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: $\begin{align} &\frac{dy}{dx}=2y^2\\ &\frac{dy}{y^2} = 2\, dx\\ -&\frac{1}{y} = 2x + c\\ -&\frac{1}{2} = c\\ -&\frac{1}{y}=2x-\frac{1}{2}\\ &\frac{1}{y}=-2x+\frac{1}{2}\\ &y=\frac{1}{-2x+\frac{1}{2}}\\ &y=\frac{2}{-4x+1}\\ y(8)=-2/31\end{align}$