Toronto Math Forum
MAT2442013S => MAT244 MathTests => Quiz 5 => Topic started by: Sabrina (Man) Luo 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^2y3x^24y,\\
&dy/dt=2xy^2+6xy
\end{aligned}
\right.\end{equation*}

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 $H = \psi$.

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.

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.