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.
@Ian Fowler Any solution will be ok even if it uses either lambert W function or some non - elementary function. Until the solution is computable it is okay for me.
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.
@Ian Fowler
Thanks for pointing out the typo error The updated solution is here.