# Tricky First Order D.E

• May 19th 2011, 08:23 AM
StaryNight
Tricky First Order D.E
I'm asked to solve the following D.E using an integrating factor:

http://latex.codecogs.com/gif.latex?...%20xdy%20=%200

The problem is the equation is not only dependant on y but also ln(y), so I can't see how the usual method will work. How would I go about solving this?
• May 19th 2011, 08:36 AM
Jester
Let $u = \ln y$ so your ODE becomes $\left(x^2+u\right)dx + xdu = 0$. See how that goes.
• May 19th 2011, 08:37 AM
FernandoRevilla
$\dfrac{1}{M}\left(\dfrac{\partial N}{\partial x}-\dfrac{\partial M}{\partial y}\right)=\ldots=-\dfrac{1}{y}$

This means that the equation has an integrating factor that depends on $y$

Edited: Sorry, I didn't see Danny's post.
• May 19th 2011, 08:54 AM
Ackbeet
When I see a problem like this, and I want to find out what integrating factor there might be, I, with the help of Danny there, have worked out a decently general method. It won't solve every first-order, but it will do many. It works like this: assume an integrating factor of the form

$h(x^{m}y^{n}),$

where h is an unknown function. The idea is that you hope to be able to obtain a differential equation that governs the behavior of h. You multiply the DE by this integrating factor thus:

$h(x^{m}y^{n})\,y(x^{2}+\ln(y))\,dx+h(x^{m}y^{n})\, x\,dy=0.$

You then utilize the conditions for exactness:

$\frac{\partial }{\partial y} \left[h(x^{m}y^{n})\,y(x^{2}+\ln(y))\right]=\frac{\partial }{\partial x} \left[h(x^{m}y^{n})\,x\right].$

So you take your derivatives and boil it all down. The hope is to get a separable DE for h. I get the following:

$\frac{h'(x^{m}y^{n})}{h(x^{m}y^{n})}=-\frac{x^{2}+\ln(y)}{x^{m}y^{n}(nx^{2}+n\,\ln(y)-m)}.$

You have some freedoms here with m and n. I can see that if I choose m = 0, the equation simplifies drastically. You should be able to finish from here.

So here's a decently general method for solving quite a few integrating factor-type nonlinear problems.
• May 19th 2011, 08:58 AM
StaryNight
Thanks a lot, I've got this solved now.