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Homework Help: DE question

  1. Nov 24, 2008 #1
    1. The problem statement, all variables and given/known data
    Solve: http://img384.imageshack.us/img384/60/clipboard01az7.jpg [Broken]

    2. Relevant equations
    DE stuff

    3. The attempt at a solution
    I started by doing substitution: x=U+3/2 and y=W-1/2

    So, it gave me http://img522.imageshack.us/img522/3172/clipboard02ly8.jpg [Broken],[/URL] but http://img167.imageshack.us/img167/8297/clipboard03db9.jpg [Broken],[/URL] so http://img175.imageshack.us/img175/3223/clipboard04xt2.jpg [Broken].[/URL]

    Therefore, http://img237.imageshack.us/img237/2264/clipboard05iu8.jpg [Broken]

    And, http://img377.imageshack.us/img377/615/clipboard06ez2.jpg [Broken].[/URL] Substitution z=W/U. Giving, http://img139.imageshack.us/img139/5639/clipboard07hb7.jpg [Broken] (*).

    And, http://img399.imageshack.us/img399/2373/clipboard08qj2.jpg [Broken]

    Rearranging gives http://img114.imageshack.us/img114/1830/clipboard09tm2.jpg [Broken].[/URL] Substitute in (*)

    http://img140.imageshack.us/img140/8541/clipboard10bz2.jpg [Broken] => http://img140.imageshack.us/img140/6706/clipboard11ao0.jpg [Broken] => http://img135.imageshack.us/img135/8163/clipboard12ee3.jpg [Broken]

    Dividing 1/(1+z2) and -z/(1+z2) and integrating gives:

    http://img521.imageshack.us/img521/3638/clipboard13nj5.jpg [Broken].[/URL] But now I'm stuck :( because I can't solve this for z and go back to y and x :( maybe I made an arithmetical mistake or maybe there is a better method?
    Thank you
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Nov 24, 2008 #2
    I don't know how correct the method you detailed is but a lot of times for a differential equation, an implicit solution is sufficient.
  4. Nov 25, 2008 #3
    Surely there is a way to solve it explicitely
  5. Nov 25, 2008 #4


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    Science Advisor

    Why "surely". You don't need to solve
    [tex]arctan z-\frac{1}{2}ln(z^2+ 1)= ln(U)+ C[/tex]

    z= W/U so that is
    [tex]arctan W/U-\frac{1}{2}ln(\frac{W^2}{U^2}+ 1)= ln(U)+ C[/tex]
    and U= x- 3/2, W= y+ 1/2 so
    [tex]arctan\frac{y+ 1/2}{x-3/2}- \frac{1}{2}ln(\frac{(y+1/2)^2}{(x-3/2)^2}= ln(x- 3/2)+ C[/tex]
    That's a perfectly good solution. (Assuming your integration was correct.)
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