## Singular Points

1. The problem statement, all variables and given/known data

Find the singular points for:

x' = ax - bxy
y' = bxy - cy

2. Relevant equations

3. The attempt at a solution

ax - bxy = 0
bxy - cy = 0

implies

x = 0 or y = a/b

y = 0 or x = c/b

the the critical points are (o,o) or (c/b, a/b)

why is (0, a/b) and (c/b, 0) NOT a critical point??
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 why is (0, a/b) and (c/b, 0) NOT a critical point??
Because, $x'$ and $y'$ are not 0 for those values of x and y. Substitute them to check.
 Recognitions: Gold Member Science Advisor Staff Emeritus Why would you think they would be? They clearly don't satisfy the equations you give. When you have a system of non-linear equations with several different solutions, the x and y are not independent- you can't just combine any value of x with any value of y: it is the specific x, y pair that satisfies the equations.

## Singular Points

Oh I see my mistake.

I thought that they were dependent.
 I have been working with the equations: x' = rx - sxy/(1+tx) y' = sxy/(1+tx) - wy I find that the only singular point is (0,0) is this correct? I got another point, ( w/(s-w) , r/s ), however, that cannot be a critical point because when substituting, x' and y' fail to be 0.

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 Quote by rad0786 I have been working with the equations: x' = rx - sxy/(1+tx) y' = sxy/(1+tx) - wy I find that the only singular point is (0,0) is this correct I got another point, ( w/(s-w) , r/s ), however, that cannot be a critical point because when substituting, x' and y' fail to be 0.
Since (w/(s-w), r/s) does not satisfy the equations for a critical point, how did you get that?

The equations for a critical point are x'= rx- sxy/(1+ tx)= 0 and y'= sxy/(1+ tx)- wy= 0. The first gives sxy/(1+ tx)= rx and the second sxy/(1+ tx)= wy. Since the left sides of both equations are the same, rx= wy. Every point on the line y= (r/w)x (or, if w= 0, the line x= 0) is a singular point.

 Quote by HallsofIvy Since (w/(s-w), r/s) does not satisfy the equations for a critical point, how did you get that?
x'= rx- sxy/(1+ tx)= 0

x(r-sy/(1 + tx)) = 0

x= 0 and (r-sy/(1 + tx)) = 0

since x = 0 r-sy = 0

therefore, y = r/s

similarly we do that for y'= sxy/(1+ tx)- wy= 0 and we get y = 0 and x = w/(s-w)

 Quote by HallsofIvy The equations for a critical point are x'= rx- sxy/(1+ tx)= 0 and y'= sxy/(1+ tx)- wy= 0. The first gives sxy/(1+ tx)= rx and the second sxy/(1+ tx)= wy. Since the left sides of both equations are the same, rx= wy. Every point on the line y= (r/w)x (or, if w= 0, the line x= 0) is a singular point.

when I substitute y= (r/w)x into x'= rx- sxy/(1+ tx), x' is NOT zero. So how could this line be a line of singular points?
 also, what are the differences between equilibrium solutions and singular points?