roadjester
  • roadjester
How the heck do I solve this diff. eq?
Differential Equations
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
roadjester
  • roadjester
\[\large ({{y \over x} + e^{-xy}})dx + dy =0\]
anonymous
  • anonymous
First, multiply through by \(xe^{xy}\):$$(ye^{xy}+x)dx+xe^{xy}dy=0$$Do you know how to solve from here?
roadjester
  • roadjester
It doesn't look seperable, I still have y in the exponent of the dx term

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • 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)
roadjester
  • roadjester
you did partial differentiation...but I'm not quite sure of what
anonymous
  • 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!
roadjester
  • roadjester
not yet
roadjester
  • roadjester
only homogenous diff. eq, and substitution
anonymous
  • anonymous
Well, we know it's not homogeneous:$$(\frac{y}x+e^{-xy})dx+dy=0\\\frac{dy}{dx}=-[\frac{y}x+e^{-xy}]\\$$
roadjester
  • roadjester
I get how you got that...but how does writing it as dy/dx help?
anonymous
  • anonymous
It's to show how it's not homogeneous.
roadjester
  • roadjester
how can you tell it's not homogeneous?
roadjester
  • roadjester
\[\large y(1)=0\]
anonymous
  • anonymous
Does \(\frac{ty}{tx}+e^{-(tx)(ty)}=\frac{y}x+e^{-xy}\)?
roadjester
  • roadjester
I don't know? That was an Intial condition that I forgot to type...
anonymous
  • anonymous
Do you know how to test whether an ODE is homogeneous?
roadjester
  • roadjester
ODE? I'm guesing DE is diff. eq.
anonymous
  • anonymous
ordinary differential equation
roadjester
  • roadjester
no clue
anonymous
  • anonymous
So you haven't learned about homogeneous differential equations?
roadjester
  • roadjester
homogeneous yes, ordinary, not a term we've used
anonymous
  • anonymous
Ordinary is on contrast with partial differential equations, where you have partial derivatives.
roadjester
  • roadjester
you realize that I'm totally confused right? this is only my second class meeting
anonymous
  • anonymous
:-p ok well tell me what you've been taught so far.
roadjester
  • roadjester
just separating the terms and substitution if the equation can't be solved on the spot
anonymous
  • anonymous
I don't think it can be solved using either technique. Maybe you need to learn how to solve exact equations first.
anonymous
  • anonymous
Use the substitution \(u(x)=xy(x).\) Then \(u^\prime=xy^\prime+y\). This results in an easy equation for \(u(x).\)
anonymous
  • anonymous
Hope this helps.
1 Attachment
anonymous
  • anonymous
... are you seriously just posting links to what Mathematica or Wolfram|Alpha spits out?
anonymous
  • anonymous
Yes.

Looking for something else?

Not the answer you are looking for? Search for more explanations.