# Differential Equation - should be simple, but?

1. Mar 7, 2007

### QuasarRage

1. The problem statement, all variables and given/known data

I am studying for the exams, and not terribly practiced on DE's yet. From a mock-test I found this DE - it be solved? A complete solution would be appreciated so I can understand the answer and compare it to my notes please...

dy/dx - 2xy = ( y ln (y) ) / x

3. The attempt at a solution

I have attempted so far:

a) Separate variables.
I cannot, I get stuck at something like dy + ( -2XY -(Y ln(y)) ) dx = 0, which I tried to make dy*x/y + (-2x^2 -lny )dx = 0 and cannot find a way to separate further

b) Find a multiplicator (is it how is it called? sorry I am not a native English speaker and not yet very familiar with English mathematics terminology), anyway, something to multiply it with so that it can become directly differentable. I don't know how to go about it at the moment.

c) Find a suitable transformation. No luck...

Thanks for reading, thanks for the solution if you provide it...

Last edited: Mar 7, 2007
2. Mar 7, 2007

### Dick

Write it as:

(x/y)*dy-(2*x^2+ln(y))*dx=0 and think about it again. "Multiplicator"="Integrating factor".

3. Mar 7, 2007

### QuasarRage

Thanks for the answer. I actually reached that point, but unless I am a complete dummy, it still is not an exact DE.

....M(x,y).............N(x,y)
(2x^2 + lny)dx + (-x/y)dy = 0

dM/dy = 1/y != dN/dx = -1/y so it is not exact.

.So how do I tackle the minus? Is it still a matter of finding a correct integrating factor? Please take it a step further for me, this is actually the point I am stuck at, it is so frustrating I cannot describe it

Last edited: Mar 7, 2007
4. Mar 7, 2007

### Dick

You are right. I missed the sign. Back to the drawing board.

5. Mar 7, 2007

### mybsaccownt

i hope it's not bad form to re-use my post from a similar type of question:

from this point:

yes, you can convert it to exact by trying:

Last edited: Mar 7, 2007
6. Mar 8, 2007

### dextercioby

SO you have the equation

$$\frac{dy}{dx}=2xy+\frac{y}{x}\ln y$$

Make the substitution
$$y(x)=e^{p(x)}$$

The resulting ODE is

$$\frac{dp}{dx}=\frac{p}{x}+2x$$

For $x\neq 0$, we get

$$\frac{1}{x}\frac{dp}{dx}-\frac{p}{x^2}=2$$

which is the same as

$$\frac{d}{dx}\left(\frac{p(x)}{x}\right)=2$$

The rest is trivial.

7. Mar 8, 2007

### QuasarRage

Thank you very much dick, mybsaccownt and dextercioby for the help, I have succeded in solving it both ways proposed (without looking at your solutions while I was doing it ofc :) )

My solution was: -2x + y + (lny)/x = c

Thanks again very much for your help.

For dextercioby's solution, what hinted us to this transformation pls? Was it the presence of lny in the equation?

Last edited: Mar 8, 2007
8. Mar 9, 2007

### dextercioby

Yes, of course, that way, i got rid of a nasty function of "y".