2
$\begingroup$

I saw this equation a book that has two solution: one Sinh[] and another one is Cosh[] but I can't find them with Mathematica 13?!!

  Clear["Global`*"]
  eqn = {(y'[t])^2 - (b*a^2/3) y[t]^2 == -k* a^2};
  sol = DSolve[{eqn}, y[t], t]

b,a are positive value and k is taking +1 and -1.

$\endgroup$
5
  • $\begingroup$ Do you have a reference for the solution? Can you include the full solution you saw to see if it is functionally equivalent to the one produced by Mathematica? $\endgroup$
    – MarcoB
    Jun 1, 2022 at 15:14
  • $\begingroup$ Sqrt[(3 k)/b] Cosh[ Sqrt[b/3] c (t - C[1])] for the case k>0 $\endgroup$ Jun 1, 2022 at 15:19
  • $\begingroup$ Sqrt[(3 |k|)/b] Sinh[ Sqrt[b/3] c (t - C[1])] for the case k<0 $\endgroup$ Jun 1, 2022 at 15:20
  • $\begingroup$ Armin, what is the definition for the constant c (small c, not the C[1]) in the solution above? $\endgroup$
    – MarcoB
    Jun 1, 2022 at 15:49
  • 1
    $\begingroup$ @MarcoB I think c should be a and the second solution should/could be written Sqrt[(-3 k)/b] Sinh[Sqrt[b/3] a (t - C[1])] $\endgroup$
    – Michael E2
    Jun 1, 2022 at 15:54

2 Answers 2

5
$\begingroup$

Let k -> -kk

Clear["Global`*"]

eqn = {(y'[t])^2 - (b*a^2/3) y[t]^2 == -k*a^2} /. k -> -kk;

sol = DSolve[{eqn}, y[t], t] // ExpToTrig // Simplify

(*   {{y[t] -> (1/(2 b))((1 - 3 b kk) Cosh[1/3 Sqrt[b] (Sqrt[3] a t + 3 C[1])] + 
(1 + 3 b kk) Sinh[1/3 Sqrt[b] (Sqrt[3] a t + 3 C[1])])},

{y[t] -> (1/(2 b))((1 - 3 b kk) Cosh[1/3 Sqrt[b] (Sqrt[3] a t - 3 C[1])] - 
  (1 + 3 b kk) Sinh[1/3 Sqrt[b] (Sqrt[3] a t - 3 C[1])])}}   *)
$\endgroup$
4
$\begingroup$

If we multiply the arbitrary constant C[1] by I, FullSimplify can do the job if passed the correct assumptions:

FullSimplify[
  sol /. C[1] -> I C[1], 
  Assumptions -> a > 0 && b > 0 && {k, t, C[1]} ∈ Reals
]

enter image description here

$\endgroup$
3
  • 1
    $\begingroup$ why the solution is complex ?!! $\endgroup$ Jun 1, 2022 at 15:40
  • 1
    $\begingroup$ Hard to say what DSolve is doing under the hood, but if it's solving as a separable equation with Solve, then I is introduced immediately e.g. Solve[eqn, y'[t]]. This doesn't mean the solution is complex, as the I could be distributed into the radical. $\endgroup$
    – Greg Hurst
    Jun 1, 2022 at 15:50
  • $\begingroup$ I'm more interested in how you discover this solution in Mathematica. Naively I would think, if it can do it for I C[1] it should also be able to do it for k C[1], where k is any complex number, including 1. $\endgroup$ Jun 2, 2022 at 7:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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