How can $e^a = 0$ when integrating? I was doing a question which asked me to turn the following into polar equation:$$(y + x − x(x^2 + y^2))\frac{dy}{dx} = y − x − y(x^2 + y^2)$$
With a lot of messy algebra I can get it down to:
$$\frac{1}{2}\ln \left\lvert\frac{{r^{2} - 1}}{r^2}\right\rvert = \theta + k.$$
Which is definitely correct. From there:
$$\pm{e^{2\theta+2k}}=\frac{r^2-1}{r^2}$$
let $e^{2k}=A$
$$\pm{}Ae^{2\theta}=1-\frac{1}{r^2}$$
$$1\mp{Ae^{2\theta}}=\frac{1}{r^2}$$
$$r^2=\frac{1}{1\mp{Ae^{2\theta}}}$$
As $ A>0$ and $e^{2k}\neq0$, where $C\neq0$
$$r^2=\frac{1}{1+Ce^{2\theta}}$$
At this point, I'm asked to sketch the graph for the different possible values of C, which I assumed meant for $C>0$ and $C<0$, however apparently $C=0$ is valid. I'm struggling to grasp why it's possible. Would that mean that $k=-\infty$? If anyone can explain this / correct any issues with my working, I'd greatly appreciate it
 A: Let's look at a simpler (but otherwise analogous) example:
$$
\frac{dy}{dx} = y.
$$
The usual method of solution is to separate variables, integrate, and solve for $y$:
$$
\frac{dy}{y} = dx,\qquad
\ln|y| = x + C,\qquad
y = Ae^{x},\quad A = \pm e^{C}.
$$
(The choice of signs arises when dropping the absolute value from $y$.) Now, $e^{C}$ cannot be zero, but the $A$ in the formula $y = Ae^{x}$ can certainly be zero. Why did our method not see this solution?
Our first step was to divide by $y$. If $A = 0$, then $y \equiv 0$, and our very first step becomes nonsensical. In other words, our method of solution implicitly omitted the case $A = 0$.
Nonetheless, it's trivial to verify that $y = 0$ solves the ODE.

In your situation, you've lost the constant solution $r = 1$ at the first expression following "after a lot of messy algebra...":
$$
\ln \left\lvert\frac{\sqrt{r^{2} - 1}}{r}\right\rvert = \theta + k.
$$
Nonetheless, $r = 1$, a.k.a. $x^{2} + y^{2} = 1$, presumably satisfies your ODE. (It doesn't appear to be a solution as typed, however....)
A: The basic point, I think is this.  A family of solutions of an analytic differential equation, given by an analytic function with a parameter, will be valid on any connected open set of values of the independent variable and the parameter, as long as you stay in the region of analyticity.  So even if you assumed $C > 0$ in your derivation of the solution, the final result  will be good for all $C$, as long as you avoid singularities.   
