Question

Solve $xdy=[y+{\cos }^2\left(\frac{y}{x}\right)]dx$

Answer 1

$dxy=[y+{\cos }^2\left(\frac{y}{x}\right)]dx$

$\frac{y}{x}=t$

$y=tx$

$\frac{dy}{dx}=x\frac{dt}{dx}+t-----\left(1\right)$

$\frac{dy}{dx}=\frac{y}{x}+\frac{{\cos }^2\left(\frac{y}{x}\right)}{x}-----\left(2\right)$

$\Rightarrow x\frac{dt}{dx}+t=t+\frac{{\cos }^{2}t}{x}-----from\left(1\right)$

$\Rightarrow x\frac{dt}{dx}=\frac{{\cos }^{2}t}{x}$

$\Rightarrow \frac{dt}{{\cos }^{2}t}=\frac{dx}{x}$

$\Rightarrow {\sec }^{2}tdt=\frac{dx}{x}$

Integrating both sides we get

$\tan t=\left(n\left|x\right|\right)+\left(n\left|c\right|\right)$

$c=$content

$\tan t=\left(n\left|cx\right|\right)$

$t={\tan }^{-1}\left(n\left|cx\right|\right)$

$\frac{y}{x}={\tan }^{-1}\left(n\left|cx\right|\right)$

$y=x{\tan }^{-1}\left(n\left|cx\right|\right)$