Hi I'm doing work over the summer for my Differential Equations Module. Finding the equilibrium points here is important for all follow on questions and wanted to check to see if I'm wayyy out? Please could someone let me know if this is okay or whether I need to go back to the drawing board? Thank you:)
The question:
Determine the number and location of the equilibrium points of the system below
$\dot{x}=yx^2 -x$
$\dot{y}=-xy-x^2y+4y^2+4xy^2$
My answer: I ended up with the equilibrium points $(x_e,y_e)=(0,0), (2+2\sqrt{2},\frac{-1+\sqrt{2}}{2}), (2-2\sqrt{2},\frac{-1-\sqrt{2}}{2})$
I did this by finding that when $\dot{x}=0$ then $x=0$ or $xy=1 \Rightarrow x=\frac{1}{y}$ and then subbing into $\dot{y}$
EDIT: Thank you for all the help I've gone over it and I'm not sure how I managed to mess up the $\dot{y}$ factorisation so badly!
I checked and changed my work and got the following (for future reference):
when $x=0, \dot{y}=0=4y^2 \Rightarrow y=0$ and when $x=\frac{1}{y}, \dot{y}=0=-1-\frac{1}{y}+4y^2+4y$ and multiplied this by $y$
To this I found the factor $(y+1)$ and used algebraic division to find other factors. I got $\dot{y}=0=(y+1)(2y+1)(2y-1)$ and then found the corresponding $x$ values
My final equilibrium points are $(x_e,y_e)=(0,0), (-1,-1), (2,\frac{1}{2}), (-2,-\frac{1}{2})$
You forgot some solution, for example: $$(x,y)=-(1,1)$$ You can check the number of solution since you have a third degree equation in $y$. So it gives three solutions.
$$-xy-x^2y+4y^2+4xy^2=0$$ Since we have $xy=1$ $$-1-x+4y^2+4y=0$$ $$-1-\dfrac 1 y+4y^2+4y=0$$ $$-(y+ 1) +4y^2(y+1)=0$$ $$(y+ 1)(4y^2-1)=0$$ $$\implies y=-1 \implies x=-1$$ Another solution is: $$y=\pm \dfrac 12 \implies x=\pm 2$$ So the Equilibrium Points are: $$S= \{(2,\dfrac 12),(-2,-\dfrac 12),(0,0),(-1,-1) \}$$ You applied the method correctly.