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Homogeneous equation

  1. Sep 14, 2004 #1
    (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
     
  2. jcsd
  3. Sep 14, 2004 #2
    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]
    [tex] (x^2 + y^2)dx + (2xy)dy = 0 [/tex]


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

    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.



    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.
     
    Last edited: Sep 15, 2004
  4. Sep 15, 2004 #3
    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]
     
  5. Sep 15, 2004 #4
    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
     
  6. Sep 15, 2004 #5
    you have your slashes going the wrong dirrection in the [ \tex ]
     
  7. Sep 15, 2004 #6
    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]
    tries [ tex ] x [ /tex ]*
    [tex] x [/tex]
     
  8. Sep 16, 2004 #7
    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 ) (
     
  9. Sep 16, 2004 #8
    Funny how this thread says otherwise ;)
     
  10. Sep 16, 2004 #9
    sorry, guess i got mixed up. But hey atleast i solved the d.e. right... :bugeye:
     
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