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

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

roadjester
 2 years ago
Best ResponseYou've already chosen the best response.0It doesn't look seperable, I still have y in the exponent of the dx term

oldrin.bataku
 2 years ago
Best ResponseYou've already chosen the best response.0If 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)

roadjester
 2 years ago
Best ResponseYou've already chosen the best response.0you did partial differentiation...but I'm not quite sure of what

oldrin.bataku
 2 years ago
Best ResponseYou've already chosen the best response.0Have 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!

roadjester
 2 years ago
Best ResponseYou've already chosen the best response.0only homogenous diff. eq, and substitution

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

roadjester
 2 years ago
Best ResponseYou've already chosen the best response.0I get how you got that...but how does writing it as dy/dx help?

oldrin.bataku
 2 years ago
Best ResponseYou've already chosen the best response.0It's to show how it's not homogeneous.

roadjester
 2 years ago
Best ResponseYou've already chosen the best response.0how can you tell it's not homogeneous?

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

roadjester
 2 years ago
Best ResponseYou've already chosen the best response.0I don't know? That was an Intial condition that I forgot to type...

oldrin.bataku
 2 years ago
Best ResponseYou've already chosen the best response.0Do you know how to test whether an ODE is homogeneous?

roadjester
 2 years ago
Best ResponseYou've already chosen the best response.0ODE? I'm guesing DE is diff. eq.

oldrin.bataku
 2 years ago
Best ResponseYou've already chosen the best response.0ordinary differential equation

oldrin.bataku
 2 years ago
Best ResponseYou've already chosen the best response.0So you haven't learned about homogeneous differential equations?

roadjester
 2 years ago
Best ResponseYou've already chosen the best response.0homogeneous yes, ordinary, not a term we've used

oldrin.bataku
 2 years ago
Best ResponseYou've already chosen the best response.0Ordinary is on contrast with partial differential equations, where you have partial derivatives.

roadjester
 2 years ago
Best ResponseYou've already chosen the best response.0you realize that I'm totally confused right? this is only my second class meeting

oldrin.bataku
 2 years ago
Best ResponseYou've already chosen the best response.0:p ok well tell me what you've been taught so far.

roadjester
 2 years ago
Best ResponseYou've already chosen the best response.0just separating the terms and substitution if the equation can't be solved on the spot

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

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

oldrin.bataku
 2 years ago
Best ResponseYou've already chosen the best response.0... are you seriously just posting links to what Mathematica or WolframAlpha spits out?
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