Two different solutions for one differential equation (population model)
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Niklas Kurz
on 29 May 2021
Commented: Sulaymon Eshkabilov
on 3 Jun 2021
I'll try solving the ODE:
Substituting
Transforming to:
Solving I get:
Finally, after back substitution:
complete solution:
what's equivalent to:
Now same stuff with MATLAB:
syms u(t); syms c1 c2 u0 real;
D = diff(u,t,1) == c1*u-c2*u^2;
k2 = u;
cond = k2(0) == u0;
S = dsolve(D,cond);
pretty(S)
Receiving:
I was hoping these expressions have some equivalence so I was plotting them:
c1 = 4; c2 = 2; u0 = 1;
syms t
P1 = (c1)/(1-exp(-c1*t)+c1/u0*exp(-c1*t));
fplot(P1)
hold on
P2 = -(c1*(tanh(atanh((c1 - 2*c2*u0)/k1) - (c1*t)/2) - 1))/(2*c2);
fplot(P2)
but no luck there. I know that's again a quite complex question, but on MathStack one told me these solutions are equvialent, so I don't see a reason for the dissonance.
3 Comments
Sulaymon Eshkabilov
on 3 Jun 2021
Most welcome. We learn by making mistakes.
Please just keep it. So others can learn.
Accepted Answer
Sulaymon Eshkabilov
on 29 May 2021
Besides k1, in your derivations, there are some errs. Here are the corrected formulation in your derivation part:
c1 = 4; c2 = 2; u0 = 1;
syms t
P1 = c1/(c2 - exp(-c1*t)*(c2 - c1/u0)); % Corrected one!
fplot(P1, [0, pi], 'go-')
hold on
P2 = -(c1*(tanh(atanh((c1 - 2*c2*u0)/c1) - (c1*t)/2) - 1))/(2*c2);
S = eval(S);
fplot(S, [0, pi], 'r-')
Good luck.
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