What is the solution for 1/16 (d^2x)/(dt^2)=-4x ?

The solution for 1/16 (d^2x)/(dt^2)=-4x is x=c_{1}cos(8t)+c_{2}sin(8t)

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1/16 (d^2x)/(dt^2)=-4x

\frac{1}{16} \frac{d ^{2} x}{d t ^{2}} = - 4 x

x = c_{1} cos (8 t) + c_{2} sin (8 t)

Solution steps

\frac{1}{16} \frac{d ^{2} x}{d t ^{2}} = - 4 x

Solve linear ODE:

x = c_{1} cos (8 t) + c_{2} sin (8 t)

x = c_{1} cos (8 t) + c_{2} sin (8 t)

y^{^{''}} + 16 y = 0, y (\frac{π}{12}) = 2, y^{^{'}} (\frac{π}{12}) = - 4

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

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

y^{^{''}} - 10 y^{^{'}} + 25 y = 0, y (0) = 4, y^{^{'}} (0) = 3

y^{^{''}} + 9 y = 0, y (0) = 5, y^{^{'}} (0) = 2

What is the solution for 1/16 (d^2x)/(dt^2)=-4x ?
The solution for 1/16 (d^2x)/(dt^2)=-4x is x=c_{1}cos(8t)+c_{2}sin(8t)