What is the solution for 5/8 (d^2x)/(dt^2)+40x=0 ?

The solution for 5/8 (d^2x)/(dt^2)+40x=0 is x=c_{1}cos(8t)+c_{2}sin(8t)

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5/8 (d^2x)/(dt^2)+40x=0

\frac{5}{8} \frac{d ^{2} x}{d t ^{2}} + 40 x = 0

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

Solution steps

\frac{5}{8} \frac{d ^{2} x}{d t ^{2}} + 40 x = 0

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^{^{''''}} - 12 y^{^{''}} + 36 y = 0

y^{^{''}} + 1 y^{^{'}} + 0 y = 0

y^{^{''}} + k (1 + k) y = 0

y^{^{''}} + y^{^{'}} - 30 y = 0, y (0) = 1, y^{^{'}} (0) = 38

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

What is the solution for 5/8 (d^2x)/(dt^2)+40x=0 ?
The solution for 5/8 (d^2x)/(dt^2)+40x=0 is x=c_{1}cos(8t)+c_{2}sin(8t)