The given differential equation is:
y dx−(x+2y2)dy=0
⇒dxdy=x+2y2y
⇒dxdy+(−1y)x=2y ...(i)
This is a linear differential equation of the form
dxdy+Px=Q,where P=−1y and Q=2y
∴I.F.=e∫ Pdy=e∫ −1ydy=elog y−1=1y
So, required solution will be:
x.(I.F) = ∫Q(I.F)dyx.1y=∫ 2 dy+C
(using:x(I.F.) = ∫ Q (I.F.) dy+C)
⇒xy=2y+C
which is the required solution.