Question

Find the general solution of the differential equation \[x \cos \left( \frac{y}{x} \right)\frac{dy}{dx} = y \cos\left( \frac{y}{x} \right) + x .\]

Answer 1

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

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

\[\text { This is a homogeneous differential equation } . \]

\[\text { Putting }y = vx and \frac{dy}{dx} = v + x\frac{dv}{dx}, \text { we get }\]

\[v + x\frac{dv}{dx} = \frac{vx \cos v + x}{x \cos v}\]

\[\Rightarrow v + x\frac{dv}{dx} = \frac{v \cos v + 1}{\cos v}\]

\[ \Rightarrow x\frac{dv}{dx} = \frac{v \cos v + 1 - v \cos v}{\cos v}\]

\[ \Rightarrow x\frac{dv}{dx} = \frac{1}{\cos v}\]

\[ \Rightarrow \cos v dv = \frac{1}{x}dx\]

\[\text { Integrating both sides, we get }\]

\[\int\cos v \ dv = \int\frac{1}{x}dx\]

\[ \Rightarrow \sin v = \log \left| x \right| + \log\left| C \right|\]

\[\text { Putting v }= \frac{y}{x}, we get\]

\[\sin\frac{y}{x} = \log \left| Cx \right|\]

\[\text { which is the general solution of the given differential equation } .\]