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Homework Help: Separable FODE

  1. Feb 13, 2010 #1
    1. The problem statement, all variables and given/known data
    The following is an explanation from my tutorial. I do not understand it.

    [tex]{\frac{d}{dx}(y_1{{y_2}'}-y_2{{y_1}'})+P(x)(y_1{{y_2}'}-y_2{{y_1}'})=0[/tex]

    Overlooking for the moment that P(x) may be undefined at certain values of x(so-called-singular points of the equation), we recognize this equation to be a separable first-order differential equation for the function [tex]y_1{{y_2}'}-y_2{{y_1}'}[/tex] that can be integrated to give:

    [tex]y_1{{y_2}'}-y_2{{y_1}'}= C_{12}(exp(-{\int{P(x')dx'}})[/tex]

    [tex]y_1{{y_2}'}-y_2{{y_1}'}= C_{12}\phi[P][/tex]

    Where [tex]C_{12}[/tex] is an integration constant that depends, possibly, on the choices of functions y_1 and y_2, and phi[P] is a functional, that is, a function that depends upon another function, in this case, P(x). The expression on the left hand side is called the Wronskian of the functions y_1 and y_2.

    I do not understand how they separate this equation. Here is what I do. I first separate like this:
    [tex]{\frac{d}{dx}(y_1{{y_2}'}-y_2{{y_1}'})=-P(x)(y_1{{y_2}'}-y_2{{y_1}'})[/tex]

    Then I multiply both sides by dx. Next I divide both sides by [tex]y_1{{y_2}'}-y_2{{y_1}'}[/tex]

    Then I have:

    [tex]{\int{1}d}=-{\int{P(x)dx}[/tex]

    What am I doing wrong here?
     
  2. jcsd
  3. Feb 14, 2010 #2

    rock.freak667

    User Avatar
    Homework Helper

    You can't just divide by y1y2'-y2y1' and have it cancel out from d/dx(y1y2'-y2y1')

    Let's do it your way

    [tex]\frac{d}{dx}(y_1y_2'-y_2y_1')+P(x)(y_1y_2'-y_2y_1')=0[/tex]

    [tex]\Rightarrow \frac{d}{dx}(y_1y_2'-y_2y_1')= -P(x)(y_1y_2'-y_2y_1')[/tex]

    now if we divide by y1y2'-y2y1' we will get


    [tex]\frac{1}{y_1y_2'-y_2y_1'} \frac{d}{dx}(y_1y_2'-y_2y_1') = -P(x)[/tex]

    Now because the term on the left is a tad bit confusing, let's simplify it and let z=y1y2'-y2y1'. So our equation becomes

    [tex]\frac{1}{z} \frac{dz}{dx}= -P(x) \Rightarrow \frac{1}{z} dz = -P(x)dx[/tex]

    now you can integrate both sides. I am sure you can get z isolated on the left side.

    Then replace 'z'.


    Alternatively what you can do is multiply both sides of the original differential equation by an integrating factor of e∫P(x) dx.
     
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