What is the solution for y^{''}+4y^'+17/4 y=0 ?

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

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y^{''}+4y^'+17/4 y=0

y^{^{''}} + 4 y^{^{'}} + \frac{17}{4} y = 0

y = e^{- 2 t} (c_{1} cos (\frac{t}{2}) + c_{2} sin (\frac{t}{2}))

Solution steps

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

Solve linear ODE:

y = e^{- 2 t} (c_{1} cos (\frac{t}{2}) + c_{2} sin (\frac{t}{2}))

y = e^{- 2 t} (c_{1} cos (\frac{t}{2}) + c_{2} sin (\frac{t}{2}))

\frac{\partial}{\partial x} (1 - cos (x))

\int cos (\frac{π}{2}) d x

- 19.6 t^{2} + 9 t + 3

\frac{d}{d x} (40320)

f (x) = \frac{x ^{3}}{6}

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