I am having difficulties rigorously proving the following:
Show that the initial value problem $$y'=e^{-y^2}(y^{2014}-y^{2013}), \quad y(0)=1/2$$ possesses a bounded solution defined on the whole of $\mathbb{R}$.
Context: I have managed to show that the solution is defined on $\mathbb{R}$ by inductively using the existence theorem. I have also sketched out the phase portrait, which allowed me to conjecture that the solution is bounded between $0$ and $1$ and that it is decreasing, since the function $f(x,y)=e^{-y^2}(y^{2014}-y^{2013})$ is nonpositive for all $y\in [0,1]$ and has zeros at $0$ and $1$. But I think my argument still has to be refined a bit.
Any help would be greatly appreciated.
The IVP $$ y'=\mathrm{e}^{-y^2}(y^{2014}-y^{2013}), \quad y(0)=\frac{1}{2} \tag{1} $$ enjoys uniqueness, since its flux function $\,f(x,y)=\mathrm{e}^{-y^2}(y^{2014}-y^{2013})\,$ is $C^1$. (In fact, it is $C^\infty$.) If $\varphi: I\to\mathbb R, \,$ is a solution, where $I$ is an open interval, then $$ 0<\varphi(x)<1, \quad \text{for all $x\in I$}. $$ For if $\varphi(x_0)=1$, for some $x_0\in I$, then $\varphi$ would also satisfy the IVP $$ y'=\mathrm{e}^{-y^2}(y^{2014}-y^{2013}), \quad y(x_0)=1, $$ which possesses as unique solution the $\psi(x)=1$, hence we would have that $\varphi(x)\equiv1$. We can similarly show that $\varphi(x)\ne 0$, for all $x\in I$.
Observe also that $\varphi$ is strictly decreasing.
Next, we use the fact that every IVP, which enjoys uniqueness, possesses a maximally defined solution. (That is, one which can not be extended further as a solution.) Assume $\varphi: I\to\mathbb R$ is the maximally defined solution, and its maximal domain is $I=(a,b)$. If $b<\infty$, let $$ y_+=\lim_{x\to b}\varphi(x). $$ Clearly, as $\varphi$ is decreasing, this limit exists, and $$ 0<\varphi(x)<1 \quad\Longrightarrow\quad 0\le y_+<1. $$ If now $\theta: (b-\eta,b+\eta)\to\mathbb R$ satisfies the IVP $$ y'=\mathrm{e}^{-y^2}(y^{2014}-y^{2013}), \quad y(b)=y_+, $$ then $$ \tilde\varphi(x)=\left\{ \begin{array}{ccc} \varphi(x) & \text{if} & x\in(a,b),\\ \theta(x) & \text{if} & x\in [b,b+\eta),\\ \end{array} \right. $$ is also a solution of (1) which contradict the maximality of $(a,b)$. We act similarly if $a>-\infty$.