A) \[y=cx\,{{e}^{-x}}\]
B) \[x=cy{{e}^{-x}}\]
C) \[y+{{e}^{x}}=cx\]
D) \[y{{e}^{x}}=cx\]
Correct Answer: D
Solution :
Given \[ydx-xdy=xydx\] Þ \[\frac{ydx-xdy}{xy}=dx\] Þ \[d\left[ \ln \left( \frac{x}{y} \right) \right]=dx\] Integrating both sides, we get \[\ln \frac{x}{y}+\ln c=x\] Þ \[y{{e}^{x}}=cx\].
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