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
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....)