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}))

y^{^{''}} + 0 y + 64 y = 0, y (0) = 1, y^{^{'}} (0) = 0

\frac{d ^{2} y}{d x ^{2}} - 8 \frac{d y}{d x} = 0

y^{^{''}} + 8 y^{^{'}} + 25 y = 0, y^{^{'}} (0) = 2

\frac{d ^{2} y}{d x ^{2}} + 6 \frac{d y}{d x} + 5 y = 0

y^{^{'''}} + 3 y^{^{''}} - 18 y^{^{'}} - 40 y = 0

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))