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प्रश्न
उत्तर
\[\left( x - y \right) e^\frac{x}{x - y} = a\]
\[\text{ Taking \log on both sides, we get }\]
\[\log\left( x - y \right) + \frac{x}{x - y} = \log a\]
\[ \Rightarrow \frac{1 - \frac{dy}{dx}}{x - y} + \frac{x - y - x\left( 1 - \frac{dy}{dx} \right)}{\left( x - y \right)^2} = 0\]
\[ \Rightarrow \frac{1 - \frac{dy}{dx}}{x - y} + \frac{x\frac{dy}{dx} - y}{\left( x - y \right)^2} = 0\]
\[ \Rightarrow \frac{\left( x - y \right)\left( 1 - \frac{dy}{dx} \right) + x\frac{dy}{dx} - y}{\left( x - y \right)^2} = 0\]
\[ \Rightarrow \frac{x - x\frac{dy}{dx} - y + y\frac{dy}{dx} + x\frac{dy}{dx} - y}{\left( x - y \right)^2} = 0\]
\[ \Rightarrow x - x\frac{dy}{dx} - y + y\frac{dy}{dx} + x\frac{dy}{dx} - y = 0\]
\[ \Rightarrow y\frac{dy}{dx} + x = 2y\]
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