What is the solution for 24/32 x^{''}+6x=0,x(0)=0.5,x^'(0)=0 ?

The solution for 24/32 x^{''}+6x=0,x(0)=0.5,x^'(0)=0 is x=0.5cos(2^{1/2+1}t)

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24/32 x^{''}+6x=0,x(0)=0.5,x^'(0)=0

\frac{24}{32} x^{^{''}} + 6 x = 0, x (0) = 0.5, x^{^{'}} (0) = 0

x = 0.5 cos (2^{\frac{1}{2} + 1} t)

Solution steps

\frac{24}{32} x^{''} (t) + 6 x = 0

Solve linear ODE:

x = 0.5 cos (2^{\frac{1}{2} + 1} t)

x = 0.5 cos (2^{\frac{1}{2} + 1} t)

What is the solution for 24/32 x^{''}+6x=0,x(0)=0.5,x^'(0)=0 ?
The solution for 24/32 x^{''}+6x=0,x(0)=0.5,x^'(0)=0 is x=0.5cos(2^{1/2+1}t)