Question

If the solution of $\frac{16}{\pi }\frac{x}{x^2+y^2}dy=(\frac{y}{x^2+y^2}-1)dx$ satisfies$y\left(0\right)=1$ ,then the value of $\frac{16}{\pi }y\left(\pi /4\right)$ is:

Answer 1

$\frac{x}{x^2+y^2}dy=(\frac{y}{x^2+y^2}-1)dx\Rightarrow \frac{xdy-ydx}{x^2+y^2}=-dx\Rightarrow \frac{d\left(y/x\right)}{1+{\left(y/x\right)}^2}=-dx\Rightarrow {\tan }^{-1}\frac{y}{x}+x=c$
For $x=0\Rightarrow c=\frac{\pi }{2}$
Thus
$\frac{y}{x}=\tan (\frac{\pi }{2}-x)=\cot x\Rightarrow \frac{16}{\pi }y\left(\frac{\pi }{4}\right)=4$