MAT244-2013S > Quiz 5

Day Section's Quiz Problem 2


Sabrina (Man) Luo:
(2) Find an equation of the form H(x,y)=c satisfied by solutions to

Alexander Jankowski:
To determine $H(x,y)$, we proceed as follows:
\begin{equation*} \frac{dy}{dx} = \frac{-2xy^2 + 6xy}{2x^2y - 3x^2 - 4y} \Longrightarrow (2xy^2 - 6xy)dx + (2x^2y - 3x^2 - 4y)dy = 0. \end{equation*}
Let $M(x,y) = 2xy^2 - 6xy$ and $N(x,y) = 2x^2y - 3x^2 - 4y$. It turns out that
\begin{equation*} M_y(x,y) = 4xy - 6x = N_x(x,y). \end{equation*}
Thus, the differential equation is exact. Suppose that there is a function $\psi(x,y)$ that satisfies the equations $\psi_x(x,y) = M$ and $\psi_y(x,y) = N$. We have
\begin{equation*} \psi_x(x,y) = 2xy^2 - 6xy \Longrightarrow \psi(x,y) = x^2y^2 - 3x^2y + g(y). \end{equation*}
Then, we try to determine $g$:
\begin{equation*} \psi_y(x,y) = 2x^2y - 3x^2 + g'(y) = 2x^2y - 3x^2 - 4y \Longleftrightarrow g'(y) = -4y \Longrightarrow g(y) = -2y^2. \end{equation*}
We conclude that
\begin{equation*} \psi(x,y) = x^2y^2 - 3x^2y - 2y^2 = c. \end{equation*}
To confirm this, I have attached (1) a stream plot of the system and (2) a contour plot of $H = \psi$.

Iven Poon:
I already written my solution down on Monday, and I planned to post on Thursday when Professor Ivrii uploaded the questions...

Anyway, I would still like to share how I did this question.

Victor Ivrii:
On the plot of Alexander you see unusual equilibrium point (0,0). The reason of its strange appearance (neither center nor saddle) is because it is degenerated critical point of $H(x,y)$ and a lot of funny things can happen then.


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