What is the solution for y^{''}+4y^'+29y=0,y(0)=4 ?

The solution for y^{''}+4y^'+29y=0,y(0)=4 is y=e^{-2t}(4cos(5t)+c_{2}sin(5t))

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y^{''}+4y^'+29y=0,y(0)=4

y^{^{''}} + 4 y^{^{'}} + 29 y = 0, y (0) = 4

y = e^{- 2 t} (4 cos (5 t) + c_{2} sin (5 t))

Solution steps

y^{''} (t) + 4 y^{'} (t) + 29 y = 0

Solve linear ODE:

y = e^{- 2 t} (4 cos (5 t) + c_{2} sin (5 t))

y = e^{- 2 t} (4 cos (5 t) + c_{2} sin (5 t))

2 y^{^{''}} + 5 y^{^{'}} - 3 y = 0, y (0) = 22

y^{^{''}} + \frac{1}{r} y^{^{'}} = 0

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

y^{^{''}} = 49 y

y^{^{''}} - 5 y^{^{'}} + 6 y = 0, y (0) = 1, y^{^{'}} (0) = \frac{7}{4}

What is the solution for y^{''}+4y^'+29y=0,y(0)=4 ?
The solution for y^{''}+4y^'+29y=0,y(0)=4 is y=e^{-2t}(4cos(5t)+c_{2}sin(5t))