Help for a hyperbolic trig equation

You're very welcome.

Ignore everything above.

Move the inverse hyperbolic tangent term to the RHS

Square both sides

Multiple the 4 in

Subtract 4M

Divide by 4*roh*A1

What's the heck is h? This significantly changes changes the whole complexity of the question.

Faizan: I don't mean to flog the point but you missed the 2nd part of my response.

In line 15 you have: e^2y = B where B = [sqrt(m) +t] / [sqrt(m) - t]

Then on line 17 you have : 2y = B

This is incorrect as line 15 implies that: 2y = ln(B)

So everything that follows doesn't work.

Wasi: I have discovered another problem:

line 15 is O.K.

I assume p is a density, A is a cross-sectional? area and x is a length?

At any rate, assuming m,p,A are positive line 15 does not have a real solution for positive x as [sqrt(m) +sqrt(m+pAx] / [sqrt(m) - sqrt(m+pAx)] will be negative since

sqrt(m+pAx) > sqrt(m).

This leads directly to ln(negative) which is not real.

Or you can look above:

e^2y can never be negative so e^2y = B(negative) has no real solution.

So I need to know more about I,m,p,A,x actually represent. It looks to me like a physics problem. I need information on the domain (possible x values). Because if x is positive then there is no real solution.

I hate to be the fly in the ointment again, but here's throwing caution to the wind.

In line 15 you have some of the form : e^2y = B.

In log form this become: 2y = ln(B),

not 2y = B which what you have in line 17

You can even take the route of e^2y = e^ln(B) , which is unnecessary, but it still leads to 2y = ln(B). So line 17 is missing the ln function.

The bottom line is you have 2y = ln( a function of y) which does not have an elementary solution. Maybe there is some kind of Lambert function solution but that's the best you have.

Wasi: You might try using the infinite Maclauren series for arc_tanh(x) and use a computer to grind out some desired level of accuracy and then go from there or the Lambert function possibility. I think those 2 are your only shot.