What is the solution for 16y^{''}+522.67y=0 ?

The solution for 16y^{''}+522.67y=0 is y=c_{1}cos((sqrt(52267)t)/(40))+c_{2}sin((sqrt(52267)t)/(40))

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16y^{''}+522.67y=0

16 y^{^{''}} + 522.67 y = 0

y = c_{1} cos (\frac{52267 t}{40}) + c_{2} sin (\frac{52267 t}{40})

Solution steps

16 y^{''} (t) + 522.67 y = 0

Solve linear ODE:

y = c_{1} cos (\frac{52267 t}{40}) + c_{2} sin (\frac{52267 t}{40})

y = c_{1} cos (\frac{52267 t}{40}) + c_{2} sin (\frac{52267 t}{40})

\frac{d ^{2} y}{d t ^{2}} + 64 y = 0

y^{^{'''}} + 6 y^{^{''}} - 37 y^{^{'}} - 90 y = 0

9 y^{^{''}} + y = 0, y (0) = - 2, y^{^{'}} (0) = 1

u, \frac{d ^{2} u}{d x ^{2}} = a u (x)

y^{^{''}} - 7 y^{^{'}} + 8 y = 0, y (0) = 2, y^{^{'}} (0) = 7

What is the solution for 16y^{''}+522.67y=0 ?
The solution for 16y^{''}+522.67y=0 is y=c_{1}cos((sqrt(52267)t)/(40))+c_{2}sin((sqrt(52267)t)/(40))