The solution of the differential equation dy/dx = y tanx , y=1 when x=0 will be-
Answer & explanation
Correct answer: option 2
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