Question

\[\left( 1 + e^{x/y} \right) dx + e^{x/y} \left( 1 - \frac{x}{y} \right) dy = 0\]

Answer 1

We have, 
\[\left( 1 + e^\frac{x}{y} \right) dx + e^\frac{x}{y} \left( 1 - \frac{x}{y} \right) dy = 0\]
\[ \Rightarrow \frac{dx}{dy} = - \frac{e^\frac{x}{y} \left( 1 - \frac{x}{y} \right)}{1 + e^\frac{x}{y}}\]
This is a homogeneous differential equation .
\[\text{ Putting }x = vy \text{ and }\frac{dx}{dy} = v + y\frac{dv}{dy},\text{ we get }\]
\[v + y\frac{dv}{dy} = - \frac{e^v \left( 1 - v \right)}{1 + e^v}\]
\[ \Rightarrow y\frac{dv}{dy} = - \frac{e^v \left( 1 - v \right)}{1 + e^v} - v\]
\[ \Rightarrow y\frac{dv}{dy} = \frac{- e^v + e^v v - v - v e^v}{1 + e^v}\]
\[ \Rightarrow y\frac{dv}{dy} = - \frac{v + e^v}{1 + e^v}\]
\[ \Rightarrow \frac{1 + e^v}{v + e^v}dv = - \frac{1}{y}dy\]
Integrating both sides, we get 
\[\int\frac{1 + e^v}{v + e^v}dv = - \int\frac{1}{y}dy\]
\[ \Rightarrow \log \left| v + e^v \right| = - \log \left| y \right| + \log C\]
\[ \Rightarrow \left| v + e^v \right| = \left| \frac{C}{y} \right|\]
\[ \Rightarrow v + e^v = \frac{C}{y}\]
\[\text{ Putting }v = \frac{x}{y},\text{ we get }\]
\[\frac{x}{y} + e^\frac{x}{y} = \frac{C}{y}\]
\[ \Rightarrow x + y e^\frac{x}{y} = C\]
\[\text{ Hence, }x + y e^\frac{x}{y} = C\text{ is the required solution }.\]