# What is the general solution of the differential equation? : y'(coshy)^2=(siny)^2

Sep 8, 2017

We have:

$y ' {\cosh}^{2} y = {\sin}^{2} y$

This is a first order separable Differential equation so we can rearrange the equation as follows:

$y ' {\cosh}^{2} \frac{y}{\sin} ^ 2 y = 1$

So now we can "seperate the variables" to get:

$\int \setminus {\cosh}^{2} \frac{y}{\sin} ^ 2 y \setminus \mathrm{dy} = \int \setminus \mathrm{dx}$

The LHS integral is non-trivial and cannot be solved using analytical methods or expressed in terms of elementary functions, and therefore the full DE solution requires a numerical techniques to solve.

$y ' {\cosh}^{2} y = {\sinh}^{2} y$

Then again we have a separable DE which this time yields:

$\int \setminus {\cosh}^{2} \frac{y}{\sinh} ^ 2 y \setminus \mathrm{dy} = \int \setminus \mathrm{dx}$
$\therefore \int \setminus {\coth}^{2} y \setminus \mathrm{dy} = \int \setminus \mathrm{dx}$
$\therefore \int \setminus {\csch}^{2} y + 1 \setminus \mathrm{dy} = \int \setminus \mathrm{dx}$

Which we can now integrate to get:

$- \coth y + y = x + c$

Which is the GS of the modified equation.