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roadjesterBest ResponseYou've already chosen the best response.0
\[\large ({{y \over x} + e^{xy}})dx + dy =0\]
 one year ago

oldrin.batakuBest ResponseYou've already chosen the best response.0
First, multiply through by \(xe^{xy}\):$$(ye^{xy}+x)dx+xe^{xy}dy=0$$Do you know how to solve from here?
 one year ago

roadjesterBest ResponseYou've already chosen the best response.0
It doesn't look seperable, I still have y in the exponent of the dx term
 one year ago

oldrin.batakuBest ResponseYou've already chosen the best response.0
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)
 one year ago

roadjesterBest ResponseYou've already chosen the best response.0
you did partial differentiation...but I'm not quite sure of what
 one year ago

oldrin.batakuBest ResponseYou've already chosen the best response.0
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!
 one year ago

roadjesterBest ResponseYou've already chosen the best response.0
only homogenous diff. eq, and substitution
 one year ago

oldrin.batakuBest ResponseYou've already chosen the best response.0
Well, we know it's not homogeneous:$$(\frac{y}x+e^{xy})dx+dy=0\\\frac{dy}{dx}=[\frac{y}x+e^{xy}]\\$$
 one year ago

roadjesterBest ResponseYou've already chosen the best response.0
I get how you got that...but how does writing it as dy/dx help?
 one year ago

oldrin.batakuBest ResponseYou've already chosen the best response.0
It's to show how it's not homogeneous.
 one year ago

roadjesterBest ResponseYou've already chosen the best response.0
how can you tell it's not homogeneous?
 one year ago

oldrin.batakuBest ResponseYou've already chosen the best response.0
Does \(\frac{ty}{tx}+e^{(tx)(ty)}=\frac{y}x+e^{xy}\)?
 one year ago

roadjesterBest ResponseYou've already chosen the best response.0
I don't know? That was an Intial condition that I forgot to type...
 one year ago

oldrin.batakuBest ResponseYou've already chosen the best response.0
Do you know how to test whether an ODE is homogeneous?
 one year ago

roadjesterBest ResponseYou've already chosen the best response.0
ODE? I'm guesing DE is diff. eq.
 one year ago

oldrin.batakuBest ResponseYou've already chosen the best response.0
ordinary differential equation
 one year ago

oldrin.batakuBest ResponseYou've already chosen the best response.0
So you haven't learned about homogeneous differential equations?
 one year ago

roadjesterBest ResponseYou've already chosen the best response.0
homogeneous yes, ordinary, not a term we've used
 one year ago

oldrin.batakuBest ResponseYou've already chosen the best response.0
Ordinary is on contrast with partial differential equations, where you have partial derivatives.
 one year ago

roadjesterBest ResponseYou've already chosen the best response.0
you realize that I'm totally confused right? this is only my second class meeting
 one year ago

oldrin.batakuBest ResponseYou've already chosen the best response.0
:p ok well tell me what you've been taught so far.
 one year ago

roadjesterBest ResponseYou've already chosen the best response.0
just separating the terms and substitution if the equation can't be solved on the spot
 one year ago

oldrin.batakuBest ResponseYou've already chosen the best response.0
I don't think it can be solved using either technique. Maybe you need to learn how to solve exact equations first.
 one year ago

ljensenBest ResponseYou've already chosen the best response.0
Use the substitution \(u(x)=xy(x).\) Then \(u^\prime=xy^\prime+y\). This results in an easy equation for \(u(x).\)
 one year ago

oldrin.batakuBest ResponseYou've already chosen the best response.0
... are you seriously just posting links to what Mathematica or WolframAlpha spits out?
 one year ago
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