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i have this exercise :

we consider the following model : $$ \begin{cases} x'& = x(4-x-y)\\ y'&=y(2+2\alpha-y-\alpha x) \end{cases} $$

a) Find the critical point $P$ does not depend on $\alpha$ and having coordinates strictly positive.

-> The critical points are : $(0,0),(0,2 +2 \alpha),(2,2),(4,0)$ then P = $(2,2)$

b) Assuming $\displaystyle\frac{dy}{dx}$,solve the system for the value $ \alpha_0$ of $\alpha$ and such that $P$ is not hyperbolic,

Draw the phase portrait for $\alpha = \alpha_0$

c) what is the nature of $P$ for $ 0 <\alpha <\alpha_0 $ and $ \alpha_0 <\alpha $, Draw the shape of the phase portraits for these values ​​of $\alpha $ "

Can someone tell me how to solve b), why they use $\displaystyle\frac{dy}{dx}$?

Please help me

Thank you

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I have no idea what "assuming $dy/dx$" might possibly mean, but $P$ is hyperbolic for $\alpha>1$. – user8268 Sep 18 '13 at 8:30
It's not a book it is an exercise derived from an examination subject – Vrouvrou Sep 18 '13 at 15:44
@user8268 , how you have found this ? – Vrouvrou Sep 19 '13 at 14:29

For system of the shape $$\begin{cases} x' = f(x,y)\\ y' = g(x,y) \end{cases}$$ If one had a solution such one could write $y(t) = \tilde y(x(t))$ then one would have that $$\frac{d y}{d t} = \frac{d \tilde y}{d x}\frac{d x}{d t}$$
So, with an abuse of notation one writes $$ \frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{f(x,y)}{g(x,y)}$$ So assumming $\frac{d y}{d x}$ could simply mean that $g(x,y) \ne 0$ for a solution, and thus the previous is well defined. With this trick you turn a PDE into a ODE (provided the above works). Now you have a fraction of polynomial, which maybe you can integrate and get something nice, hence they say''solve''.

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