(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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 dont 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

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Integration by substitution

**Physics Forums | Science Articles, Homework Help, Discussion**