# Thread: using the homogeneous equation...

1. ## using the homogeneous equation...

ok so here is the problem...

(x^2+ xy)dx - (2xy + y^2)dy = 0 let y = ux

ok the problem seems easy enough... if y = ux then dy = udx + xdu

so now im ready to substitute...

when i solved it i got something like

ln|x| = (2u^2 + u)/(-u^3-2u^2+u+1)

so where did i go wrong...

(x^2 +ux^2)dx - [(2ux^2 + (ux)^2)](udx + xdu) from here i used algebra to simplify?

2. You started out just fine. With both substitutions $\displaystyle y=ux$ and $\displaystyle x=vy,$ you get a horrendous integration that looks like

$\displaystyle \displaystyle-\int\frac{u(2+u)}{1+u-2u^{2}-u^{3}}\,du$ or

$\displaystyle \displaystyle\int\frac{v(v+1)}{v^{3}+v^{2}-2v-1}\,dv.$

Both integrals are pretty nasty. Not really sure what to do from here. Ideas, anyone else?

3. Originally Posted by slapmaxwell1
ln|x| = (2u^2 + u)/(-u^3-2u^2+u+1)

Check the numerator. It should be:

$\displaystyle \ln \left |{x}\right |=\displaystyle\int \dfrac{u^2+2u}{1+u-2u^2-u^3}\;du$

Fernando Revilla

Edited: Sorry, I didn't see Ackbeet's post

4. i tried i couldnt finish the problem on my exam, so thats what i left..i even ran it through maxima, and maxima wouldnt integrate it...its the only problem that gave me fits...thanks for checking my work.

5. Mathematica Version 4 and WolframAlpha won't integrate it, either. You're not alone! You're probably good if you reduce the DE to integrals - but that's about as far as you can go with this one, I think, unless there's a clever substitution or an integrating factor to get the equation exact. It's not linear or Bernoulli in x or y.

6. Maybe General can help us out. He's elected not to allow visitor messages or PM's. We'll just have to hope he sees this thread. That fairly likely, though.

7. i will redo the problem and try to make it exact...that just might work. thanks for the tip..