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

The solution for y^{''''}+y=0 is

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d y}{d x} = \frac{y ^{2}}{2 x + 3}

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y^{^{''}} - 4 y^{^{'}} + 3 y = t, y (0) = 1, y^{^{'}} (0) = 0

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