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richyw Group Title

how do I prove that this initial value problem has infinitly many solutions? \[x'=x^{2/3}\] \[x(0)=0\] Hirsch has an example like it in the practice problems, but there are no solutions to the questions. The chapter doesn't cover this stuff at all!

  • one year ago
  • one year ago

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  1. ZeHanz Group Title
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    A differential equation like this has infinitely many solution curves, but there is only one of them going through (0,0), so it is not entirely clear what you have to prove, IMO. Could you be more specific? If you want I could help you solve the equation.

    • one year ago
  2. SWAG Group Title
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    Agreed^

    • one year ago
  3. ZeHanz Group Title
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    OK, using separation of variables and assuming x is function of t:\[x' =\frac{ dx }{ dt }=x^{ \frac{ 2 }{ 3 }}\]\[x^{-\frac{ 2 }{ 3 }}dx=dt\]Now integrate:\[\int\limits_{}^{} x^{-\frac{ 2 }{ 3 }}dx=\int\limits_{}^{} dt\]Do you know how to do this?

    • one year ago
  4. richyw Group Title
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    I have done that. The questions specifically asks me to prove it has infinitely many solutions

    • one year ago
  5. richyw Group Title
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    I initially thought this was a problem and violated the uniqueness theorem but my textbook has an identical example except with \(x'=x^{1/3}\). Unfortunately it gives me no solution, or no indication at how I should solve it.

    • one year ago
  6. richyw Group Title
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    if you google "hirsch differential equations" the textbook is available for free (second hit i think). It's similar to question 1.12.a

    • one year ago
  7. Hoa Group Title
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    when you take integral of something, the result is something + C. that means you have infinite set of solution. because they are family solution with are combination between a defined solution with ANY C. that what it means

    • one year ago
  8. richyw Group Title
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    does not make sense to me still. I cannot come up with any sort of proof for this. In my last differential equations course be basically assumed that IVP's have a unique solution. So far this is what I have got \[\frac{dx}{dt}=x^{2/3}\]\[x^{-2/3}=dt\]\[x=\frac{(t-c)^3}{27}\]where c is a constant. Now I know that \(x(t=0)=0\) so\[\frac{(-c)^3}{27}=0\]so obviously (0,0) is a solution. Where do the other infinite solutions come from?

    • one year ago
  9. richyw Group Title
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    I have found this as well, need some explaining why this is a proof!

    • one year ago
  10. ZeHanz Group Title
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    @richyw: I came up with the same solutions as you. From the screenshot I understand there is some stitching going on to make it true: There is a constant solution x(t)=0 for every t (the t-axis). Part of this line is stitched to our function, at t=c.

    • one year ago
  11. richyw Group Title
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    well you can't just stich something in. It must prove that the initial statement is true. I can't understand how it is true yet. It's a bit late for me now though. I get so damn frustrated when math profs choose useless books. absolutely worthless. Does anyone know a good online differential equations course that is at the level of Hirsch? I am going to fail my course without some better resources.

    • one year ago
  12. ZeHanz Group Title
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    Initial statement is: Infinitely many functions are solutions of x'=x^(2/3), all with x(0)=0. We construct the following: (it works for every c>0, so that's an infinite amount): For x < c, take x=0 as solution, for x>=c take x=(t-c)³/27. Now the following statements are true, for every c >0: 1. x(0)=0 (we defined it that way) 2. From t = -infty up until t=c, x'=0, which is equal to x^(2/3) (constant solution) 3. If t>=c, x=(t-c)³/27. Now also x'=x^(2/3) (we checked it) 4. for t=c, x'=0, so there is a smooth transition Conclusion: we've found an infinite amount of functions, defined for all real numbers, that satisfy the diff. eq. and the extra condition x(0)=0. Looks like a proof to me.

    • one year ago
  13. ZeHanz Group Title
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    Here is an example solution, where c=4:

    • one year ago
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