What is the solution for 16y^{''''}=y ?

The solution for 16y^{''''}=y is y=c_{1}e^{t/2}+c_{2}e^{-t/2}+c_{3}cos(t/2)+c_{4}sin(t/2)

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16y^{''''}=y

16 y^{^{''''}} = y

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

Solution steps

16 y^{(4)} (t) = y

Solve linear ODE:

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

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

3 y^{^{''}} + 10 y^{^{'}} - 8 y = 0

y^{^{'''}} + 4 y^{^{''}} - 128 y = 0

y^{^{''''}} - 2 y^{^{'''}} + 4 y^{^{''}} + 2 y^{^{'}} - 5 y = 0

y^{^{''}} - 6 y^{^{'}} + 25 y = 0, y (0) = 2

(D^{3} + 6 D^{2} + 12 D + 8) y = 0

What is the solution for 16y^{''''}=y ?
The solution for 16y^{''''}=y is y=c_{1}e^{t/2}+c_{2}e^{-t/2}+c_{3}cos(t/2)+c_{4}sin(t/2)