Differential Equation Breaks Euler Method Solving ${dy\over dx} = 2y^2$,  $y(0)=2$ analytically yields $y(8)= -2/31$, but from using Euler's method and looking at the slope field, we see that $y(8)$ should be a really large positive answer. Why?
Differential equation:
$$\begin{align}
&\frac{dy}{dx}=2y^2\\
&\frac{dy}{y^2} = 2\, dx\\
-&\frac{1}{y} = 2x + c\\
-&\frac{1}{2} = c\\
-&\frac{1}{y}=2x-\frac{1}{2}\\
&\frac{1}{y}=-2x+\frac{1}{2}\\
&y=\frac{1}{-2x+\frac{1}{2}}\\
&y=\frac{2}{-4x+1}\\
y(8)=-2/31\end{align}$$

As you found, the solution is $y={2\over 1-4x}$, which has a vertical asymptote at $x=1/4$. In the slope field, you should be able to convince yourself that such a function can indeed "fall along the slope vectors". The curve will shoot up to infinity  as you approach $x=1/4$ from the left. To the right of $x=1/4$ the curve "comes from below".
The graph of $y={2\over 1-4x}$ over $[0,1]$ is shown below:

