Correct me if I am wrong but I said that 2) was False because for a unique solution to exist: f must be continuous in x and lipschitz in y and here we don't know if it is lipschitz so the answer is F.

But now I have problem for the 3) I guess it's answer 3 since it should be lipschitz in y(x) and continuous in x from the theorem I stated above.

is this the right way to go?

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For 2. the correct answer is "False", but your explanation is not correct. Indeed, if $f$ is Lipschitz, the there exists a unique solution, but there are cases of uniqueness also with $f$ only continuous. Lipschitzianity is a sufficient condition for (existence and) uniqueness.

For 3., as you said, the correct answer is (iii).

  • $\begingroup$ is my reasoning for iii correct? $\endgroup$
    – SoHCahToha
    Jun 2 '17 at 13:14
  • $\begingroup$ Yes, it is. Since the r.h.s. is locally Lipschitz continuous w.r.t. $y$, $y'$ and $y''$, uniformly w.r.t. the $x$ variable, the Cauchy problem admits a unique solution. $\endgroup$
    – Rigel
    Jun 2 '17 at 13:51

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