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

The solution for y^{'''}-8y=0 is y=c_{1}e^{2t}+e^{-t}(c_{2}cos(sqrt(3)t)+c_{3}sin(sqrt(3)t))

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

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

y = c_{1} e^{2 t} + e^{- t} (c_{2} cos (3 t) + c_{3} sin (3 t))

Solution steps

y^{'''} (t) - 8 y = 0

Solve linear ODE:

y = c_{1} e^{2 t} + e^{- t} (c_{2} cos (3 t) + c_{3} sin (3 t))

y = c_{1} e^{2 t} + e^{- t} (c_{2} cos (3 t) + c_{3} sin (3 t))

\frac{d}{d x} (ln (2 - 5 x))

\frac{d}{d x} (arctan (x) - x^{2})

\int sin (7 x) ln (sin (7 x)) d x

y^{^{'}} = 2 (1 + y^{2}) x

\frac{d}{d x} (2 x - 1 - \frac{1}{x})

What is the solution for y^{'''}-8y=0 ?
The solution for y^{'''}-8y=0 is y=c_{1}e^{2t}+e^{-t}(c_{2}cos(sqrt(3)t)+c_{3}sin(sqrt(3)t))