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))

\frac{d}{d x} (3 e^{x} + 8 x^{2} ln^{2} (x))

f (x) = \frac{5}{x ^{11}}

\frac{1}{s ^{2} + 4 s + 13}

\frac{d}{d x} (e^{12 x})

\int \frac{1}{2 x ^{2} + 5 x - 7} d x

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))