Find the particular solution of the differential equation given by x^2 (d y)/(d x)-x y=x^2 cos^2(y/(2 x)), given that when x=1, y=(π)/2.

$$\frac{dy}{dx} = \frac{y}{x} + \cos^2\left(\frac{y}{2x}\right)$$Put $\displaystyle y = vx$ so that $\displaystyle \dfrac{dy}{dx} = v + x\dfrac{dv}{dx}$$$\Rightarrow v + x\frac{dv}{dx} = v + \cos^2\left(\frac{v}{2}\right)$$$$\Rightarrow \int \sec^2\left(\frac{v}{2}\right) dv = \int \frac{1}{x}\, dx$$$$\Rightarrow 2\tan\left(\frac{v}{2}\right) = \log|x| + C$$$$\Rightarrow 2\tan\left(\frac{y}{2x}\right) = \log|x| + C$$$$2\tan\frac{\pi}{4} = \log 1 + C \Rightarrow C = 2 \Rightarrow 2\tan\left(\frac{y}{2x}\right) = \log|x| + 2$$