# Theorem for existence of a solution and existence of a unique solution.

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?

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.
• 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. Jun 2 '17 at 13:51