What is the solution for 3/4 x^{''}+72x=0,x(0)= 1/3 ,x^'(0)=0 ?

The solution for 3/4 x^{''}+72x=0,x(0)= 1/3 ,x^'(0)=0 is x= 1/3 cos(4sqrt(6)t)

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3/4 x^{''}+72x=0,x(0)= 1/3 ,x^'(0)=0

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

x = \frac{1}{3} cos (46 t)

Solution steps

\frac{3}{4} x^{''} (t) + 72 x = 0

Solve linear ODE:

x = \frac{1}{3} cos (46 t)

x = \frac{1}{3} cos (46 t)

y^{^{''}} - 2 y + y^{^{'}} = 0

y^{^{''}} + 81 y = 0, y (0) = 1, y^{^{'}} (0) = 7

2 y^{^{''}} + 4 y^{^{'}} + 6 y = 0

\frac{d ^{2} y}{d x ^{2}} - 4 \frac{d}{d x} (4) + 4 y = 0

- 0.0375 x^{^{''}} + 0.4535 x^{^{'}} + 3.34 x = 0

What is the solution for 3/4 x^{''}+72x=0,x(0)= 1/3 ,x^'(0)=0 ?
The solution for 3/4 x^{''}+72x=0,x(0)= 1/3 ,x^'(0)=0 is x= 1/3 cos(4sqrt(6)t)