The given differential equation is: $e^x$ tan y dx + $(1-e^x) \sec^2 y$ dy = 0 ⇒ $e^x$ tan y dx = - $(1-e^x) \sec^y$ dy ⇒ $e^x$ tan y dx = $(e^x-1) \sec^y$ dy ⇒ $e^x - (e^x-1)$ dx = $\frac{\sec^2 y}{\tan y}$ dy On integrating on both sides, we get ∫ $e^x - (e^x-1)$ dx = ∫ $\frac{\sec^2 y}{\tan y}$ dy ... (1) Let $I_1$ = ∫ $\frac{\sec^2 y}{\tan y}$ dy Put tany = t ⇒ $\sec^2$ y dy = t ∴ ∫ $\frac{\sec^2 y}{\tan y}$ dy = ∫ $\frac{dt}{t}$ = log |t| = log tan y ... (2) Let $I_2$ = ∫ $\frac{e^x}{e^x-1}$ dx Put $e^x$ - 1 = u ∴ $e^x$ dx = du ∫ $\frac{e^x}{e^x-1}$ dx = ∫ $\frac{du}{u}$ = log u = log $(e^x-1)$ ... (3) From i , ii and iii , we get log tan y = log ($e^x$ - 1) + log C ⇒ log tan y = log C ($e^x$ - 1) ⇒ tan y = C ($e^x$ - 1) The solution of the given differential equation is tan y = C ($e^x$ - 1).