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

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

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

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

x = - \frac{1}{4} cos (46 t)

Solution steps

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

Solve linear ODE:

x = - \frac{1}{4} cos (46 t)

x = - \frac{1}{4} cos (46 t)

\frac{1}{16} x^{^{''}} + x^{^{'}} + 4 x = 0

y^{^{''}} - 4 y^{^{'}} - 5 y = 0, y (1) = 0, y (1) = 2

(d^{3} + d^{2} - 9 d - 9) y = 0

x^{^{''}} + 9 x = 0

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

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