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)+3/4 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{3}{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{3}{4} sin (2 t))

y = e^{- \frac{t}{2}} (- cos (2 t) + \frac{3}{4} sin (2 t))

81 y^{^{''}} + 18 y^{^{'}} + y = 0

16 y^{^{''''}} + 40 y^{^{''}} + 25 y = 0

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

16 y^{^{''}} - 16 y^{^{'}} - 5 y = 0

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

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)+3/4 sin(2t))