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Differential equations: change of variables

  1. May 1, 2007 #1
    can someone explain 'change of variables' to me? when do you use it, and why? (not to mention HOW you use it!) I'm taking an intro to DE course, and the textbook only mentions the idea in an exercise, not in the text itself. my prof never covered it in class, either.

    for example, i was trying to solve this problem:
    x^2y'' + 5xy' + (4 + pi^2)y = ln x
    y(1) = y(e) = 0

    a solution's manual told me to change variables:
    let t = ln x

    after a few lines, they simplified the original equation into:
    y'' + 4y' + (4 + pi^2)y = t

    can someone explain this to me?
  2. jcsd
  3. May 2, 2007 #2


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    Staff Emeritus
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    The best I can say about WHEN to use "change of variables" is "when it works!"- and you determine that by trying it and seeing if it works! In other words, it is one more tool you can have in your "tool chest".

    As for HOW to do- use the chain rule. If y is a function of x and x is a function of t, then dy/dx= (dy/dt)(dt/dx). To get the second derivative do that again: d2y/dx2= d/dx [(dy)/dt)(dt/dx)].

    In this particular case, t= ln x so dt/dx= 1/x. by the chain rule, dy/dx= (dy/dt)(dt/dx)= (1/x)(dy/dt). Differentiating again, d/dx(dy/dx)= d/dx((1/x)dy/dt). By the product rule, that is d(1/x)/dx(dy/dt)+ (1/x)(d(dy/dt)/dx)= (-1/x2)(dy/dt)+ (1/x)[(1/x)d(dy/dt)/dt)]= (1/x2)[d2y/dt2- dy/dt].
    That is: dy/dx= (1/x)(dy/dt) and
    d2y/dx2= (1/x2)(d2y/dt2- dy/dt). Put those into your differential equation and see what happens.
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