(a) Show that all the successive approximations for the problem $y'=y^2$, $y(0) = 1$, exist for all real $x$.

(b) Find a solution of the initial value problem in (a). On what interval does it exist?

(c) Assuming there is just one solution of the problem in (a), indicate why the successive approximations found in (a) can not converge to a solution for all real $x$.

This problem is from Introduction to differential equations by Coddington. He has not yet used Lipschitz continuity and stuff, just the method of successive approximations.

My attempt: Using the method of successive approximations we get a sequence of polynomials and so they are continuous and exist for all $x$. Now in second part the solution i get after solving the ivp is $y=\frac{1}{1-x}$. So according to me the interval on which this solution exists is $(-\infty,1) \cup (1,\infty)$. But the answer is $(-\infty,1)$ and solution ddoes not exist for $x \geq 1$. How did we get this?

I would be really thankful if someone could give a detailed solution since I am very confused.

  • $\begingroup$ what are the difficulties you are facing with this problem? please, add what you tried and the obstacles you encountered. $\endgroup$ – the_candyman Aug 30 '15 at 9:14
  • $\begingroup$ i have made an edit.please have a look $\endgroup$ – RagingBull Aug 30 '15 at 9:18

Note that the question asks on what $\it interval$ does the solution exist and $(-\infty,1)\cup (1,\infty)$ is not an interval. Since your initial condition occurs at $x = 0$ you need to choose the interval that contains $0$, that is, $(-\infty, 1)$. Clearly $y$ loses continuity at $x= 1$, so this is the largest interval you can choose.

  • $\begingroup$ No problem, my pleasure. $\endgroup$ – K. Miller Aug 30 '15 at 14:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.