What is the solution for 16(d^4y)/(dx^4)-y=0 ?

The solution for 16(d^4y)/(dx^4)-y=0 is y=c_{1}e^{x/2}+c_{2}e^{-x/2}+c_{3}cos(x/2)+c_{4}sin(x/2)

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16(d^4y)/(dx^4)-y=0

16 \frac{d ^{4} y}{d x ^{4}} - y = 0

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

Solution steps

16 \frac{d ^{4} y}{d x ^{4}} - y = 0

Solve linear ODE:

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

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

What is the solution for 16(d^4y)/(dx^4)-y=0 ?
The solution for 16(d^4y)/(dx^4)-y=0 is y=c_{1}e^{x/2}+c_{2}e^{-x/2}+c_{3}cos(x/2)+c_{4}sin(x/2)