MATLAB: Using the Solve command

solve

Equations:
df/dt= 4f(t) – 3f(t)p(t)
dp/dt= -2p(t) + f(t)p(t)
Question: Figure out the critical points of the system, that is, those points (f;p) such that f'=p'=0 simultaneously. If we happen to start at one of these points, then there's no change since f'= 0 and p'= 0, so the population will just sit there forever. Use the solve command
Code:
[f, p] = dsolve('Df = 4*f - 3*f*p', 'Dp = -2*p + f*p', 'f(0) = 0', 'p(0) = ');
I am getting an error with this: Error in dsolve (line 193) sol = mupadDsolve(args, options);
Error in Project_5_2 (line 5) [f, p] = dsolve('Df = 4*f – 3*f*p', 'Dp = -2*p + f*p', 'f(0) = 0', 'p(0) = ');

Best Answer

You DID make some effort here. :)
Essentially, this problem does not really ask you to use dsolve. It never asked you to solve the differential equations, but to use solve to locate all of the critical points of the system. Really, you could do this on paper.
A critical point for this problem has f'=p'=0. So when will you have both f'==0, AND p'==0?
f'(t) = 4*f(t) - 3*f(t)*p(t)
SET IT TO ZERO! READ THE QUESTION. It told you exactly what to do here.
4*f - 3*f*p == 0
Likewise, when is p'(t)==0?
-2*p + f*p == 0
Can you solve this problem for f and p? Pencil and paper will suffice, as I said. The first reduces to:
f*(4-3*p) == 0
So either f == 0, OR p == 4/3.
Likewise the second relation tells us that p==0, OR f==-2. But for both of these equations to simultaneously equal zero, there are only two solutions. Clearly f=p=0 is one pair. But the other solution occurs when f=-2 AND p=4/3.
For those who are paper challenged, or who need to use solve to solve the problem as it is homework...
syms f p
[f,p] = solve(4*f - 3*f*p == 0,-2*p + f*p == 0,f,p)
f =
 0
 2
p =
   0
 4/3
So, we have two solutions, the same two that I predicted above.
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