How to solve this differential equation?

1. Oct 2, 2011

1. The problem statement, all variables and given/known data
(xy+2xyln^2y+ylny)dx + (2x^2lny + x)dy = 0

3. The attempt at a solution

well, I've tried my best to solve it and I've filled almost two papers trying to solve it with introducing new variables and substituting and then plugging them in but It hasn't gotten solved yet. any ideas would be appreciated.

2. Oct 2, 2011

snipez90

Try to turn it into an exact differential equation?

3. Oct 3, 2011

I've tried my best to do that already. Do you know any integrating factors or substitutions to turn it into an exact differential equation?

4. Oct 3, 2011

NewtonianAlch

Divide through by dx?

You would be looking for an integrating factor if it was a first-order linear ODE, I think.

5. Oct 3, 2011

NewtonianAlch

Also if you do partial differentiation with both sides (y,x), they are not exact.

6. Oct 3, 2011

I checked the answers at the end of the book and it gives a hint to take xlny=t. the final answer should be 2x^2 + (2xlny+1)^2 = c and x=0.

any ideas?

7. Oct 4, 2011

any ideas?

8. Oct 4, 2011

Dickfore

Is the equation:

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

9. Oct 4, 2011

Dickfore

If this is the general solution, then differentiating it, gives:
$$4 x \, dx + 2 (2 \, x \, \ln{y} + 1) \, 2 \left( \ln{y} \, dx + \frac{x}{y} \, dy \right) = 0$$
and now the arbitrary constant $C$ is gone. Canceling a common factor of 4 and multiplying by $y$ to get rid of the fractions, we get:
$$x \, y \, dx + (2 \, x \, \ln{y} + 1) (y \, \ln{y} \, dx + x \, dy) = 0$$

Multiply out and collect the terms in front of $dx$ and $d y$ to see if you get the same equation as you quoted.