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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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    @AndréNicolas: Yes, that's ok! Thank you.2011-12-29

1 Answers 1

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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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    @nepal do click on the tick sign near the question :)2011-12-29