We have:

\begin{equation} \frac{2y^{2}}{9-x^{2}} + y \dv{y}{x} + \frac{3y}{2-x} = 0 \end{equation}

We want to get rid of things; let’s begin by dividing the whole thing by \(y\).

\begin{equation} \frac{2y}{9-x^{2}} + \dv{y}{x} + \frac{3}{2-x} = 0 \end{equation}

Finally, then, moving the right expression to the right, we have:

\begin{equation} \frac{2y}{9-x^{2}} + \dv{y}{x} = \frac{-3}{2-x} \end{equation}

In this case, we have functions:

\begin{equation} \begin{cases} P(x) = \frac{2}{9-x^{2}}\\ Q(x) = \frac{-3}{2-x}\\ \end{cases} \end{equation}

Considering the system:

\begin{equation} \begin{cases} \dv{x}{t} = -2x+y+(1-\sigma)z \\ \dv{y}{t} = 3x-y \\ \dv{z}{t} = (3-\sigma y)x-z\\ \end{cases} \end{equation}

with the initial locations \((x_0, y_0, z_0)= (-1,1,2)\).

We notice first that the top and bottom expressions as a factor in \(x\) multiplied by \(y\), which means that our system is not homogenous. Let’s expand all the expressions first.

\begin{equation} \begin{cases} \dv{x}{t} = -2x+y+(1-\sigma)z \\ \dv{y}{t} = 3x-y \\ \dv{z}{t} = 3x-\sigma yx-z\\ \end{cases} \end{equation}

Intersects:

\begin{equation} f(x) = (x+c)^{2} \end{equation}

\begin{equation} h(x) = c x \end{equation}

Doesn’t Intersect:

\begin{equation} g(x) = c e^{\frac{x^{4}}{4}}} \end{equation}

\begin{align} &h_1(x)-h_2(x) = c_1x-c_2x \\ \Rightarrow\ & 0 = c_1x-c_2x \\ \Rightarrow\ & 0 = x(c_1-c_2) \end{align}

\begin{align} &g_1(x)-g_2(x) = c_1e^{\frac{x^{4}}{4}} - c_2e^{\frac{x^{4}}{4}} \\ \Rightarrow\ & 0 = \qty(c_1 - c_2)e^{\frac{x^{4}}{4}} \\ \Rightarrow\ & 0 = e^{\frac{x^{4}}{4}}(c_1-c_2) \end{align}

\begin{align} & f_1(x)-f_2(x)=(x+c_1)^{2}-(x+c_2)^{2} \\ \Rightarrow\ & 0 = (x+c_1)^{2}-(x+c_2)^{2} \\ \Rightarrow\ & 0 = 2x(c_1-c_2)+{c_1}^{2}+{c_2}^{2} \end{align}

We are given a set of expressions:

\begin{equation} \begin{cases} \dv{x}{t} = \frac{x}{w}\\ \dv{y}{t} = \frac{xz}{w} \\ \dv{z}{t} = \beta y \\ \dv{w}{t} = \beta y \end{cases} \end{equation}

We are asked to analyze the solutions to this system, its periodicity, etc.

Stability Analysis

The immediate thing to do is to shove all of this into a Jacobian matrix—not for linearnalization, but to check how the slope changes. We will take the eigenvalues of the matrix at the critical points of the function, which will tell us whether or not the functions converge or diverge from those points.

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01/09/2023