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Homework Help: VERY simple question regarding substitution

  1. Sep 27, 2010 #1
    1. The problem statement, all variables and given/known data

    I'm not even going to get to the real problem. I'm just having a basic mental block with how to do the substitution. I just need to know how to convert this ODE into terms of [tex]z[/tex]

    2. Relevant equations

    [tex]x^2y''+xy'+4(x^4-1)y = 0[/tex]
    [tex]x^2 = z[/tex]

    3. The attempt at a solution

    I have some vague idea. I think I'm supposed to use the chain rule. This is rather embarrassing in the first place.

    [tex]\frac{dy}{dx} = \frac{dy}{dz}\frac{dz}{dx}[/tex]

    I don't know where to go from here. I've been looking around at some of the chain rule instructions that are out there online, but I have not been able to link it to what I need to do here. In fact, the chain rule in general has been a large weakness for me, and I'm surprised that I have gotten to graduate classes and am still mixing it up so frequently.
    Last edited: Sep 27, 2010
  2. jcsd
  3. Sep 27, 2010 #2


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    well remember that y'=dy/dx = (dy/dz)(dz/dx)

    so you need to use the product rule on y'=(dy/dz)(dz/dx) to get y''
  4. Sep 27, 2010 #3
    With that in mind, does [tex]y(x)[/tex] become a function of [tex]z[/tex]? Here's my attempt:

    [tex]y' = y'(z) \cdot 2x[/tex]
    [tex]y'' = y''(z) \cdot 2x + 2y'(z)[/tex]
  5. Sep 27, 2010 #4


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    Well y would not become a function of z it seems and Wolfram Alpha is giving me a solution that contains the Bessel functions.
  6. Sep 27, 2010 #5
    If it's not a function of Z, then what is it? See, this is where I am getting confused.

    As I said before, I don't need help in actually solving the Bessel Function part of it (I already know how to do that, and how to derive the Bessel Functions). I'm hung up on the substitution.

    I've always stumbled on this aspect of how to apply the chain rule. I keep mixing up things. If anyone can provide assistance to just stepping me the way through this Calculus 1 topic I should have known quite well 6 years ago, that would be greatly appreciated. I'm clearly looking at this basic concept in the wrong way, and I'm sure that a short explanation would set me straight from here on out for all future substitution problems.
  7. Sep 27, 2010 #6


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    Are you sure you need to use a substitution?

    I think you can just make into an Euler-Cauchy equation and then solve for the particular integral.

    http://en.wikipedia.org/wiki/Cauchy–Euler_equation" [Broken]
    Last edited by a moderator: May 4, 2017
  8. Sep 28, 2010 #7


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    [tex]\frac{dy}{dx} = \frac{dy}{dz}\frac{dz}{dx}[/tex]
    Yes, and since [itex]z=x^2[/itex], [itex]dz/dx= 2x[/itex] so that [itex]dy/dx= 2x dy/dz[/itex].

    Then [itex]d^2y/dx^2= d/dx(dy/dx)= d/dx(2x dy/dz)[/itex][itex]= 2 dy/dz+ 2x(d/dx(dy/dx)= 2 dy/dx+ 4x^2 d^2y/dz^2[/itex]

    Putting those into your equation,
    [tex]x^2 d^2y/dx^2+ x dy/dx+ (x^4- 1)y= x^2(2 dy/dz+ 4x^2 d^2y/dz^2)+ x(2x dy/dz)+ (x^4- 1)y[/tex]
    [tex]= 4x^4 d^2y/dz^2+ 3x^2 dy/dz+ (x^4- 1)y[/tex]

    Of course, [itex]x^2= z[/itex] and [itex]x^4= z^2[/itex] so that becomes
    [tex]4z^2 d^2y/dz^2+ 3z dy/dz+ (z^2- 1)y= 0[/tex]
  9. Sep 28, 2010 #8
    Thank you very much for the reply. This is starting to make more sense. However, the final equation, after substitution, does not seem to effectively put the equation into the same form as Bessel's differential equation.

    I suppose, I actually have to ask now! How would the coefficients be manipulated so it gets into the form of

    [tex]x^2y'' + xy' + (x^2-\nu^2)y = 0[/tex]
  10. Sep 28, 2010 #9
    Thanks HallsofIvy. I managed to get the answer. I think, after plugging in the substitution, the final differential equation is slightly different, and does indeed match Bessel's differential equation, except in terms of z.

    I got:

    [tex]z^2 \ddot y + z\dot y + (z^2-1)y = 0[/tex]

    In terms of z, I get the following:

    [tex]y(z) = c_1J_1(z) + c_2Y_1(z)[/tex]

    Substituting back x:

    [tex]y(z) = c_1x^2J_1(x) + c_2x^2Y_1(x)[/tex]

    Mathematica returns something like [tex]2c_2x^2Y_1(x)[/tex] for the second term, which doesn't make sense, since the 2 is redundant, given it's simply linearly dependent on the regular [tex]c_2x^2Y_1(x)[/tex]
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