Chapter 3.6, Problem 22P

In this problem we indicate how to find the asymptote suggested by Figure $\mathrm{3.6.3}$.

a) Show that if $x$ and $y$ satisfy the initial conditions $x\left(0\right)=3$ and $y\left(0\right)=0$, then the constant $c$ in Eqs. $\left(15\right)$ is zero. Then show that Eqs. $\left(15\right)$ can be written in the form

${\left(x-y\right)}^2+x^2-\frac{2}{3}{x}^3=0$.

b) From the first of Eqs. $\left(12\right)$, recall that $x-y=x\text{'}$. Use this fact in ${\left(x-y\right)}^2+x^2-\frac{2}{3}{x}^3=0$ to obtain the differential equation ${\left(x\text{'}\right)}^2=\frac{2}{3}{x}^3-x^2$ or $x\text{'}=\frac{1}{3}x\sqrt{6x-9}$. Why must the positive square root be chosen?

c) Show that the solution of the equation $x\text{'}=\frac{1}{3}x\sqrt{6x-9}$ that satisfies the initial condition is

$x=\frac{3}{2}+\frac{3}{2}{\tan }^2\left(\frac{1}{2}t+\frac{1}{4}\pi \right)$. Hint: In solving $x\text{'}=\frac{1}{3}x\sqrt{6x-9}$, you may find it helpful to use the substitution $s^2=6x-9$.

d) Use the result of part (c) to show that the solution has a vertical asymptote at $t=\frac{\pi }{2}$. Compare this result with the graph in Figure $\mathrm{3.6.3}$.

e) From $x=\frac{3}{2}+\frac{3}{2}{\tan }^2\left(\frac{1}{2}t+\frac{1}{4}\pi \right)$, there is another vertical asymptote at $t=-\frac{3\pi }{2}$. What is the significance of this asymptote?

Equation- $15$: $H\left(x,y\right)=x^2-xy+\frac{1}{2}{y}^2-\frac{1}{3}{x}^3=c$

Equation- $12$: The system of equations is $\frac{dx}{dt}=x-y,\frac{dy}{dt}=2x-y-x^2$.