0
$\begingroup$

I need help with the following differential equation. I do not know how I can calculate this.

$\alpha'(t) = \alpha_1^{3/2} - \alpha^{3/2}(t) \qquad \text{with} \qquad \alpha(0)=0$

$\endgroup$
  • 4
    $\begingroup$ Is this question about the software Mathematica. If so, what have you tried? If not, perhaps you want to post at Mathematics StackExchange $\endgroup$ – Bob Hanlon Jun 16 '17 at 13:35
2
$\begingroup$

If you do wish to solve this problem with Mathematica, the approach is as follows.

a[t] /. Flatten@DSolve[{a'[t] == a1^(3/2) - a[t]^(3/2)}, a[t], t]
(* InverseFunction[(2 Sqrt[3] ArcTan[(1 + (2 Sqrt[#1])/Sqrt[a1])/Sqrt[3]] + 
    2 Log[Sqrt[a1] - Sqrt[#1]] - Log[a1 + Sqrt[a1] Sqrt[#1] + #1])/(3 Sqrt[a1]) &]
    [-t + C[1]] *)

Apply the boundary condition to obtain the constant of integration.

C[1] -> Simplify[%[[0, 1]][0]]
(* C[1] -> π/(3 Sqrt[3] Sqrt[a1]) *)

Addendum

The implicit solution derived by @Nasser also can be obtained by

%%[[0, 1]][a[t]] == First[%%] /. %
(* (2 Sqrt[3] ArcTan[(1 + (2 Sqrt[a[t]])/Sqrt[a1])/Sqrt[3]] + 
    2 Log[Sqrt[a1] - Sqrt[a[t]]] - Log[a1 + Sqrt[a1] Sqrt[a[t]] + a[t]])/(3 Sqrt[a1]) 
    == π/(3 Sqrt[3] Sqrt[a1]) - t *)
| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.