What is the solution for the homogeneous equation (x^2 + y^2)dx + (2xy)dy = 0?

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Discussion Overview

The discussion revolves around solving the homogeneous differential equation (x^2 + y^2)dx + (2xy)dy = 0. Participants explore various methods for finding solutions, including exact differential equations and different approaches to expressing the solution.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant proposes a solution of y = sqrt((kx^5 + x^2)/3) with k defined in terms of c, but expresses confusion over the teacher's solution.
  • Another participant suggests that the method of exact differential equations is relevant, noting that their own application of this method yields a different result.
  • A third participant identifies the equation as an exact homogeneous differential equation and outlines a process to find the solution, arriving at xy^2 + (x^3)/3 = C.
  • Several participants discuss the proper formatting for LaTeX, indicating issues with displaying equations correctly in the forum.
  • One participant acknowledges that their solution can be verified by implicit differentiation, reinforcing its correctness.

Areas of Agreement / Disagreement

There is no consensus on the solution to the differential equation, as multiple participants propose different answers and methods. The discussion reflects a range of interpretations and approaches without resolving which is correct.

Contextual Notes

Participants express uncertainty regarding the formatting of LaTeX in the forum, which may affect the clarity of their mathematical expressions. Additionally, there are unresolved aspects regarding the exactness of the differential equation and the implications of the proposed solutions.

amb123
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(x^2 + y^2)dx + (2xy)dy = 0

I get y = sqrt((kx^5 + x^2)/3) Where k = c2 cubed, and c2 = ln(c) so k = 3ln(c)

But, the answer the teacher gave is (x^2)(y^3) - x - ln(y) = c I can't come up with anything remotely close. I know this isn't in a pretty LaTeX form, but I am new and haven't figured it out yet. Also, I'm not sure how to get my computer to read LaTeX, so if there is a program I need to d/l, can someone link me?

Any help would be great !
A
 
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just see the sticky thread on using tex basically you just put tex *forward slash* tex in square brackets around
around equations and use *backward slash* sqrt instead of sqrt

QUOTE -
[tex]x [\tex]<br /> [tex](x^2 + y^2)dx + (2xy)dy = 0[/tex]<br /> <br /> <br /> I get [tex]y = \sqrt((kx^5 + x^2)/3)[/tex] Where k = c2 cubed, and c2 = ln(c) so k = 3ln(c)<br /> <br /> But, the answer the teacher gave is [tex](x^2)(y^3) - x - ln(y) = c[/tex] I can't come up with anything remotely close. <br /> <br /> <br /> <br /> to answer your question I think it uses the method of exact differential equations just look it up on the web SOS would be a good start. However when I Use exact differential equations I get a totally different answer but it works when I sub things back in.[/tex]
 
Last edited:
Thats funny I have not had a problem with latex before. I will insert spaces to show the simple things I sometimes type and it still does not work

[ tex ] x [ /tex ]
[tex]x[/tex]
 
i get a different answer from both of you

(x^2 + y^2)dx + (2xy)dy = 0
Is an exact homogenous d.e.

So you take the anti-partial derivative in respect to y of your dy term:
xy^2 + h(x).
this is your solution, but you need to figure out what that left over function of x is

to do that you first take the partial of that in respect to x
y^2 + h’(x).

you know this function must be equal to your dx term since the d.e. is exact. So…

y^2 + h’(x) = x^2 + y^2
h’(x) = x^2
anti-differentiate
h(x) = (x^3)/3 + C

so your solution is
xy^2 + (x^3)/3 = C
 
you have your slashes going the wrong dirrection in the [ \tex ]
 
That was the answer I got but did not mention. It is right because you can implicitly differentiate and get back to the original differential equation.

I thought you used backslashes in latex but to go into tex mode the standard VB code is to use a forward slash the same way as bold font colour ect.

*tries [ tex ] x [ \tex]*
[tex]x [\tex]<br /> tries [ tex ] x [ /tex ]*<br /> [tex]x[/tex][/tex]
 
The spaces were there so it didn't think I was doing latex and not show you what I was trying to convey. Proper usage is (tex) (\tex) with ] [ instead of ) (
 
Proper usage is (tex) (\tex) with ] [ instead of ) (

Funny how this thread says otherwise ;)
 
sorry, guess i got mixed up. But hey atleast i solved the d.e. right... :bugeye:
 

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