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2nd order, linear, homogeneous proof

  1. Oct 9, 2014 #1

    I was wondering if someone could provide either a bit of intuition or a mathematical proof (or both) as to why if the Wronskian (W(f,g)) does not equal 0 for all t in an interval, then the linear combinations of the two functions f and g encompass ALL solutions. Is there any particular reason that this can be known to include all possible solutions for the second order, linear, homogeneous differential equation?

    Any insight would be greatly appreciated!
  2. jcsd
  3. Oct 11, 2014 #2
    Any help at all is appreciated!
  4. Oct 14, 2014 #3
    The equation has two constants of integration, and two possible solutions: y = c1*y1 + c2*y2, where y1(t0) = 1, y1'(t0) = 0, y2(t0) = 0, y2'(t0) = 1 for independent-variable value t = t0. It's easy to show that y1 and y2 have a nonzero Wronskian.

    As to how the Wronskian behaves in general, consider that this differential equation has general form p2*y'' + p1*y'+ p0*y = 0, a form which both f and g satisfy.

    Take the derivative of the Wronskian, W(f,g), and plug in the differential equation's values for f'' and g''. You will find a differential equation for W: W' = (something) * W. You can then solve it for W, though you will have an integral involving p2, p1, and p0. You will now be able to have an idea of what is necessary to make the Wronskian zero if it had been nonzero somewhere.
  5. Oct 15, 2014 #4


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    You are leaving out quite a lot here! For one thing "the linear combinations of the two functions f and g encompass ALL solutions" to what problem?! I presume you are referring to a second order linear differential equation. It can be shown that the set of all solutions to an "nth order" linear differential equation form an "n dimensional" vector space so if you can find n independent solutions, they form a basis for the vector space of all solutions. We use the Wronskian to show that the solutions are independent. For if f and g are solutions to a given second order differential equation then they are "independent" if and only if af(x)+ bg(x)= 0, for all x the a= b= 0. Of af(x)+ bg(x)= 0, for all x, then af'(x)+ bg'(x)= 0. That will have a unique solution, which would obviously be a= b= 0, if and only if the determinant of coefficients
    [tex]\left|\begin{array}{cc}f(x) & g(x) \\ f'(x) & g'(x)\end{array}\right| \ne 0[/tex]

    That is, of course, the Wronskian.
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