I am provided with the following image of the direction field of $\dfrac{dy}{dx} = y^2 - x^2$:
Plotted in yellow is the -1-isocline: $-1 = y^2 - x^2$.
The solution states that the lower-right asymptote for this isocline is the line $y = -x - 1$. However, as I understand it, the formula for the asymptote of a hyperbola centered at the origin and with foci on the x-axis is $y = \pm \dfrac{bx}{a}$. Therefore, shouldn't the formula for the lower-right asymptote be $y = -x$?
I would greatly appreciate it if people could please take the time to clarify this for me.
EDIT:
Indeed, even WolframAlpha states that the asymptotes are $y = x$ and $y = -x$! So I'm very confused as to whether I'm misunderstanding the solution or the solution is incorrect?
EDIT2:
$$\dfrac{dy}{dx} = y^2 - x^2 \tag 1$$ The -1-isocline is the curve where the curves solutions of $(1)$ have a slope $=-1$. $$-1 = y^2 - x^2 \tag 2$$ We must no confuse the asymptotes of the curve $(2)$ with the asymptote of the curves solutions of equation $(1)$.
The respective asymptotes are the same only if the slopes are equal, that is $=-1$.
Considering the curve $(2)$ which is an hyperbola, the two asymptotes are $y=x$ and $y=-x$.
The slope of $y=x$ is $+1$ while the slope of the solutions of $(1)$ is $-1$. This means that the curves solutions of $(1)$ tend to be orthogonal to the asymptote $y=x$ of the hyperbola at infinity. Thus the asymptote $y=x$ of the hyperbola is NOT asymptote of the curves solutions of $(1)$. On the contrary, they tends to be orthogonal at infinity.
On the other hand, the slope of $y=-x$ is $-1$ which is the same as the slope of the solutions of $(1)$. The asymptote $y=-x$ of the hyperbola is also asymptote of the curves solutions of $(1)$.
NOTE : Equation $(1)$ is a Riccati's ODE. It is analytically solvable in terms of Bessel functions. The explicit general solution is complicated.
IN ADDITION :
If we express the asymptotic solution of (1) on the form: $$y=-x+a_0+\sum_{n=1}\frac{a_n}{x^n}$$ puting it into (1) leads to : $a_0=0\quad;\quad a_1=\frac{1}{2}\quad;\quad a_2=0 \quad;\quad a_3=\frac{3}{8}\quad;\quad ... $ $$y\simeq -x+\frac{1}{2x}+\frac{3}{8x^3}+...$$ So, $y\simeq -x-1$ is not correct.