Question

Find all the critical points of the system

$$\begin{gathered} \frac{\mathbf{d}\mathbf{x}}{\mathbf{d}\mathbf{t}}\mathbf{=}{\mathit{x}}^{\mathbf{2}}\mathbf{-}\mathbf{1} \\ \frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathit{x}\mathit{y} \end{gathered}$$

and the solution curves for the related phase plane differential equation. Thereby proving that two trajectories lie on semicircles. What are the endpoints of the semicircles?

Short answer

The result is

$$\begin{gathered} for y>0,\left|x\right|>1,y=e^c\sqrt{x^2-1} \\ for y>0,\left|x\right|<1,y=e^c\sqrt{1-x^2} \\ for y<0,\left|x\right|>1,y=-e^c\sqrt{x^2-1} \\ for y<0,\left|x\right|<1,y=-e^c\sqrt{1-x^2} \end{gathered}$$

And the end points are (-1,0) (1,0).

Step-by-step solution

Find critical points

For a critical point put the system equal to 0.

$$\begin{gathered} x^2-1=0 \\ xy=0 \\ {x}^2=1 \\ x=\pm 1 \\ y=0 \end{gathered}$$

Find the value of y

Now,

$\begin{array}{rcl}\frac{dy}{dx}& =& \frac{xy}{x^2-1}\\ \int \frac{1}{y}dy& =& \int \frac{x^2}{x^2-1}\\ \ln \left|y\right|& =& \frac{1}{2}\ln \left|t\right|+c \left(\text{bysubtitution}\right)\\ \ln \left|y\right|& =& \ln {\left|t\right|}^{\frac{1}{2}}+c\\ y& =& e^c{\left|t\right|}^{\frac{1}{2}}\\ y& =& e^c\sqrt{x^2-1}\end{array}$

Prove results that endpoints are semicircles.

Now there are two cases when y > 0 and y < 0 for both cases $x^2-1<0 \mathrm{and} x^2-1>0$then $\left|x\right|<1 and \left|x\right|>1$

$$\begin{gathered} fory>0,\left|x\right|>1,y=e^c\sqrt{x^2-1} \\ fory>0,\left|x\right|<1,y=e^c\sqrt{1-x^2} \\ fory<0,\left|x\right|>1,y=-e^c\sqrt{x^2-1} \\ fory<0,\left|x\right|<1,y=-e^c\sqrt{1-x^2} \end{gathered}$$

The trajectories that possibly lie on semicircles. If we square the equation then

$y^2=e^{2c}(1-x^2)$

This will be the equation of circle only if ${\mathit{e}}^{\mathbf{2}\mathbf{c}}\mathbf{=}\mathbf{1}$. Therefore, only two solutions lie on the semicircle, $\mathit{y}\mathbf{=}\sqrt{{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{1}}\mathbf{,}\mathit{y}\mathbf{=}\mathbf{-}\sqrt{{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{1}}$.

Therefore, it is a circle of radius 1 center at origin the endpoints are$\left(-1,0\right), \left(1,0\right)$.