What is the solution for y^{''}+2y^'+2y=0,y(0)=-2,y^'(0)=-4 ?

The solution for y^{''}+2y^'+2y=0,y(0)=-2,y^'(0)=-4 is y=e^{-t}(-2cos(t)-6sin(t))

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y^{''}+2y^'+2y=0,y(0)=-2,y^'(0)=-4

y^{^{''}} + 2 y^{^{'}} + 2 y = 0, y (0) = - 2, y^{^{'}} (0) = - 4

y = e^{- t} (- 2 cos (t) - 6 sin (t))

Solution steps

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

Solve linear ODE:

y = e^{- t} (- 2 cos (t) - 6 sin (t))

y = e^{- t} (- 2 cos (t) - 6 sin (t))

(- x + 2) \cdot y^{^{'}} - y = e^{- 2 x}

\frac{d y}{d x} = a (y^{2} + 1)

\frac{d}{d x} (\frac{x ^{2}}{1 - x ^{2}})

\frac{d y}{d x} = (1 + x) (1 + y^{2})

\frac{e ^{- s}}{s - 5}

What is the solution for y^{''}+2y^'+2y=0,y(0)=-2,y^'(0)=-4 ?
The solution for y^{''}+2y^'+2y=0,y(0)=-2,y^'(0)=-4 is y=e^{-t}(-2cos(t)-6sin(t))