Proving statement about the ODE $y'=y+y^4$ 
Is there a solution to the problem $$\left\{\begin{matrix}
y'=y+y^4\\ 
y(x_0)=y_0
\end{matrix}\right.$$
  which is defined on $\mathbb{R}$? ($x_0,y_0$ might be any real numbers)

It's easy to prove that for all $(x,y)\in\mathbb{R}^2$ there exists an open interval $I$ (with $x_0\in I$) where the problem has a unique solution. However is the maximal interval always $(-\infty,\infty)$? I know that the answer is no, but that's just because I found a solution for particular values of $x_0$ and $y_0$ and checked its domain. But is there a way to prove that $I$ need not be $I=(-\infty,\infty)$ without actually solving the problem for a certain initial condition? In other words, how can I prove that the unique solution need not be defined on $\mathbb{R}$?
 A: actual solutions, three regions: $y > 0,$ then $0 > y > -1,$ then $y < -1.$ I should emphasize that your ODE is autonomous, no explicit dependence on $x,$ which means that every solution curve is a sideways translate of one of those in the picture. It is sort of accidental that a solution for $y > 0$ also provides the exact correct formula for a solution with $y < -1.$ That is why there are two curves in light green, with matching vertical asymptote. 

it is first order, so solutions cannot intersect or cross each other. There are two constant solutions, $y=0$ and $y=-1.$ Between those, solution curves  descend from horizontal asymptote at $y=0$ to that at $y=-1.$
Above $0,$ $y$ increases. Once, say, $y > 2,$ we have $y' > 1 + y^2$ and $y > \tan (x - x_0).$ Blowup in finite time; there is a vertical asymptote. 
Similar for $y < -1$ if you run time backwards. Very roughly, the plot of a large number of solution curves can be rotated $180^\circ$ to give something  similar to the original. 
Oh, well. You can also change variables to get good estimates on when $y$ goes to $\infty.$ We are free to move right or left, all solution curves are translates of each other. In considering $y > 0,$ we are free to demand $y(0) = 1.$ Next, take
$$ y = \tan w, $$ so $w(0) = \pi/4$ and
$$ y' = \sec^2 w \; w' $$
From $y' = y + y^4, $ I got
$$ w' = \cos w \sin w + \frac{\sin^4 w}{\cos^2 w}.  $$
For $w$ between $\pi/4$ and $\pi/2,$ we know $\sin w \geq 1$ while $\cos w \leq 1.$ So, as a first try, $w' > 1,$ so we know that $w = \pi/2$ occurs with $x < \pi/2 - \pi / 4 = \pi / 4.$ Much greater precision is available, just solve the numerical ODE for $w$ with $w(0) = \pi/4,$ find when $w = \pi/2.$ In fact, since this thing is solvable in closed form, we find $y(0) = 1$ means $y$ goes to infinity, meaning vertical asymptote, at $$ x = \frac{1}{3} \log 2 \approx 0.231049 $$

A: Actually, there is a whole family of solutions of this equation that are defined on the whole real line. Look carefully at the equation itself
$$\frac{dy}{dx} = y + y^4$$ and write it in the form
$$\frac{dy}{dx} = y(1 + y^3)$$ As the equation is locally Lipschitz everywhere, because it is polynomial, locally everywhere there exists a unique solution to it and so two solutions cannot cross. Now, observe that the constant functions $y(x) \equiv 0$ and $y(x) \equiv -1$ are actually the unique solutions to the inital value problems
\begin{align}
\frac{dy}{dx} &= y + y^4\\
y(0) &= 0
\end{align} and
\begin{align}
\frac{dy}{dx} &= y + y^4\\
y(0) &= -1
\end{align}
respectively. Since two different solutions of the equations cannot cross unless they coincide everywhere, any solution that satisfies the initial value problem 
\begin{align}
\frac{dy}{dx} &= y + y^4\\
y(x_0) &= -y_0
\end{align}
for $(x_0, y_0) \in \mathbb{R} \times [-1, 0]$ will be trapped in the region $\mathbb{R} \times [-1, 0]$ for all $x$ from their maximal interval of definition and therefore that maximal interval is the whole $\mathbb{R}$. 
In conclusion, all solutions $y=y(x)$ of the differential equation
$$\frac{dy}{dx} = x + x^4$$ such that $-1 \leq y(x_0) \leq 0$ for some $x_0 \in \mathbb{R}$ are defined as differentiable functions $y : \mathbb{R} \to \mathbb{R}$ with the property that $-1 \leq y(x) \leq 0$ for all $x \in \mathbb{R}$. 
A: This is a Bernoulli equation, and can thus be solved directly. Setting $u(x)=y(x)^{-3}+1$ results in the equation
$$
u'(x)=-3y(x)^{-4}y'(x)=-3u(x)\implies u(x)=u(x_0)e^{-3(x-x_0)}
$$
so that
$$
y(x)=\frac{y(0)e^{x-x_0}}{\sqrt[3]{1+y_0^3(1-e^{3(x-x_0)})}}
$$
where the cube root is extended as odd function to negative values.
The solution has a singularity in finite time if the denominator has a root, which is the case if $1+y_0^{-3}>0$, that is, $y_0>0$ or $y_0<-1$.
