Isocline in dynamical system

Question

I solved differential equation and now should draw graph. So, we have:$$y'=\sqrt{3+y^2}$$ $$x'=x^2+x$$ I get isocline $x=-1$ (from $x'=0$). Is it correct or maybe here is more?

Answer 1

Solving both equations we get the following:

$$y = \sqrt{3}\sinh(t+C)$$

$$ t + C = \int \frac{1}{x^2+x}dx = \int \frac{\frac{1}{x^2}}{1+\frac{1}{x}}dx = -\log\left(1+\frac{1}{x}\right)$$

Then plugging in, we get these nice graphs which we can plot:

$$y = \frac{\sqrt{3}}{2}\left(\frac{C}{1+\frac{1}{x}} - C^{-1}\left(1+\frac{1}{x}\right)\right)$$

where this $C$ must be positive. Or equivalently,

$$ y = \frac{\sqrt{3}}{2}\left(\frac{Cx^2-C^{-1}(1+x)^2}{x^2+x}\right)$$

Of course, this is not valid when $x=0$ or $x=-1$. Those are asymptotes of this family of functions.