How to find the general solution of #xy dy/dx= (x^2-1)/(y-1)# ?

Question

4012 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-14T08:56:51

The general solution is #y^2/6(2y-3)=x^2/2-ln(|x|)+C#

The ODE is

#xydy/dx=(x^2-1)/(y-1)#

The variables are separable

#=>#, #y(y-1)dy=(x^2-1)/xdx#

#=>#, #(y^2-y)dy=(x-1/x)dx#

#=>#, #int(y^2-y)dy=int(x-1/x)dx#

#y^3/3-y^2/2=x^2/2-ln(|x|)+C#

#y^2/6(2y-3)=x^2/2-ln(|x|)+C#