3.3.6.6 Example 6 $y^{\prime }=\frac{1-\cos \left(2y\right)}{x^2}$

Solve

$$\begin{array}{rl}{y}^{\prime }& =\frac{1-\cos \left(2y\right)}{x^2}\\ y\left(\mathrm{\infty }\right)& =\frac{10}{3}\pi \end{array}$$

The ode becomes

$$\begin{array}{rl}\int \frac{dy}{1-\cos \left(2y\right)}& =\int \frac{dx}{x^2}\\ -\frac{1}{2\tan y}& =-\frac{1}{x}+c\end{array}$$

Applying IC

$$\begin{array}{rl}-\frac{1}{2\tan \left(\frac{10}{3}\pi \right)}& =c\\ c& =-\frac{1}{2\sqrt{3}}\end{array}$$

Hence solution (1) becomes

$$\begin{array}{rl}-\frac{1}{2\tan y}& =-\frac{1}{x}-\frac{1}{2\sqrt{3}}\\ \cot \left(y\right)& =\frac{2}{x}+\frac{1}{3}\sqrt{3}\end{array}$$

If we want explicit solution then

$$y=\mathrm{arccot}(\frac{2}{x}+\frac{1}{3}\sqrt{3})+n\pi$$

By checking few $n$, it turns out that $n=3$ is the one needed such that IC are satisfied. Hence

$$y=\mathrm{arccot}(\frac{2}{x}+\frac{1}{3}\sqrt{3})+3\pi$$