The solution of the differential equation dy/dx = y tanx , y=1 when x=0 will be-

sec x

The given differential equation is dy/dx = y tanx

⇒ dy/y = tan x dx

Integrating both sides, we get:

∫dy/y = ∫tan x dx

⇒ log y = log (sec x) + log C

⇒ log y = log (C sec x)

⇒  y =  (C sec x)....................(eq. 1)

now y= 1 when x= 0

⇒1 = C × sec(0)

⇒ C = 1

now substituting C =1 in equation 1, we get:

y = sec x