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Confused what method to use on this Diff EQ! 2nd order

  1. Feb 25, 2006 #1
    ello ello!
    I havn't seen an example of this type of problem. It has another variabler in the equation, i tried to divide by it, but its still in there. Here is the problem:
    Find y as a function of x if
    http://cwcsrv11.cwc.psu.edu/webwork2_files/tmp/equations/ee/448e92b4900790fd8de8a4207ce40d1.png [Broken]
    http://cwcsrv11.cwc.psu.edu/webwork2_files/tmp/equations/c9/0bc438e10b7ec37bb81942786780e41.png [Broken]
    y = ?
    So i divided by x^2 and got:
    y''+2y'/x -30y/x^2 = x;

    This kind of looks like an integrating factor problem but not. ANy suggestions or can someone explain to me whats going on? Thanks! :biggrin:
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Feb 25, 2006 #2
    I've not looked at these type of equations before either but i'll speculate an answer if it's alright with the sire.

    You're told the values of y' and y for when x is 1 so substitute them in with x as 1 to get:

    (1^2)y'' + 2(1x5) - (30x6) = 1
    y'' = 171

    Then integrate to get: y' = 171x + c

    but you're told that this is equal to 5 when x is 1 so that...

    5 = 171 + c
    c = -166
    therefore y' = 171x - 166

    then integrate again and find the constant again and you should get that..

    y = 171(x^2)/2 -166x + 173/2

    Don't hold me to that, i'm just frying logic
  4. Feb 25, 2006 #3
    p.s. i'm new here. Where is everyone getting their maths type?
  5. Feb 25, 2006 #4
    Rooftop, its LaTex.
    And i subbmited that answer and it wasn't right, i have infinately many tries so i'm not worried about submitting wrong answers. Anyone else know?
  6. Feb 25, 2006 #5
    thanks for the heads up...

    btw... you wouldn't happen to be doing some kind of radical calculus crossword puzzle would you?

    edit: wait..... i don't seem to be finding any LaTex. I may need a bigger heads up...
    Last edited: Feb 25, 2006
  7. Feb 25, 2006 #6


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    RooftopDuvet: y"= 171 only when x= 1. y'= 171x+ C only if y"= 171 for all x.

    mr_coffee: You are asking the same kind of questions about the same kinds of problems over and over: This is an "Euler type" equation (also called "equipotential" equation). Remember what I told you about them last time?
  8. Feb 25, 2006 #7
    Sorry Ivey, i post too much! For it to be a Euler Type doesn't i have to "be centerd around" a value?
    like http://tutorial.math.lamar.edu/AllBrowsers/3401/EulerEquations_files/eq0001M.gif [Broken]
    around Xo = 0. These type of differential equations are called Euler Equations. But mine doesn't loook like that because its not homogenous, and it doens't say its centered around anything, do i have to modify it or somthing?

    I've solved Euler Equations beofre like this one:
    http://cwcsrv11.cwc.psu.edu/webwork2_files/tmp/equations/26/a0c710e74525576a0593cd3a5bf7881.png [Broken]
    and got it right, but it was equated to 0, not x
    Last edited by a moderator: May 2, 2017
  9. Feb 25, 2006 #8
    Worst comes to worst you can always use my favorite method: Use the substitution u = ln x. That reduces the problem to a 2nd order linear with constant coefficients.

    Last edited by a moderator: May 2, 2017
  10. Feb 26, 2006 #9
    topsquark, what are u subbing? u = ln x? where do u put that?
  11. Feb 26, 2006 #10


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    Science Advisor

    An "Euler type equation" has x to a power equal to the order of the derivative (that's why it's also called "equipotential"). Your example,
    x2y"+ 2xy'- 30y= x3 is exactly of that kind. As for the substitution, u= ln x, I've also told you about that before.

    If u= ln x, by the chain rule, dy/dx= (du/dx)(dy/du)= (1/x)dy/du and
    d2y/dx2= (-1/x2)dy/du+ (1/x2)d2y/du2. Also x= eu so x3= e3u Your equation becomes
    d2y/du2+ dy/du- 30u= e3u.
    That should be easy.
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