How the heck do I solve this diff. eq?

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How the heck do I solve this diff. eq?

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

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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
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
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?
It's to show how it's not homogeneous.
how can you tell it's not homogeneous?
\[\large y(1)=0\]
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...
Do you know how to test whether an ODE is homogeneous?
ODE? I'm guesing DE is diff. eq.
ordinary differential equation
no clue
So you haven't learned about homogeneous differential equations?
homogeneous yes, ordinary, not a term we've used
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
:-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
I don't think it can be solved using either technique. Maybe you need to learn how to solve exact equations first.
Use the substitution \(u(x)=xy(x).\) Then \(u^\prime=xy^\prime+y\). This results in an easy equation for \(u(x).\)
Hope this helps.
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... are you seriously just posting links to what Mathematica or Wolfram|Alpha spits out?
Yes.

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