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Homework Help: The Wronskian and diff.equations

  1. Aug 29, 2010 #1

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

    I have differential equation

    [tex]y'' + p(x)\cdot x' + q(x)y = 0[/tex] which have two solutions [tex]y_1(x)[/tex] and [tex]y_2(x)[/tex] where [tex]y_1(x) \neq 0[/tex]

    show that [tex]y_2(t) = y_1(t)\int_{t_0}^{t} \frac{1}{y_1(s)^2} e^{-\int_{t_0}^s p(r) dr} ds[/tex] is also a solution.

    2. Relevant equations

    I know that wronskian for this is defined

    [tex]w(y_1,y_2)(t) = c \cdot e^{-\int p(t) dt}[/tex] according to Abels identity theorem.

    3. The attempt at a solution

    My question do I use this property of theorem and properties of the Wronskian to derive c?
    Last edited: Aug 29, 2010
  2. jcsd
  3. Aug 29, 2010 #2
    Susanne -

    The solution specified for y2 is the general form of a second solution to second order differential equation when the roots to its characteristic equation are repeated. I am not sure what to do when you have functions as coefficients, but perhaps if you can show your second order equation can be reduced to a second order equation with constant coefficients (with repeated roots), then you may assume the solution y2 = F(t)*y1 (F a function to be determined). Twice differentiating y2 and substituting it back into your original second order equation, you should obtain a new equation in terms of F', and F''. Hence you may let g = F' and obtain a first order, separable equation. Solve for g, then back-substitute F' and integrate to solve for F. Finally, since y2 = F(t)*y1, and since F(t) is known, you should get the answer specified for y2. Hope this helps.

    - Sam
  4. Aug 29, 2010 #3
    does this have any relation to reduction of order tecnique ?
  5. Aug 29, 2010 #4
    Yes, I believe the name of the method I described above is reduction of order. Sorry, I forgot its name.
  6. Aug 29, 2010 #5
    Why this isn't in my Calculus book. I don't know :(

    Found this is a different textbook, but thanks ;)

    Found this note here:


    You seem to me more intelligent than me and therefore is the ansatz mentioned

    [tex]y_2(x) = y_1(x) \cdot v(x) [/tex] can this be describe as linearcombination of first solution and the abtr function v(x)??
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