Question

\[x\frac{dy}{dx} + x \cos^2 \left( \frac{y}{x} \right) = y\]

Answer 1

We have,

\[x\frac{dy}{dx} + x \cos^2 \left( \frac{y}{x} \right) = y\]

\[\frac{dy}{dx} + \cos^2 \left( \frac{y}{x} \right) = \frac{y}{x}\]

\[ \Rightarrow \frac{dy}{dx} = \frac{y}{x} - \cos^2 \left( \frac{y}{x} \right)\]

Putting `y = vx,` we get

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

\[ \therefore v + x\frac{dv}{dx} = v - \cos^2 \left( v \right)\]

\[ \Rightarrow x\frac{dv}{dx} = - \cos^2 v\]

\[ \Rightarrow \sec^2 v\ dv = - \frac{1}{x}dx\]

Integrating both sides, we get

\[\int sec^2 v\ dv = - \int\frac{1}{x}dx\]

\[ \Rightarrow \tan v = - \log x + C\]

\[ \Rightarrow \tan \frac{y}{x} = - \log x + C\]