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I try to solve this equation : $$(1+x^{2})\sinh x=2x\cosh x.$$ When I use this equivalence $e^{2x}=\frac{1+x}{1-x},$ this result (i.e. $x=0$) is obtained.
But two other solutions are obtained using this. How to find these solutions?

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3  
Wolfram Alpha "knows" some very obscure named functions. If there were a solution in terms of these, it is fairly likely that it would produce it. The fact that it only gives a numerical result is evidence (but not proof) that the other solutions cannot be so expressed. The phenomenon should not be viewed as surprising. "Most" equations that mix algebraic and transcendental functions do not have closed form solutions. –  André Nicolas Dec 29 '11 at 11:34
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Don't you mean $e^{2x}=\frac{(1+x)^2}{(1-x)^2}$? Because $$\sinh(x)=\frac{e^x-e^{-x}}{2},\qquad\cosh(x)=\frac{e^x+e^{-x}}{2}$$ your equation becomes $$(1+x^2)\left(\frac{e^x-e^{-x}}{2}\right)=2x\left(\frac{e^x+e^{-x}}{2}\right),$‌​$ and multiplying both sides by $2e^x$ (which is never 0), we have $$(1+x^2)(e^{2x}-1)=2x(e^{2x}+1),$$ or rearranging, $$x^2(e^{2x}-1)-2x(e^{2x}+1)+(e^{2x}-1)=0$$ $$(e^{2x}-1)(x^2-2x+1)=4x$$ $$e^{2x}=\frac{(x+1)^2}{(x-1)^2}$$ –  Zev Chonoles Dec 29 '11 at 11:37
    
@ ZevChonoles : yes, I was wrong, but in this stage, $ln\frac{x+1}{x-1}=x$ must be solved,yes? –  nepal Dec 29 '11 at 12:06
    
@nepal: It turns out that yes, though in principle we might need to consider $\ln|\frac{x+1}{x-1}|$. Nice reduction, though from the original equation the root $x=0$ is already clear, since $\sinh 0=0$. What's more, each side of original equation is an odd function ($f(-x)=-f(x)$) so the fact that roots are symmetric about the origin is clear. –  André Nicolas Dec 29 '11 at 15:06
    
@AndréNicolas: Yes, that's ok! Thank you. –  nepal Dec 29 '11 at 19:26

1 Answer 1

up vote 1 down vote accepted

Given

$(1+x^2)\sinh x=2x\cosh x$.

$\sinh x=\frac{e^{x}-e^{-x}}{2}$ and $\cosh x=\frac{e^{x}+e^{-x}}{2}$.

Therefore, $(1+x^2) \frac{e^{x}-e^{-x}}{2}=2x(\frac{e^{x}+e^{-x}}{2})$.

Thus, $(1+x^2)({e^{x}-e^{-x}})=2x({e^{x}+e^{-x}})$.

Thus, $e^{x}(x^2-2x+1)= e^{-x}(x^2+2x+1)$.

Thus, $e^{2x}=\frac{(x+1)^2}{(x-1)^2}$.

Hence, $e^{x}=\frac{(1+x)}{(1-x)}$ or $e^{x}=-\frac{(1+x)}{(1-x)}$.

First case has $x=0$ as a root. Now, consider

Case 1:

$f(x)=e^{x}-\frac{(1+x)}{(1-x)}$.

$f(-2)=e^{-2}-\frac{-1}{3}=-0.19799805$ and $f(-1)=e^{-1}-\frac{0}{2}=0.367879441$. Hence, a root lies between $-2$ and $-1$. You can use Newton-Raphson to get the answe quickly, or, as befits the tag, use old-school bisection method.

So, $f(-1.5)=e^{-1.5}-\frac{-0.5}{2.5}=0.42313016$. Since the signs of $f(-1.5)$ and $f(-2)$ are opposite, a real root lies between 1$-1.5$ and $-2$. Go on iterating like this and you have your negative root $x=-1.54\cdots$.

Now,2 $f(1)$ is not defined. Plus, for $x>1,$ the function is roughly $O(e^x)$and it doesn't have any more real roots greater than $0$.

Case 2:

$f(x)=e^{x}+\frac{(1+x)}{(1-x)}$.

Clearly, one can see that the function is roughly $O(e^x)$ for $x<(-1)$ and that it doesn't have any real roots less than $-1$.

$f(2)=e^{2}+\frac{3}{-1}=4.3890561$ and $f(1.5)=e^{1.5}+\frac{2.5}{-0.5}=-0.51831093$. Since signs of $f(1.5)$ and $f(2)$ are opposite, a real root lies between $1.5$ and $2$. So there you go with bisection method or any other method, and you will get the positive root $x=1.54\cdots$.

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Thank you all for good responces! –  nepal Dec 29 '11 at 14:15
    
@nepal do click on the tick sign near the question :) –  Nikhil Bellarykar Dec 29 '11 at 17:40

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