1
$\begingroup$

I'm trying to solve the following differential equation, but it seems Mathematica doesn't give me an answer for that. Do you guys have any clue on how to make this work?

DSolve[{y''[x] == Sinh[y[x]]}, y[x], x]

Thanks!

$\endgroup$
  • 1
    $\begingroup$ What version are you using? 11.2 returns a result in terms of the Jacobi amplitude. $\endgroup$ – J. M.'s technical difficulties Apr 17 at 16:13
  • 2
    $\begingroup$ same as JM's. On V 12.1 it works screen shot !Mathematica graphics but just gives warning about inverse function which is common $$\left\{\left\{y(x)\to -2 i \text{am}\left(\frac{1}{2} \sqrt{(-c_1-2) (x+c_2){}^2}|\frac{4}{c_1+2}\right)\right\},\left\{y(x)\to 2 i \text{am}\left(\frac{1}{2} \sqrt{(-c_1-2) (x+c_2){}^2}|\frac{4}{c_1+2}\right)\right\}\right\}$$ and always it is best to give version number when asking such questions. Mathematica changes from one version to another. $\endgroup$ – Nasser Apr 17 at 16:15
  • 1
    $\begingroup$ Related Get a symbolic solution from DSolve $\endgroup$ – Artes Apr 17 at 17:04
2
$\begingroup$

From the ordinary differential equation given in the question, it is easily seen that y==0 is a solution for all x. But that is too not the solution shown by Mathematica or Nasser.

input and output

The message is not an error message. It shows that the solution has a periodicity of some kind. This is the pure function type of solution so implicitly Mathematica already complains about missing boundary conditions. That avoids both the message and the representation as a pure function for the solution.

Have a look at the documentation of DSolve. Even more complicated at JacobiAmplitude.

A really nice help in Mathematica is:

WolframAlpha["y''[x]\[Equal]Sinh[y[x]]"]

This provides the mathematical terms and some of the boundary conditions that led to solutions. So the term is a family of solutions figured by Nasser even if y[0] and y'[0] are given.

enter image description here

| 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.