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

The solution for y^{'''}-y=0 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=0

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

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 = 0

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}))

\int \frac{x}{2 + 2 x ^{2}} d x

\frac{1 + x}{4 - x}

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\int e^{c o s (27 t)} sin (27 t) d t

\frac{d}{d x} (\frac{x ^{2} - 8 x + 12}{( x - 4 ) ^{2}})

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