### Author Topic: Day Section's Quiz Problem 2  (Read 2701 times)

#### Sabrina (Man) Luo

• Jr. Member
•  • Posts: 9
• Karma: 4 ##### Day Section's Quiz Problem 2
« on: April 03, 2013, 09:12:44 AM »
(2) Find an equation of the form H(x,y)=c satisfied by solutions to
\begin{equation*}
\left\{\begin{aligned}
&dx/dt=2x^2y-3x^2-4y,\\
&dy/dt=-2xy^2+6xy
\end{aligned}
\right.\end{equation*}
« Last Edit: April 03, 2013, 10:16:56 AM by Victor Ivrii »

#### Alexander Jankowski

• Full Member
•   • Posts: 23
• Karma: 19 ##### Re: Day Section's Quiz Problem 2
« Reply #1 on: April 03, 2013, 10:38:23 AM »
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 $H = \psi$.
« Last Edit: April 06, 2013, 10:52:55 AM by Alexander Jankowski »

#### Iven Poon

• Newbie
• • Posts: 4
• Karma: 2 ##### Re: Day Section's Quiz Problem 2
« Reply #2 on: April 03, 2013, 02:17:15 PM »
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 ##### Re: Day Section's Quiz Problem 2
« Reply #3 on: April 03, 2013, 04:36:43 PM »
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.