## roadjester 4 years ago How the heck do I solve this diff. eq?

$\large ({{y \over x} + e^{-xy}})dx + dy =0$

2. anonymous

First, multiply through by $$xe^{xy}$$:$$(ye^{xy}+x)dx+xe^{xy}dy=0$$Do you know how to solve from here?

It doesn't look seperable, I still have y in the exponent of the dx term

4. anonymous

If only there were some function $$f$$ s.t. $$\frac{\partial f}{\partial x}=ye^{xy}+x$$, $$\frac{\partial f}{\partial y}=xe^{xy}$$... (I didn't solve it using separation of variables)

you did partial differentiation...but I'm not quite sure of what

6. anonymous

Have you studied exact differential equations yet? If we integrate both sides of our original equation, assuming that there is some function $$f$$ whose partial derivatives match whatever we've multiplied by $$dx$$, $$dy$$ respectively, we end up with $$f(x,y)=c$$... now, we merely need to determine what $$f$$ is!

not yet

only homogenous diff. eq, and substitution

9. anonymous

Well, we know it's not homogeneous:$$(\frac{y}x+e^{-xy})dx+dy=0\\\frac{dy}{dx}=-[\frac{y}x+e^{-xy}]\\$$

I get how you got that...but how does writing it as dy/dx help?

11. anonymous

It's to show how it's not homogeneous.

how can you tell it's not homogeneous?

$\large y(1)=0$

14. anonymous

Does $$\frac{ty}{tx}+e^{-(tx)(ty)}=\frac{y}x+e^{-xy}$$?

I don't know? That was an Intial condition that I forgot to type...

16. anonymous

Do you know how to test whether an ODE is homogeneous?

ODE? I'm guesing DE is diff. eq.

18. anonymous

ordinary differential equation

no clue

20. anonymous

So you haven't learned about homogeneous differential equations?

homogeneous yes, ordinary, not a term we've used

22. anonymous

Ordinary is on contrast with partial differential equations, where you have partial derivatives.

you realize that I'm totally confused right? this is only my second class meeting

24. anonymous

:-p ok well tell me what you've been taught so far.

just separating the terms and substitution if the equation can't be solved on the spot

26. anonymous

I don't think it can be solved using either technique. Maybe you need to learn how to solve exact equations first.

27. anonymous

Use the substitution $$u(x)=xy(x).$$ Then $$u^\prime=xy^\prime+y$$. This results in an easy equation for $$u(x).$$

28. anonymous

Hope this helps.

29. anonymous

... are you seriously just posting links to what Mathematica or Wolfram|Alpha spits out?

30. anonymous

Yes.