What is the solution for y^{''}+8y^'+24y=0 ?

The solution for y^{''}+8y^'+24y=0 is y=e^{-4t}(c_{1}cos(2^{1+1/2}t)+c_{2}sin(2^{1+1/2}t))

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y^{''}+8y^'+24y=0

y^{^{''}} + 8 y^{^{'}} + 24 y = 0

y = e^{- 4 t} (c_{1} cos (2^{1 + \frac{1}{2}} t) + c_{2} sin (2^{1 + \frac{1}{2}} t))

Solution steps

y^{''} (t) + 8 y^{'} (t) + 24 y = 0

Solve linear ODE:

y = e^{- 4 t} (c_{1} cos (2^{1 + \frac{1}{2}} t) + c_{2} sin (2^{1 + \frac{1}{2}} t))

y = e^{- 4 t} (c_{1} cos (2^{1 + \frac{1}{2}} t) + c_{2} sin (2^{1 + \frac{1}{2}} t))

y^{^{'''}} - 9 y^{^{''}} + 25 y^{^{'}} - 17 y = 0

y^{^{'''}} - 6 y^{^{''}} + 5 y + 12 y = 0

10 x^{^{''}} + 25 x^{^{'}} = 0

x^{^{''}} + 8 x^{^{'}} + 11 x = 0, x (0) = 0, x^{^{'}} (0) = 3

y^{^{''}} - 8 y^{^{''}} - 9 y = 0

What is the solution for y^{''}+8y^'+24y=0 ?
The solution for y^{''}+8y^'+24y=0 is y=e^{-4t}(c_{1}cos(2^{1+1/2}t)+c_{2}sin(2^{1+1/2}t))