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