What is the solution for 3/4 (d^2x)/(dt^2)+72x=0,x(0)= 1/4 ?

The solution for 3/4 (d^2x)/(dt^2)+72x=0,x(0)= 1/4 is x= 1/4 cos(4sqrt(6)t)+c_{2}sin(4sqrt(6)t)

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3/4 (d^2x)/(dt^2)+72x=0,x(0)= 1/4

\frac{3}{4} \frac{d ^{2} x}{d t ^{2}} + 72 x = 0, x (0) = \frac{1}{4}

x = \frac{1}{4} cos (46 t) + c_{2} sin (46 t)

Solution steps

\frac{3}{4} \frac{d ^{2} x}{d t ^{2}} + 72 x = 0

Solve linear ODE:

x = \frac{1}{4} cos (46 t) + c_{2} sin (46 t)

x = \frac{1}{4} cos (46 t) + c_{2} sin (46 t)

y^{^{'''}} - 11 y^{^{''}} + 28 y^{^{'}} = 0

(d^{4} + 2 d^{2} + 1) y = 0

\frac{d ^{3} y}{d x ^{3}} - \frac{d ^{2} y}{d x ^{2}} = 0

y^{^{'''}} - y^{^{''}} + 25 y^{^{'}} - 25 y = 0

y^{^{''}} - 16 y^{^{'}} - 63 y = 0

What is the solution for 3/4 (d^2x)/(dt^2)+72x=0,x(0)= 1/4 ?
The solution for 3/4 (d^2x)/(dt^2)+72x=0,x(0)= 1/4 is x= 1/4 cos(4sqrt(6)t)+c_{2}sin(4sqrt(6)t)