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Analysis graduate school past entrance exam

  1. Jul 30, 2010 #1
    Hello there. I'm studying for graduate school entrance exam (nagoya university), and analysis is part of it. I've learned calculus on my undergraduate course but since I didn't get differential equations, I'm kinda learning it by myself right now.

    The questions are from past problems which are publicly available. I can solve some already (like second order homogenous linear ones with constant coefficients), but I'm stuck on some. I would really appreciate some help, even if only hints or keywords so that I can search more about it myself.

    1. The problem statement, all variables and given/known data

    We are supposed to find the general solution y for the given differential equation below:

    xy'' + y' - x = 0

    However we are to follow the following procedure:

    a) By substituting p = y', express p' in terms of p and x
    b) Make the equation in (a) homogeneous by setting p=xu. Then solve the resulting differential equation and express u in terms of x
    c) Using the result in (b) and the fact that p = xu = y', give the general solution for the original differential equation (express y in terms of x)

    (the original question is in Japanese. This is my best translation attempt)

    2. The attempt at a solution

    (a)

    if p = y', then clearly p' = y''. So

    xy'' + y' - x = 0 implies
    xp' + p - x = 0
    xp' = x - p
    p' = (x - p)/x

    My question here is, isn't the equation homogeneous already? if f(x, p) = (x-p)/x, then clearly f(tx, tp) = f(x, p) (my understanding of the definition of homogeneous)

    (b)

    If we substitute p = xu, then p' = u + xu', so

    p' = (x - p)/x implies
    u + xu' = (x - xu)/x
    u' = (1-2u)/x

    Which isn't homogeneous at all! What I'm I doing wrong here?

    Note that I can solve the differential equation by deviating from the hinted procedure. From

    xp' + p - x = 0

    And the fact that:

    [tex]\frac{d}{dx}(xp) = x\frac{d}{dx}p + p
    [/tex]

    We can write

    [tex]\frac{d}{dx}(xp) - x = 0
    [/tex]

    [tex]\frac{d}{dx}(xp) = x
    [/tex]

    [tex]xp = \frac{1}{2}x^2 + C
    [/tex]

    [tex]p = \frac{1}{2}x + \frac{C}{x}
    [/tex]

    [tex]y' = \frac{1}{2}x + \frac{C}{x}
    [/tex]

    [tex]y = \frac{1}{4}x^2 + C ln|x| + D
    [/tex]

    How do I solve it by following the procedure? Thanks!
     
  2. jcsd
  3. Jul 31, 2010 #2

    lanedance

    User Avatar
    Homework Helper

    there are differenet meanings to homogenous, i've usually used the follwoing for a 2nd order DE, say you have
    [tex] a_2(x)y'' + a_1(x)y' + a_0(x)y = f(x) [/tex]

    then the defintion of homogenous is that f(x) = 0

    this means if you think of the differential equation as an operator, L, it is linear in y, so let
    [tex] \hat{L} = a_2(x)\frac{\partial^2}{\partial y^2} +a_1(x)\frac{\partial}{\partial y} + a_0(x)y [/tex]

    it is linear as
    [tex] \hat{L} (y_1 + y_2) = \hat{L} y_1 + \hat{L} y_2 [/tex]

    which is not the case with the x term

    if you write the original DE as
    [tex] xy'' + y' = x[/tex]

    it is not homgoenous due to the x term,

    after the variable change you have
    [tex] xp' + p = x[/tex]

    which is not homogenous for the same reason...
     
  4. Jul 31, 2010 #3

    lanedance

    User Avatar
    Homework Helper

    so starting from
    [tex] xp' + p = x[/tex]

    then as you say subsititute p = xu, then p' = u + xu', so
    [tex] x(u + xu') + xu = x[/tex]

    which gives
    [tex] x^2 u' + 2xu = x[/tex]

    then
    [tex] xu' + 2u = 1[/tex]

    so by the defintion I gave, this is still not homogenous, but it is very close, if you let u = v+1/2, then u' = v' and you get
    [tex] xv' + 2v = 0[/tex]
     
  5. Jul 31, 2010 #4

    lanedance

    User Avatar
    Homework Helper

    Last edited: Jul 31, 2010
  6. Jul 31, 2010 #5
    Thanks for the reply lanedance. Your explanation from the viewpoint of operator is interesting.

    Anyway I tried following the substitution as is, and after playing around a bit I can transform it into a separable form...

    u' = (1-2u)/x
    du/dx = (1/x)/(1/(1-2u))
    [tex]\frac{1}{x}dx + \frac{1}{2u-1}du = 0[/tex]

    Integrating both sides and performing back substitutions, we will get the same result but with a more tedious process.
     
    Last edited: Jul 31, 2010
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