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I've already tried to simplify it as: $$e^{2x}(x-1)-x=0$$

And that gives me: $e^{2x}=\frac{x}{x-1}$

so $2x = \ln(\frac{x}{x-1}) => 2x = \ln(x)-\ln(x-1)$

But then I am stuck. If anyone has a trick or anything?

Thank you in advance.

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    i think it is only possibe by a numerical method – 2017-01-07
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    @A.Molendijk No you did not. – 2017-01-07

3 Answers 3

3

Divide both sides by $2$ and recall the hyperbolic sine function:

$$x\frac{e^x-e^{-x}}2-\frac{e^x}2=0$$

$$x\sinh(x)-\frac{e^x}2=0$$

Now we employ Newton's method:

$$x_{n+1}=x_n-\frac{x_n\sinh(x_n)-\frac{e^{x_n}}2}{\sinh(x_n)+x_n\cosh(x_n)-\frac{e^{x_n}}2}$$

With $x_0=1$,

$x_1=1.135335283$

$x_2=1.119569316$

$x_3=1.119322977$

$x_4=1.119322917$

And so that is the solution near $x_0=1$. With $x_0=-0.5$,

$x_1=-0.530772575$

$x_2=-0.530045575$

$x_3=-0.530045160$

$x_4=-0.530045160$

And that is the desired solution near $x=-0.5$

2

with a substitution it looks nicer setting $$t=e^x$$ then we have $$t^2(\ln(t)-1)-\ln(t)=0$$ and now see the Newton-Raphson method

2

Since $e^x-e^{-x}=2\sinh x$ and $e^x=\sinh x+\cosh x$, we can rewrite the equation as $2x\sinh x-\sinh x-\cosh x=0$ or $$ 2x=1+\coth x $$ (after noticing that $x=0$ is not a solution).

If we consider $f(x)=2x-1-\coth x$, we have $$ f'(x)=2-(1-\coth^2x)=1+\coth^2x $$ which is positive. So the function is increasing in both the intervals $(-\infty,0)$ and $(0,\infty)$.

Since $$ \lim_{x\to-\infty}f(x)=-\infty,\quad \lim_{x\to0^-}f(x)=\infty,\quad \lim_{x\to0^+}f(x)=-\infty,\quad \lim_{x\to\infty}f(x)=\infty $$ we know that the equation has two solutions.