What is the solution for 4y^{''}+4y^'+17y=0,y(0)=-1,y^'(0)=-2 ?

The solution for 4y^{''}+4y^'+17y=0,y(0)=-1,y^'(0)=-2 is y=e^{-t/2}(-cos(2t)-5/4 sin(2t))

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4y^{''}+4y^'+17y=0,y(0)=-1,y^'(0)=-2

4 y^{^{''}} + 4 y^{^{'}} + 17 y = 0, y (0) = - 1, y^{^{'}} (0) = - 2

y = e^{- \frac{t}{2}} (- cos (2 t) - \frac{5}{4} sin (2 t))

Solution steps

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

Solve linear ODE:

y = e^{- \frac{t}{2}} (- cos (2 t) - \frac{5}{4} sin (2 t))

y = e^{- \frac{t}{2}} (- cos (2 t) - \frac{5}{4} sin (2 t))

x^{^{''}} + \frac{x}{a} = 0

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

15 y^{^{''}} + 26 y^{^{'}} + 7 y = 0, y (0) = 8, y^{^{'}} (0) = 1

y^{^{''}} + 3 y^{^{'}} + 2 y = 0, y (0) = 1, y^{^{'}} (0) = - 2

10 y^{^{''''}} + 13 y^{^{''}} + 4 y = 0

What is the solution for 4y^{''}+4y^'+17y=0,y(0)=-1,y^'(0)=-2 ?
The solution for 4y^{''}+4y^'+17y=0,y(0)=-1,y^'(0)=-2 is y=e^{-t/2}(-cos(2t)-5/4 sin(2t))