Question
यदि $(x - y){e^{x/(x - y)}} = k$, तब
$\log (x - y) + \frac{x}{{x - y}} = \log k$
==> $(x - y) - (x - y)\frac{{dy}}{{dx}} + (x - y) - x + \frac{{dy}}{{dx}} = 0$
==> $y\frac{{dy}}{{dx}} + x = 2y$.
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