ex tan y dx + (2−ex)sec2 y dy
∫

ex

(ex−2)

dx = ∫

sec2y

tany

dy
now, ∫

f′(x)

f(x)

dx = log |f (x)| + c
log |(ex−2)| - log |tan y| + log |c| = 0
log |

c(ex−2)

tany

| = 0

c(ex−2)

tany

= e0 = 1
tan y = (c (ex - 2))
now, x = 0 , y =

π

4

1 = (x (1 - 2))
c = - 1
so
tan y = (- 1 (ex - 2))
y = tan−1 (2 - ex)
OR

dy

dx

+ 2y tan x = sin x
comparing with the standard form

dy

dx

+ Py = Q
P = 2 tan x , Q = sin x
Now,
I.F. = e∫Pdx = e∫2tanxdx = e2‌log|secx| = sec2 x
y sec2 x = sec x + c
0 = 2 + c ⇒ c = - 2 (Since when x =

π

3

, y = 0)
y sec2 x = sec x - 2