Integration by substitution for dy/dx=(x+2y)/(3y-2x)

[Deathfish]

Homework Statement

Original problem is differential equation dy/dx=(x+2y)/(3y-2x)
This is part of solving differential equation.

x(dv/dx) = (1+4v-3v^2)/(3v-2)

so one way of solving, I take out the negative sign

x(dv/dx) = -((3v^2-4v-1)/(3v-2)) , separate and bring over
-∫((3v-2)/(3v^2-4v-1)) dv = ∫1/x dx

which gets me
-(1/2)ln(3v^2-4v-1)=ln(x)+C

however, if i don't take out negative sign,
x(dv/dx) = (1+4v-3v^2)/(3v-2), separate and bring over
∫((3v-2)/(1+4v-3v^2)) = ∫1/x dx

which gets me
-(1/2)ln(-3v^2+4v+1)=ln(x)+C result is opposite sign from above.

checked again and again and i can't seem to find out why even though the working seems ok, the answer is very different...

i get 3y^2-4yx-x^2=C for one method
and x^2+4xy-3y^2=C for the other method

 

[Dick]

It doesn't make any difference whether you take the sign out or not. Eg. 3y^2-4yx-x^2=1 is the same solution as x^2+4xy-3y^2=(-1).