Question

y e^x/y dx = (xe^x/y + y) dy

Answer 1

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