A community for students. Sign up today!
Here's the question you clicked on:
 0 viewing
 one year ago
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
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!

This Question is Closed

ZeHanz
 one year ago
Best ResponseYou've already chosen the best response.2A 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.

ZeHanz
 one year ago
Best ResponseYou've already chosen the best response.2OK, 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?

richyw
 one year ago
Best ResponseYou've already chosen the best response.0I have done that. The questions specifically asks me to prove it has infinitely many solutions

richyw
 one year ago
Best ResponseYou've already chosen the best response.0I 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.

richyw
 one year ago
Best ResponseYou've already chosen the best response.0if you google "hirsch differential equations" the textbook is available for free (second hit i think). It's similar to question 1.12.a

Hoa
 one year ago
Best ResponseYou've already chosen the best response.0when 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

richyw
 one year ago
Best ResponseYou've already chosen the best response.0does 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 \[\frac{dx}{dt}=x^{2/3}\]\[x^{2/3}=dt\]\[x=\frac{(tc)^3}{27}\]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?

richyw
 one year ago
Best ResponseYou've already chosen the best response.0I have found this as well, need some explaining why this is a proof!

ZeHanz
 one year ago
Best ResponseYou've already chosen the best response.2@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 taxis). Part of this line is stitched to our function, at t=c.

richyw
 one year ago
Best ResponseYou've already chosen the best response.0well 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.

ZeHanz
 one year ago
Best ResponseYou've already chosen the best response.2Initial 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=(tc)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=(tc)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.

ZeHanz
 one year ago
Best ResponseYou've already chosen the best response.2Here is an example solution, where c=4:
Ask your own question
Ask a QuestionFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.