What is the solution for y^{''}+4y^'+24y=0 ?

The solution for y^{''}+4y^'+24y=0 is y=e^{-2t}(c_{1}cos(2sqrt(5)t)+c_{2}sin(2sqrt(5)t))

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y^{''}+4y^'+24y=0

y^{^{''}} + 4 y^{^{'}} + 24 y = 0

y = e^{- 2 t} (c_{1} cos (25 t) + c_{2} sin (25 t))

Solution steps

y^{''} (t) + 4 y^{'} (t) + 24 y = 0

Solve linear ODE:

y = e^{- 2 t} (c_{1} cos (25 t) + c_{2} sin (25 t))

y = e^{- 2 t} (c_{1} cos (25 t) + c_{2} sin (25 t))

x \to 1 lim (x^{2} + 5 x - 6)

\frac{\partial}{\partial y} (x sin (y))

\frac{d}{d x} (\frac{( 1 + x )}{( 1 - x )})

\frac{\partial}{\partial x} (5 x y (15 - x - y))

15 e^{x^{2}}

What is the solution for y^{''}+4y^'+24y=0 ?
The solution for y^{''}+4y^'+24y=0 is y=e^{-2t}(c_{1}cos(2sqrt(5)t)+c_{2}sin(2sqrt(5)t))