Two different solutions for one differential equation (population model)

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.

[Torsten]

P1 = c1/(c2*(1-exp(-c1*t))+c1/u0*exp(-c1*t))

In P2, k1 must be c1.

[Niklas Kurz]

Oh my gosh, all the trouble and questionings were caused by a little typo. I really cherish that hint. It's worth a reply for sure hence I can accept and close the question. Or should I delete it since it's not others people buiseness?

[Sulaymon Eshkabilov]

Most welcome. We learn by making mistakes.

Please just keep it. So others can learn.

Answer 1

[Sulaymon Eshkabilov]

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.