Question

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

Answer 1

We have:

`y' cosh^2y=sin^2y`

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

`y' cosh^2y/sin^2y = 1`

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

`int \ cosh^2y/sin^2y \ dy = int \ 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.

If however, the equation is incorrect and should instead read:

`y' cosh^2y=sinh^2y`

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

`int \ cosh^2y/sinh^2y \ dy = int \ dx`
`:. int \ coth^2y \ dy = int \ dx`
`:. int \ csch^2y+1 \ dy = int \ dx`

Which we can now integrate to get:

`-cothy+y = x + c`

Which is the GS of the modified equation.