Discussion
Consider the initial value problem (IVP)
\(y' = e^{-y^{2}} + 1,\ y(0) = 0.\)
I. IVP has a unique solution on \(\mathbb{R}\).
II. Every solution of IVP is bounded on its maximal interval of existence.
Which one of the following is correct?
Answer: B
I is TRUE and II is FALSE.
I: \(F(y) = e^{-y^{2}} + 1\) has bounded derivative \(F'(y) = -2ye^{-y^{2}}\) (its maximum absolute value is \(\sqrt{2/e}\)), so \(F\) is globally Lipschitz. By Picard-Lindelof the IVP has a unique solution defined on all of \(\mathbb{R}\).
II: \(y' = e^{-y^{2}} + 1 \ge 1\), so \(y(x) \ge x\) for \(x \ge 0\) and \(y(x) \le x\) for \(x \le 0\). The solution is unbounded on its maximal interval \(\mathbb{R}\). Options calling II true, or I false, are wrong.
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