Question

Solve the following differential equation:- \[x \cos\left( \frac{y}{x} \right)\frac{dy}{dx} = y \cos\left( \frac{y}{x} \right) + x\]

Answer 1

We have,

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

\[ \Rightarrow \frac{dy}{dx} = \frac{y \cos\left( \frac{y}{x} \right) + x}{x \cos \left( \frac{y}{x} \right)} . . . . . \left( 1 \right)\]

Clearly this is a homogeneous equation,

Putting y = vx

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

\[\text{Substituting }y = vx\text{ and }\frac{dy}{dx} = v + x\frac{dv}{dx}\text{ in (1) we get}\]

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

\[ \Rightarrow v + x\frac{dv}{dx} = \frac{vx \cos \left( v \right) + x}{x \cos \left( v \right)}\]

\[ \Rightarrow v + x\frac{dv}{dx} = \frac{v \cos \left( v \right) + 1}{\cos \left( v \right)}\]

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

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

\[ \Rightarrow x\frac{dv}{dx} = \frac{1}{\cos \left( v \right)}\]

\[ \Rightarrow \cos \left( v \right) dv = \frac{1}{x}dx\]

Integrating both sides, we get

\[\int\cos \left( v \right) dv = \int\frac{1}{x}dx\]

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

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