Solve\begin {align*} y^{\prime } & =\frac {1-\cos \left ( 2y\right ) }{x^{2}}\\ y\left ( \infty \right ) & =\frac {10}{3}\pi \end {align*}
The ode becomes\begin {align*} \int \frac {dy}{1-\cos \left ( 2y\right ) } & =\int \frac {dx}{x^{2}}\\ -\frac {1}{2\tan y} & =-\frac {1}{x}+c \end {align*}
Applying IC\begin {align*} -\frac {1}{2\tan \left ( \frac {10}{3}\pi \right ) } & =c\\ c & =-\frac {1}{2\sqrt {3}} \end {align*}
Hence solution (1) becomes\begin {align*} -\frac {1}{2\tan y} & =-\frac {1}{x}-\frac {1}{2\sqrt {3}}\\ \cot \left ( y\right ) & =\frac {2}{x}+\frac {1}{3}\sqrt {3} \end {align*}
If we want explicit solution then \[ y=\operatorname {arccot}\left ( \frac {2}{x}+\frac {1}{3}\sqrt {3}\right ) +n\pi \] By checking few \(n\), it turns out that \(n=3\) is the one needed such that IC are satisfied. Hence\[ y=\operatorname {arccot}\left ( \frac {2}{x}+\frac {1}{3}\sqrt {3}\right ) +3\pi \]