Question

\[\frac{y}{x}\cos\left( \frac{y}{x} \right) dx - \left\{ \frac{x}{y}\sin\left( \frac{y}{x} \right) + \cos\left( \frac{y}{x} \right) \right\} dy = 0\]

Answer 1

\[\frac{y}{x}\cos \left( \frac{y}{x} \right)dx - \left\{ \frac{x}{y}\sin \left( \frac{y}{x} \right) + \cos \left( \frac{y}{x} \right) \right\}dy = 0\]
\[ \Rightarrow \left\{ \frac{x}{y}\sin \left( \frac{y}{x} \right) + \cos \left( \frac{y}{x} \right) \right\}dy = \frac{y}{x}\cos \left( \frac{y}{x} \right)dx\]
\[ \Rightarrow \frac{dy}{dx} = \frac{\frac{y}{x}\cos \left( \frac{y}{x} \right)}{\frac{x}{y}\sin \left( \frac{y}{x} \right) + \cos \left( \frac{y}{x} \right)}\]
This is a homogeneous differential equation . 
\[\text{ Putting }y = vx\text{ and }\frac{dy}{dx} = v + x\frac{dv}{dx},\text{ we get }\]
\[v + x\frac{dv}{dx} = \frac{v \cos v}{\frac{1}{v}\sin v + \cos v}\]
\[ \Rightarrow x\frac{dv}{dx} = \frac{v \cos v}{\frac{1}{v}\sin v + \cos v} - v\]
\[ \Rightarrow x\frac{dv}{dx} = \frac{- \sin v}{\frac{1}{v}\sin v + \cos v}\]
\[ \Rightarrow \left( \frac{\frac{1}{v}\sin v + \cos v}{\sin v} \right)dv = - \frac{1}{x}dx\]
\[ \Rightarrow \left( \frac{1}{v} + \cot v \right)dv = - \frac{1}{x}dx\]
Integrating both sides, we get
\[\int\left( \frac{1}{v} + \cot v \right)dv = - \int\frac{1}{x}dx\]
\[ \Rightarrow \int\frac{1}{v}dv + \int \cot v dv = - \int\frac{1}{x}dx\]
\[ \Rightarrow \log \left| v \right| + \log \left| \sin v \right| = - \log \left| x \right| + \log C\]
\[ \Rightarrow \log \left| vx\sin v \right| = \log C\]
\[ \Rightarrow \left| v x \sin v \right| = C\]
\[\text{ Putting }v = \frac{y}{x},\text{ we get }\]
\[ \Rightarrow \left| y\sin \frac{y}{x} \right| = C\]
\[\text{ Hence, }\left| y\sin \frac{y}{x} \right| = C\text{ is the required solution }.\]