How to prove this cauchy problem has a unique solution on entire $\mathbb{R}$? On an old exam, I found the following problem. Given a cauchy problem $y' = e^{-y^2} (y^{2014} - y^{2013})$ and $y(x_0) = y_0$, do the following:
i) prove that there is exactly $C^1$ function on $\mathbb{R}$, which solves this cauchy problem
ii) determine if there exist such $x_0, y_0 \in \mathbb{R}$ and $a,b \in \mathbb{R}$, such that the graph of solution of this cauchy's problem goes through $(a, \frac{1}{2}), (b, \frac{3}{2})$.
I guess I should first of all choose a rectangle around point $(x_0, y_0)$, let's say $[x_0 - a, x_0 + a] \times [y_0 - b, y_0 + b]$ and somehow bound the functions $f(x,y) = e^{-y^2} (y^{2014} - y^{2013})$ and $f_y (x,y) = e^{-y^2} (-2y^{2015} - 2 y^{2014} + 2014 y^{2013} - 2013 y^{2012})$ on this rectangle. The problem is, I can't do it at all; I have no idea how to bound this function. If I could do that, I suppose, I would be able to find out the maximal interval where the solution is well defined, by $(x_0 - \alpha, x_0 + \alpha)$, where $\alpha = \min \{ a, \frac{b}{M}, \frac{1}{L} \}$, where $M$ is the upper bound of this function and $L$ is the upper bound of derivative. After that, my plan would be to try to extend the interval by choosing points moving towards the border of the interval. However, I'm already stuck at trying to bound $f$ and $f_y$, so I would be very happy if anyone can help.
As for the second part, I would also appreciate any hint that could help me approach it (but please don't tell me the solution).
 A: Hint
Question 1
Let $f(y) = e^{-y^2} (y^{2014} - y^{2013})$. As $f$ is locally Lipschitz, the Initial Value Problem $y^\prime(t) = f(y(t))$ and $y(x_0)=y_0$ has a unique maximal solution according to Picard-Lindelöf theorem.
If $y_0 \in \{0,1\}$, then the constant map $y(t)=y_0$ is the unique solution defined on $\mathbb R$.
Otherwise, we can rewrite the equation as
$$\int_{y_0}^{y(x)} \frac{e^{u^2}}{u^{2013}(u-1)} du = x-x_0$$
Now, you can notice that:

*

*If $y_0 \lt 0$ the solution is increasing to $0$ but can't reach $0$ as otherwise we would get two solutions.

*If $0 \lt y_0 \lt 1$ the solution is decreasing to $0$ but can't reach $0$.

*If $y_0 \gt 1$ the solution is increasing.

Finally, use the fact that $\int_{-1}^0 \frac{du}{f(u)}$, $\int_{0}^{1/2} \frac{du}{f(u)}$ and $\int_{2}^\infty \frac{du}{f(u)}$ are all diverging integral to prove that in each of the three cases above, the IVP is defined on $\mathbb R$.
Question 2
If a solution would pass through the two given points, if would exist $c \in \mathbb R$ such that $y(c) = 1$. This would contradict the uniqueness of the IVP.
