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प्रश्न
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उत्तर
We have,
\[x\frac{dy}{dx} - y + x \sin \left( \frac{y}{x} \right) = 0\]
\[ \Rightarrow \frac{dy}{dx} = \frac{y - x \sin \left( \frac{y}{x} \right)}{x}\]
This is a homogenoeus differential equation .
\[\text{ Putting }y = vx\text{ and }\frac{dy}{dx} = v + x\frac{dv}{dx}, \text{ we get }\]
\[v + x\frac{dv}{dx} = \frac{vx - x \sin v}{x}\]
\[ \Rightarrow x\frac{dv}{dx} = v - \sin v - v\]
\[ \Rightarrow x\frac{dv}{dx} = - \sin v\]
\[ \Rightarrow\text{ cosec }v dv = - \frac{1}{x}dx\]
Integrating both sides, we get
\[\int \text{ cosec }v dv = - \int\frac{1}{x}dx\]
\[ \Rightarrow - \int \text{ cosec }v dv = \int\frac{1}{x}dx\]
\[ \Rightarrow - \log \left|\text{ cosec }v - \cot v \right| = \log \left| x \right| + \log C\]
\[ \Rightarrow \log \left| \frac{1}{\text{ cosec }v - \cot v} \right| = \log \left| Cx \right|\]
\[ \Rightarrow \log \left|\text{ cosec }v + \cot v \right| = \log \left| Cx \right|\]
\[ \Rightarrow \log \left| \frac{1 + \cos v}{\sin v} \right| = \log \left| Cx \right|\]
\[ \Rightarrow \frac{1 + \cos v}{\sin v} = Cx\]
\[ \Rightarrow x \sin v = \frac{1}{C}\left( 1 + \cos v \right)\]
\[ \Rightarrow x \sin v = K\left( 1 + \cos v \right) ...........\left(\text{where, }K = \frac{1}{C} \right)\]
\[\text{Putting }v = \frac{y}{x},\text{ we get }\]
\[ \Rightarrow x \sin\left( \frac{y}{x} \right) = K\left[ 1 + \cos\left( \frac{y}{x} \right) \right]\]
\[\text{ Hence, }x \sin\left( \frac{y}{x} \right) = K\left[ 1 + \cos\left( \frac{y}{x} \right) \right]\text{ is the required solution }.\]
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