$e^x$ tan y dx + $(2-e^x)\sec^2$ y dy ∫ $\frac{e^x}{e^x-2}$ dx = ∫ $\frac{\sec^2 y}{\tan y}$ dy now, ∫ $\frac{f'(x)}{f(x)}$ dx = log |f (x)| + c log $\left|(e^x-2)\right|$ - log |tan y| + log |c| = 0 log $\left|\frac{c(e^x-2)}{\tan y}\right|$ = 0 $\frac{c(e^x-2)}{\tan y}$ = $e^0$ = 1 tan y = (c ($e^x$ - 2)) now, x = 0 , y = $\frac{\pi}{4}$ 1 = (x (1 - 2)) c = - 1 so tan y = (- 1 ($e^x$ - 2)) y = $\tan^{-1}$ (2 - $e^x$) OR $\frac{dy}{dx}$ + 2y tan x = sin x comparing with the standard form $\frac{dy}{dx}$ + Py = Q P = 2 tan x , Q = sin x Now, I.F. = $e^{\int P dx}$ = $e^{\int 2 \tan x dx}$ = $e^{2 \log |\sec x|}$ = $\sec^2$ x y $\sec^2$ x = sec x + c 0 = 2 + c ⇒ c = - 2 (Since when x = $\frac{\pi}{3}$ , y = 0) y $\sec^2$ x = sec x - 2