Question

How the heck do I solve this diff. eq?

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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)

Answer 5

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

Answer 6

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!

Answer 7

not yet

Answer 8

only homogenous diff. eq, and substitution

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

how can you tell it's not homogeneous?

Answer 13

\[\large y(1)=0\]

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

ordinary differential equation

Answer 19

no clue

Answer 20

So you haven't learned about homogeneous differential equations?

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

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

Answer 26

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

Answer 27

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

Answer 28

Hope this helps.

Answer 29

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

Answer 30

Yes.