What is the solution for 4y^{''}+4y^'+17y=0,y(0)=2 ?

The solution for 4y^{''}+4y^'+17y=0,y(0)=2 is y=e^{-t/2}(2cos(2t)+c_{2}sin(2t))

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4y^{''}+4y^'+17y=0,y(0)=2

4 y^{^{''}} + 4 y^{^{'}} + 17 y = 0, y (0) = 2

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

Solution steps

4 y^{''} (t) + 4 y^{'} (t) + 17 y = 0

Solve linear ODE:

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

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

x^{^{''}} + x^{^{'}} + \frac{1}{2} x = 0

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

y^{^{''}} + 4 y^{^{'}} + 3 y = 0, y (0) = x e^{x}

2 y^{^{''}} - 7 y^{^{'}} + 3 y = 0, y (0) = 0, y^{^{'}} (0) = 2

625 y^{^{''''}} = y

What is the solution for 4y^{''}+4y^'+17y=0,y(0)=2 ?
The solution for 4y^{''}+4y^'+17y=0,y(0)=2 is y=e^{-t/2}(2cos(2t)+c_{2}sin(2t))