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 t} (6 t^{2})

\frac{d}{d x} ((x + 1)^{2 x})

\int \frac{sec ^{2} ( 2 t )}{5 + 3 tan ( 2 t )} d t

\frac{d}{d x} (\frac{x ^{2} - 3 x}{2})

f (x) = 3 x^{2} - x^{3}, at x = 1

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))