What is the solution for y^{''}+2y^'+5y=0,y^'(0)=-2 ?

The solution for y^{''}+2y^'+5y=0,y^'(0)=-2 is y=e^{-t}((2+2c_{2})cos(2t)+c_{2}sin(2t))

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

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

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

Solution steps

y^{''} (t) + 2 y^{'} (t) + 5 y = 0

Solve linear ODE:

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

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

x^{^{''}} + 42 x^{^{'}} + 8 x = 0

y^{^{''}} + 2 y^{^{'}} + 9 y = 0, y (0) = 1, y^{^{'}} (0) = 0

y^{^{''}} - 4 y^{^{'}} + 12 y = 0, y (0) = 0, y^{^{'}} (0) = 1

4 y^{^{''}} - 4 y^{^{'}} + 3 y = 0

3 x^{^{''}} - 7 x^{^{'}} + 4 x = 0

What is the solution for y^{''}+2y^'+5y=0,y^'(0)=-2 ?
The solution for y^{''}+2y^'+5y=0,y^'(0)=-2 is y=e^{-t}((2+2c_{2})cos(2t)+c_{2}sin(2t))