but I honestly think it's more instructive to show them sometimes even small things can be confusing.

Prof. Grothendieck (snipped from F. W. Lawvere Aurelio Carboni) expressed a similiar view:

But, on the other hand, to keep the errors man invented informatics.

eqn = x'[t] == Sqrt[x[t]];

soln = DSolve[{eqn, x[0] == x0}, x[t], t] // FullSimplify

{{x[t] -> 1/4 (t - 2 Sqrt[x0])^2}, {x[t] -> 1/4 (t + 2 Sqrt[x0])^2}}

both solutions satisfy the initial condition

soln /. t -> 0

{{x[0] -> x0}, {x[0] -> x0}}

rules = Flatten[NestList[D[#, t] &, #, 1]] & /@ soln

{{x[t] -> 1/4 (t - 2 Sqrt[x0])^2, Derivative[1][x][t] -> 1/2 (t - 2 Sqrt[x0])}, {x[t] -> 1/4 (t + 2 Sqrt[x0])^2, Derivative[1][x][t] -> 1/2 (t + 2 Sqrt[x0])}}

The second solution is valid for (x0 >= 0) and (t >= -2 Sqrt[x0])

Simplify[eqn /. rules, {x0 >= 0, t >= -2 Sqrt[x0]}]

{t >= 2 Sqrt[x0], True}

Both solutions are valid for (x0 >= 0) and (t >= 2 Sqrt[x0])

Simplify[eqn /. rules, {x0 >= 0, t >= 2 Sqrt[x0]}]

{True, True}

Hello Frank,

Thank you for your reply. I think I will stop teaching my young nephew differential equations AND Mathematica :-) Nothing wrong with what you write/code but I taught him (and his prof. too) the theorem of existence and uniqueness. By the way I have to admit that I used another CAS for this ODE and it returned one solution (the second one with the + sign). Best to you, Jean-Michel