Q:dy/dx = (x^2 - 3y^2)/(2xy)
Both sides divide x^2,
dy/dx = [1-3(y/x)^2]/[2(y/x)]
Let u=y/x,then,y=ux
dy/dx = d(ux)/dx
(du/dx)x + u = (1-3u^2)/2u
(du/dx)x = [(1-3u^2)/2u] - u = (1-5u^2)/2u
[2u/(1-5u^2)]du = (1/x)dx
∫[2u/(1-5u^2)]du=∫(1/x)dx
LHS uses substitution,
t=1-5u^2
dt= (-10u)du
=(-1/5)∫(1/t)dt
So,LHS= (-1/5)In(1-5u^2)
RHS= ln(x)+c
In(1-5u^2) = -5In(x)-5c
1-5(y/x)^2 = e^(-5In(x)-5c)
5(y/x)^2 = 1 - [e^(-5In(x))]/e^5c]= 1- [x^(-5)]/[e^5c] = (e^5c-1)/[(e^5c)x^5]
(y/x)^2 = (e^5c-1)/5[(e^5c)x^5]
y^2 = [(e^5c-1)(x^2)]/5[(e^5c)x^5]
y = ±√[(e^5c-1)(x^2)]/5[(e^5c)x^5]
