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Higher Order Differential Equation: Substitution

  1. Nov 6, 2013 #1
    1. The problem statement, all variables and given/known data

    Solve [itex]x^{2}\times y'' - 4 \times x \times y' + 6 \times y = 0[/itex] for [itex]y(x)[/itex] by first using the substitution [itex]v = ln(x)[/itex] to obtain an equation involving [itex]y, dy/dv, d^2y/dv^2[/itex] and no [itex]x[/itex]. Solve for [itex]y(v)[/itex], then return to [itex]y(x)[/itex].

    2. Relevant equations

    NA

    3. The attempt at a solution

    I know how to solve the differential once I substitute in for [itex]v[/itex], but what I'm not getting right is the substitution for the derivatives. I know [itex]dy/dx = dy/dv \times dv/dx = dy/dv \times \frac{1}{x}[/itex], but what I can't figure out is [itex]d^2y/dx^2[/itex]. I got [itex]-dy/dv \times \frac{ln(x)}{x^2}[/itex], but that's not giving me the right answer (I checked w/ Wolfram Alpha). So, I'm a bit stuck on that.
    Any help is appreciated, thanks.
     
  2. jcsd
  3. Nov 7, 2013 #2

    ehild

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    The derivative of a product fg is (fg) ' =f 'g + f g '. (dy/dv) (1/x) is a product of the functions (dy/dv) and 1/x. Apply the chain rule again when you derive dy/dv with respect to x.

    ehild
     
  4. Nov 7, 2013 #3
    Yeah, I think I just got it. My new result for [itex]d^2y/dx^2 = dy^2/dv^2 * \frac{1}{x^2} - dy/dv * \frac{1}{x^2}[/itex]. I think that's right because the diff eq worked out pretty well from there.
     
  5. Nov 7, 2013 #4

    ehild

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    Well done!

    ehild
     
  6. Nov 7, 2013 #5
    Awesome! Thanks a bunch for your help.
     
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