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

The solution for linear y^{''''}-y=0 is y=c_{1}e^t+c_{2}e^{-t}+c_{3}cos(t)+c_{4}sin(t)

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

linear

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

y = c_{1} e^{t} + c_{2} e^{- t} + c_{3} cos (t) + c_{4} sin (t)

Solution steps

y^{(4)} (t) - y = 0

Solve linear ODE:

y = c_{1} e^{t} + c_{2} e^{- t} + c_{3} cos (t) + c_{4} sin (t)

y = c_{1} e^{t} + c_{2} e^{- t} + c_{3} cos (t) + c_{4} sin (t)

25 x^{^{''}} + 5 x^{^{'}} + 204.167 x = 0, x (0) = 28.577

y^{^{''}} + 2 y^{^{'}} + 5 y = 0, y (0) = 2, y^{^{'}} (0) = 4

\frac{d ^{2} y}{d t ^{2}} - 14 \frac{d y}{d t} + 149 y = 0

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

\frac{1}{8} y^{^{''}} + 16 y = 0

What is the solution for linear y^{''''}-y=0 ?
The solution for linear y^{''''}-y=0 is y=c_{1}e^t+c_{2}e^{-t}+c_{3}cos(t)+c_{4}sin(t)