Q10E - Fundamentals Of Differential Equations And Boundary Value Problems

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?

Step-by-Step Solution

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 1: Find critical points

For a critical point put the system equal to 0.

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

Step 2: Find the value of y

Now, 

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

Step 3: 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 $\left|x\right|<1 and \left|x\right|>1$

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

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$\left(-1,0\right), \left(1,0\right)$.