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प्रश्न
\[x\frac{dy}{dx} + x \cos^2 \left( \frac{y}{x} \right) = y\]
उत्तर
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
\[\int sec^2 v\ dv = - \int\frac{1}{x}dx\]
\[ \Rightarrow \tan v = - \log x + C\]
\[ \Rightarrow \tan \frac{y}{x} = - \log x + C\]
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