Parameterizing Equations with a and b: A Calculus Question

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In summary, the conversation discusses the process of parameterizing a set of equations involving calculus, specifically the equations $\dot{x} = -x + ay + xy^2$ and $\dot{y} = b - ay - x^2y$. The speaker shares an example that they understand, but they are having trouble applying it to the given problem. They mention using the Jacobian to find the values for $u_0$ and $v_0$, but they encounter difficulties when solving for $x_0$ and $y_0$. They eventually find a solution by setting
  • #1
Dustinsfl
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Since I no longer own a calculus book anymore, I have a simple question.
\begin{align}
\dot{x} =& -x + ay + xy^2\notag\\
\dot{y} =& b - ay - x^2y\notag
\end{align}

How can parameterize those equations for a and b?
 
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  • #2
dwsmith said:
Since I no longer own a calculus book anymore, I have a simple question.
\begin{align}
\dot{x} =& -x + ay + xy^2\notag\\
\dot{y} =& b - ay - x^2y\notag
\end{align}

How can parameterize those equations for a and b?

So I found an easy example which I understand but applying it to this problem is tough.

The example is

\begin{align}
\dot{u} &= a - u + u^2v = f(u,v)\\
\dot{v} &= b - u^2v = g(u,v)
\end{align}

\begin{align}
f(u_0,v_0) &= a - u + u^2v = 0\\
g(u_0,v_0) &= b - u^2v 0
\end{align}

So this is relatively easy to find $u_0 = b+a$ and $v_0 = \dfrac{b}{(a+b)^2}$.

Then using the Jacobian, we get $\text{tr}(A) = f_u+g_v = \dfrac{b-a}{a+b}-(a+b)^2$ and $\text{det}(A)=f_ug_v-f_vg_u = (a+b)^2$.

Then

$\text{tr}(A) > 0, \ |A| > 0, \quad (\text{tr}(A))^2\begin{cases} >\\<\end{cases} \ 4|A|\Rightarrow\text{unstable}\begin{cases} \text{node} \\ \text{spiral}\end{cases}$.

The equations then are $a-b=(a+b)^3$ and $a+b=0$.

However, when I solved for my equation I obtained:

$$
y_0 = \frac{b}{a+x_0^2}
$$
which lead to $-x_0+\dfrac{ab}{a+x_0^2}+\dfrac{x_0^2b}{a+x_0^2}=0$.

This didn't lead to fruit solutions for $x_0$ and $y_0$.

I did I overlook something when I solved mine: a mistake or technique to avoid this?
 
  • #3
Ok the parameterization has been
Solved
 

FAQ: Parameterizing Equations with a and b: A Calculus Question

What does it mean to "parameterize" an equation?

Parameterizing an equation means expressing the equation in terms of one or more parameters, such as a and b. This allows for a more general representation of the equation and allows for a wider range of solutions to be found.

How do you determine the values for a and b in a parameterized equation?

The values for a and b can be determined through various methods, such as trial and error or using known values from other equations in the system. In some cases, the values may also be given in the problem or can be solved for using calculus techniques.

Can a parameterized equation have multiple solutions?

Yes, a parameterized equation can have multiple solutions, as the values for a and b can be varied to produce different solutions. This is one of the advantages of using parameterization in equations.

How does parameterization affect the graph of an equation?

Parameterization can change the shape and behavior of a graph, as the values for a and b can affect the slope, intercepts, and other characteristics of the equation. This can be seen by graphing different values for a and b and observing the changes in the graph.

What are some real-world applications of parameterized equations?

Parameterized equations are commonly used in physics, engineering, and other fields to model and solve complex systems. They can also be used in economics, biology, and other areas to analyze and predict trends and patterns. For example, in physics, parameterized equations can be used to describe the motion of objects under the influence of forces, while in economics, they can be used to model supply and demand curves.

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