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Find an explicit or implicit solutions to the differential equation, what now?

  1. Feb 2, 2006 #1
    Hello everyone, yet another obscure problem on web work. No examples like this in the book nor did the professor go over it so i was wondering if someone can let me know what exactly they are wanting me to do!

    Find an explicit or implicit solutions to the differential equation
    http://cwcsrv11.cwc.psu.edu/webwork2_files/tmp/equations/7b/3a599a5966a2ac65a84b6575a645b61.png [Broken]
    http://cwcsrv11.cwc.psu.edu/webwork2_files/tmp/equations/8d/8616ca1dd6ce230911485b98a57fa31.png [Broken] = ?
    What i did was divide through by dx, so i got
    http://cwcsrv11.cwc.psu.edu/webwork2_files/tmp/equations/2c/f2d7bb6b17ac7be8cca5e4969f1a091.png [Broken]y' = 0;
    I let (x^2-4xy) = M, and N = x;
    My = -4x;
    Nx = 1;
    Not exact.

    Did i f this up or what? :cry:
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Feb 2, 2006 #2
    I believe this is homogeneous, try the substitution y=ux

    disregard that

    This is linear, try looking at it this way [tex] (x^2-4xy)=-x\frac{dy}{dx} [/tex]
    Last edited: Feb 2, 2006
  4. Feb 3, 2006 #3
    hey I figured it out, i wasn't usre if this is what u were talking about. But if anyone was cur8ious this is my work:
    http://img430.imageshack.us/img430/2407/lastscan1hu.jpg [Broken]

    http://img229.imageshack.us/img229/1413/lastscan23ch.jpg [Broken]

    Last edited by a moderator: May 2, 2017
  5. Feb 3, 2006 #4
    i see what you did but I just solved it by rewriting it as
    [tex] \frac {dy}{dx}-4y=-x [/tex]

    Then use the integrating factor
    [tex] e^{-4x} [/tex]

    Then multiply through to get

    [tex]e^{-4x} \frac {dy}{dx}-e^{-4x}4y=-e^{-4x}x [/tex]

    Now integrate both sides to get

    [tex]e^{-4x}y=\frac{x}{4}e^{-4x}+\frac{1}{16}e^{-4x}+C [/tex]

    multiply through by


    and I came up with

    [tex]y=\frac{x}{4}+\frac{1}{16}+Ce^{4x} [/tex]

    Just seems like a few less steps than yours...same answer though

    Do you have to use the exact solution method?
  6. Feb 3, 2006 #5
    Wow that was alot easier, this problem was under the Exact Equations problems, so yes he makes us do it this way, but really, it only evaluates the answer so I guess i wasn't forced to do it that way. Thanks!
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