I get
x*dv=(v^2-1)/(2-v) dx
I integrate both sides and get
x*v=[(v^2-1)/(2-v)]*x +C
using v=y/x
y=...=[y^2-x^2]/(2x^2-xy) + C
now what do I do from here?
you can't integrate like that!
It is a separable equation, meaning, you have to separate the variables. Only x's should be on one side, and only v's on the other. this includes the dx and dv
we reverse the sides and get
(2-v)/(v^2-1) dv = dx/x
integrate and get
1/2[ln(1-v) - 3ln(v+1)]= lnx +C
so
ln[(1-y/x)/(y/x+1)^3]=2lnx+2c
(1-y/x)/[(y/x+1)^3]=x*e^2c
we reverse the sides and get
(2-v)/(v^2-1) dv = dx/x
integrate and get
1/2[ln(1-v) - 3ln(v+1)]= lnx +C
so
ln[(1-y/x)/(y/x+1)^3]=2lnx+2c
(1-y/x)/[(y/x+1)^3]=x*e^2c
what shoud i do from here?
Lets take it from
Then we have
multiply the left side by to get
divide both sides by to get
Plug in your initial data: and solve for . you end up with