Question

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

Answer 1

We have, 
\[x \cos\left( \frac{y}{x} \right)\left( y dx + x dy \right) = y \sin \left( \frac{y}{x} \right)\left( x dy - y dx \right)\]
\[ \Rightarrow xy \cos \left( \frac{y}{x} \right) dx + x^2 \cos \left( \frac{y}{x} \right) dy = xy \sin \left( \frac{y}{x} \right) dy - y^2 \sin \left( \frac{y}{x} \right) dx\]
\[ \Rightarrow \left[ xy \cos \left( \frac{y}{x} \right) + y^2 \sin \left( \frac{y}{x} \right) \right] dx = \left[ xy \sin \left( \frac{y}{x} \right) - x^2 \cos \left( \frac{y}{x} \right) \right] dy\]
\[ \Rightarrow \frac{dy}{dx} = \frac{xy \cos \left( \frac{y}{x} \right) + y^2 \sin \left( \frac{y}{x} \right)}{xy \sin \left( \frac{y}{x} \right) - x^2 \cos \left( \frac{y}{x} \right)}\]
This is a homogeneous differential equation . 
\[\text{ Putting }y = vx\text{ and }\frac{dy}{dx} = v + x\frac{dv}{dx},\text{ we get }\]
\[v + x\frac{dv}{dx} = \frac{v x^2 \cos v + v^2 x^2 \sin v}{v x^2 \sin v - x^2 \cos v}\]
\[ \Rightarrow v + x\frac{dv}{dx} = \frac{v \cos v + v^2 \sin v}{v \sin v - \cos v}\]
\[ \Rightarrow x\frac{dv}{dx} = \frac{v \cos v + v^2 \sin v}{v \sin v - \cos v} - v\]
\[ \Rightarrow x\frac{dv}{dx} = \frac{v \cos v + v^2 \sin v - v^2 \sin v + v \cos v}{v \sin v - \cos v}\]
\[ \Rightarrow x\frac{dv}{dx} = \frac{2v \cos v}{v \sin v - \cos v}\]
\[ \Rightarrow \frac{v\sin v - \cos v}{2 v \cos v}dv = \frac{1}{x}dx\]
Integrating both sides, we get
\[\int\frac{v\sin v - \cos v}{2 v \cos v}dv = \int\frac{1}{x}dx\]
\[ \Rightarrow \int\frac{v\sin v - \cos v}{v \cos v}dv = 2\int\frac{1}{x}dx\]
\[ \Rightarrow \int\frac{v \sin v}{v \cos v}dv - \int\frac{\cos v}{v \cos v}dv = 2\int\frac{1}{x}dx\]
\[ \Rightarrow \int\tan v dv - \int\frac{1}{v}dv = 2\int\frac{1}{x}dx\]
\[ \Rightarrow \log \left| \sec v \right| - \log \left| v \right| = 2 \log \left| x \right| + \log C\]
\[ \Rightarrow \log \left| \frac{\sec v}{v} \right| = \log \left| C x^2 \right|\]
\[ \Rightarrow \frac{\sec v}{v} = C x^2 \]
\[\text{ Putting }v = \frac{y}{x},\text{ we get }\]
\[\sec \left( \frac{y}{x} \right) = \frac{y}{x} \times C \times x^2 \]
\[ \Rightarrow \sec \left( \frac{y}{x} \right) = Cxy\]
\[\text{Hence, }\sec \left( \frac{y}{x} \right) = Cxy\text{ is the required solution.}\]