I made up a problem where this ODE appears and is the solution to that, well I'm supposed to say what I've tried but I'm not even at university and never even taken calculus class. However I'm a bit familiar to integration and solving first order ODE, I made a quick google search but didn't found any ODE like this unfortunately. Basically I want the family of functions whose the tangent of the function is the same as its derivative.

It's the same to solve y=arctan(y'), no problem in worrying about the domain.

How can I "generalise" these kind of solutions?: tan(g(y,x))=y' or tan(g(y))=y'

Wolfram alfa couldn't help me. Can the solution be expressed in terms of elementary function? if not how would it look like?

  • $\begingroup$ Your question is not that clear. For example, you state the problem $\tan y = y'$. This one is easily solvable in terms of standard functions. Now, the differential equation $\tan(g(y,x)) = y'$ is very different, because now you have a function with two argument and on the RHS one of the arguments. $\endgroup$
    – Rebellos
    Jul 13 '19 at 19:35
  • $\begingroup$ Alpha does give the solution. $\endgroup$
    – user65203
    Jul 13 '19 at 19:37
  • 1
    $\begingroup$ $\tan(g(y,x))$ where $g$ is an arbitrary function is of the form $h(y,x)$. The tangent function is irrelevant. $\endgroup$
    – user65203
    Jul 13 '19 at 19:38
  • $\begingroup$ I meant y is a function of x, like y=x^2 $\endgroup$ Jul 13 '19 at 20:42

$$ \tan y = {dy\over dx}\implies \int dx = \int {\cos y \over \sin y }dy$$

So $$x+c =\ln|\sin y|\implies \sin y = Ce^x $$

So $$y= \arcsin (Ce^x)$$

  • $\begingroup$ Can tan(sqrt(x^2 + y^2)) = y' be solved as well? y is a function of variable x? $\endgroup$ Jul 13 '19 at 21:02

$$\frac{y'}{\tan y}=1$$ integrates as

$$\ln\sin y=x+c$$



There is also the degenerate solution $y=0$.

  • $\begingroup$ Can tan(sqrt(x^2 + y^2)) = y' be solved as well? y is a function of variable x? $\endgroup$ Jul 13 '19 at 21:02
  • $\begingroup$ Seems difficult, Alpha fails. Do you have any reason to try such equations ? $\endgroup$
    – user65203
    Jul 13 '19 at 21:32
  • $\begingroup$ Yeah, it has the solutions to a problem I am very curious about $\endgroup$ Jul 14 '19 at 11:06

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