Hi David,

I believe the problem here is that Mathematica offered a solution that is in fact not valid at the point (-1,1) which was a defined initial condition. So there is no contradiction of the uniqueness theorem. You will see below (and I attach the notebook) that the solution found is valid only for t <= -2 or t > 0. I find it a bit odd that Mathematica did in fact provide a solution that does not include the initial conditions.

(* equation *)
eq = y'[t] == (2 Sqrt[y[t]] - 2 y[t])/t

(* solution *)
sol = DSolve[{eq, y[-1] == 1}, y, t] // First

(* it verifies, but with restrictions *)
Reduce[eq /. sol, t, Reals]

(* t\[LessEqual]-2||t>0 *)

(* look at LHS and RHS over a range *)
backsub = eq /. sol

lhs = backsub[[1]]

rhs = backsub[[2]]

(* we see that this is NOT a solution at (-1,1) *)
Plot[{lhs, rhs}, {t, -3, 3}, PlotLegends -> {"LHS", "RHS"}]

Attachments:

Hi David,

I'm confused. In the text lead in you give dy/dt as a function of y and x, with an initial condition for y. Since you use dy/dt, and y depends on a scalar initial condition, y must be y[t], but what is x? Is this a constant in an ordinary diffeq?

In In[177] below, dy/dt becomes dy/dx. Now there is only one independent variable which is x, and an ordinary diffeq?

Am I missing something?

Best,

David