Question

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

 

Answer 1

We have,

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