please have a look at the attached picture ...and help me understand what to do next...
thank you for whatever help anyone can provide ...
plz note , i know how to linearize at the equillibrium point which is ( x= m/n , y =a/b )
please have a look at the attached picture ...and help me understand what to do next...
thank you for whatever help anyone can provide ...
plz note , i know how to linearize at the equillibrium point which is ( x= m/n , y =a/b )
you can change to the "scaled variables" Y= y/(a/b) and X= x/(m/n) by solving those for x and y: x= (m/n)X and y= (a/b)Y. Then the differential equations become
$\displaystyle \frac{dx}{dt}= \frac{m}{n}\frac{dX}{dt}= (a- b(\frac{a}{b})Y)\frac{m}{n}X$
or
$\displaystyle \frac{dX}{dt}= a(1- Y)X$
and
$\displaystyle \frac{dy}{dt}= \frac{a}{b}\frac{dY}{dt}= (m- n(\frac{m}{n}X)\frac{a}{b}Y$
or
$\displaystyle \frac{dY}{dt}= m(1- X)Y$
which have equilibrium points at (0, 0) and (1, 1).
You then get $\displaystyle \frac{dY}{dX}= \frac{m(1- X)Y}{a(1-Y)X}$
which can be separated as
$\displaystyle a\frac{Y}{1-Y}dY= m\frac{X}{1-X}dX$
Integrate that to get g(Y)= f(X).