Find the equation of a curve whose gradient is $\frac{dy}{dx}=\frac{y}{x}-{\cos }^2\frac{y}{x}\text{ where }x>0,y>0$ and which passes through the point $(1,\pi /4)$

$\text{ Put }y=vx\text{, then }\frac{dy}{dx}=v+x\frac{dv}{dx}$

$\Rightarrow v+x\frac{dv}{dx}=v-{\cos }^{2}v\Rightarrow x\frac{dv}{dx}=-{\cos }^{2}v$

$\Rightarrow \int {\sec }^{2}vdv=-\int \frac{dx}{x}\Rightarrow \tan v=-\log \left|x\right|+c$

$\Rightarrow \tan \frac{y}{x}+\log \left|x\right|=c$

$(1,\pi /4)\text{ lies on the curve }$

$\Rightarrow \tan \frac{\pi }{4}+\log 1=c\Rightarrow c=1$

The required curve is $\tan \frac{y}{x}+\log \left|x\right|=1$