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Diff eq problem

  1. Feb 5, 2005 #1
    While I seem to be able to do most of the assigned problem, I often feel like I don't have a clue what I'm doing. I can follow the procedure laid out in the book and in class, but I'm not always sure of what the answer means. I hope if you guys and gals can help me out with this problem, I'll get a better understanding of what's going on.

    The problem says:
    Given that [tex]c1e^x+c2e^{-x}[/tex] is a 2 parameter family of solutions of
    [tex]xy''-y' = 0[/tex] on the interval(-infinity, infinity), show that constants c1 and c2 cannot be found so that a member of the family satisfies the initial conditions y(0) = 0, y'(0) = 1. Explain why this does not violate theorem 4.1

    I guess part of my problem is that I'm not really clear what the difference is between a family of solutions and a unique solution. Is it just the difference between y=x+C and y=x+4 where C=4?

    But back to the problem at hand, given those initial conditions, it looks like
    c1 = 0 and c2 can't be found. Would that be the majority of my answer, take the derivative of the solution and show that c2 can't be found?

    I'm also not sure what they mean, "does not violate theorem 4.1".
    Theorem 4.1 says

    [tex] let a_{n}(x), a_{n-1}(x),.......a_1(x),a_0(x) and g(x)[/tex] be continuous on an interval I and let [tex]a_n(x) not equal 0[/tex] for every x in this interval. If [tex]x = x_0[/tex] is any point in this interval, then a solution y(x) of the IVP exists on the interval and is unique.
  2. jcsd
  3. Feb 5, 2005 #2


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    Homework Helper

    Yes, that's true. y=x+c represents a "general" solution and all the plots (one for each specific value of c) is a group of solutions which is called the "family" of solutions. y=x+4 is a "specific" solution. Not to your equation, just an example we're talking about.

    That's true. You get c1=c2 and c1=-c2 and the only way for that to be true is if c1=c2=0 which gives the NULL solutions which is not desired.

    Looks like the theorem you quoted is incomplete, to me anyway. Can't help on that one.

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