- A$y = cx\,{e^{ - x}}$
- B$x = cy{e^{ - x}}$
- C$y + {e^x} = cx$
- ✓$y{e^x} = cx$
==> $\frac{{ydx - xdy}}{{xy}} = dx$ ==> $d\left[ {\ln \left( {\frac{x}{y}} \right)} \right] = dx$
Integrating both sides, we get $\ln \frac{x}{y} + \ln c = x$ ==> $y{e^x} = cx$.
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