What is the solution for x^{'''}+x=0 ?

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

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x^{'''}+x=0

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

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

Solution steps

x^{'''} (t) + x = 0

Solve linear ODE:

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

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

y^{^{'}} = - sin (π x)

\frac{d}{d x} (ln (- 3 x + 2))

\int_{0}^{- 0.4} 150 x d x

\int x sin (\frac{x}{8}) d x

\int \frac{- 1}{e ^{3 x}} d x

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