### Author Topic: Day Section's Quiz Problem 2  (Read 5077 times)

#### Sabrina (Man) Luo

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##### 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

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##### 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
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##### 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

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.