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

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

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

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

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

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

2 y^{^{'}} + y = 3 t

\frac{\partial}{\partial x} (10 x + 5 y)

n = 1 \sum \infty cos (e^{- n})

\int x^{\frac{- 3}{2}} d x

\int 2 x - 8 d x

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