What is the solution for y^{''}+2y^'+5y=0,y(0)=2,y(0)=4 ?

The solution for y^{''}+2y^'+5y=0,y(0)=2,y(0)=4 is y=e^{-t}(2cos(2t)+c_{2}sin(2t))

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y^{''}+2y^'+5y=0,y(0)=2,y(0)=4

y^{^{''}} + 2 y^{^{'}} + 5 y = 0, y (0) = 2, y (0) = 4

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

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

2 y^{^{''}} - 7 y^{^{'}} - 4 y = 0

y^{^{''}} - 2 y^{^{'}} + 24 y = 0

35 y^{^{''''}} + 153 y^{^{''}} + 100 y = 0

y^{^{''}} - 8 y^{^{'}} - 9 y = 0, y (0) = 0, y^{^{'}} (0) = 4

(D^{4} + 2 D^{3} + D^{2}) y = 0

What is the solution for y^{''}+2y^'+5y=0,y(0)=2,y(0)=4 ?
The solution for y^{''}+2y^'+5y=0,y(0)=2,y(0)=4 is y=e^{-t}(2cos(2t)+c_{2}sin(2t))