The given differential equation (D.E) is dydx+2y tan x=sin x
Comparing it with dydx+Py=Q, we have
P=2 tanx and Q=sinx
Now the integral factor (I.F) of this type of D.E is given by
I.F=e∫P.dx=e∫2 tan x dx=e2 log |sec x|=elog sec2x=sec2x
Solution of given differential equation is :
y (I.F)=∫Q (I.F) dx+C
y.sec2x=∫sin x.sec2x dx+C
y.sec2x=∫tan x.secx dx+C
y.sec2x=sec x+C
When x=π3, then y=0
⇒0.sec2π3=secπ3+C
⇒ C=−2
Hence, the required particular solution of the differential equation is
y.sec2x=secx−2