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Is it possible to get the value of $y$ in terms of $x$ from the below equation? If so please give give me a clue how to do that :) $$y \sqrt{y^2 + 1} + \ln\left(y + \sqrt{y^2 + 1}\right) = \frac{a}{x^2}.$$

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Please verify that the LaTeX I entered gives your intended equation. –  Henry T. Horton Aug 28 '12 at 21:19
    
Here is the link to the equation in the higher quality postimage.org/image/99zftkfaj –  user38959 Aug 28 '12 at 21:19
    
Yes it is correct, thank you very much :) –  user38959 Aug 28 '12 at 21:20
    
Should the equation start $y\sqrt{y^2+1}$ or $y + \sqrt{y^2+1}$? If the answer is the latter then there is a very simple solution. If the answer is the former then the solution is very messy, involving solutions to quadric equations. –  Fly by Night Aug 28 '12 at 21:26
    
Unfortunately, it is a multiplication. Can you tell me please how to start with it? –  user38959 Aug 28 '12 at 21:29

1 Answer 1

up vote 0 down vote accepted

Since $\ \rm{asinh}(y)=\ln\left(y + \sqrt{y^2 + 1}\right)\ $ let's set $\ y:=\sinh(u)\ $ and rewrite your equation as :

$$\sinh(u) \sqrt{\sinh(u)^2 + 1} + u = \frac{a}{x^2}$$ $$\sinh(u) \cosh(u) + u = \frac{a}{x^2}$$ $$\sinh(2u) + 2u = 2\frac{a}{x^2}$$

After that I fear you'll have to solve this numerically (to get $u$ in function of $x$).

I don't see something simpler sorry...


To solve $\ \sinh(w) + w = r\ $ numerically you may :

  • use iterations (Newton-Raphson) : $\ \displaystyle w_{n+1}=w_n-\frac {\sinh(w_n)+w_n-r}{\cosh(w_n)+1}$
    (starting with $w_0=\frac r2$)
  • use reversion of series to get $w=w(r)$ : $w(r)= \frac 12 r - \frac 1{96}r^3 + \frac 1{1920}r^5 - \frac{43}{1290240}r^7 + \frac {223}{92897280}r^9 - \frac{60623}{326998425600}r^{11} + \frac{764783}{51011754393600}r^{13} - \frac {107351407}{85699747381248000}r^{15} + \mathrm{O}\bigl(r^{17}\bigr)$
  • perhaps that other methods exist in the litterature...
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I have started with it and i thought that there will be a way to avaid a numerical solution if I try to do something with these logarithms. I was trying to solve it for 2 hours and nothing so probably you are right and it is simply not possible to do. –  user38959 Aug 28 '12 at 21:48
    
@user1354439: the problem is that you have a $\ln$ in the second term and none in the first : these two terms are of different type ! I'll add some numerical methods in my answer, –  Raymond Manzoni Aug 28 '12 at 21:51
    
No really thanks i was trying to determine the path eqation of the ball thrown with taking drag force ( proportional to the square of the velocity) into account. This equation according to my knowladge does not yet exist despite many attempts to find it. I was stupid that i thought i can do it :) –  user38959 Aug 28 '12 at 22:05
    
@user1354439: too late done! :-) (and with fast convergence) –  Raymond Manzoni Aug 28 '12 at 22:09
    
My reputation is 1 :D i cant give you 1 up :( –  user38959 Aug 28 '12 at 22:15

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