# The solution for Sinh[] and Cosh[]

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.

• 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? Jun 1, 2022 at 15:14
• Sqrt[(3 k)/b] Cosh[ Sqrt[b/3] c (t - C[1])] for the case k>0 Jun 1, 2022 at 15:19
• Sqrt[(3 |k|)/b] Sinh[ Sqrt[b/3] c (t - C[1])] for the case k<0 Jun 1, 2022 at 15:20
• Armin, what is the definition for the constant c (small c, not the C[1]) in the solution above? Jun 1, 2022 at 15:49
• @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])] Jun 1, 2022 at 15:54

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])])}}   *)


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
]


• why the solution is complex ?!! Jun 1, 2022 at 15:40
• 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. Jun 1, 2022 at 15:50
• 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. Jun 2, 2022 at 7:24