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

The solution for y^{'''}=-y is y=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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y^{'''}=-y

y^{^{'''}} = - y

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

Solution steps

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

Solve linear ODE:

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

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

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