What is the solution for y^{''''}+4y=0 ?

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + y^{^{'}} - y = 0, y (- 4) = 7, y^{^{'}} (- 4) = 1

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

y^{^{''''''''}} + 8 y^{^{''''}} + 16 y = 0

y^{^{''}} + y = 0, y (\frac{π}{3}) = 6, y^{^{'}} (\frac{π}{3}) = - 12

4 y^{^{''}} + 17 y^{^{'}} + 4 y = 0

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