{2}} + k \frac{d y}{d x} - π y = 0

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