Find the root of the given equation in terms of ##u##

[chwala]

Homework Statement

##x^2\sinh 2u +2x - \sinh 2u=0##

Relevant Equations

hyperbolic equations

Hmmmm was a nice one... took me some time to figure out ...seeking alternative ways ...

My working;

##x=\dfrac{-2±\sqrt{4+4 \sinh^2 2u}}{2 \sinh 2u}##

##x=\dfrac{-2±2\sqrt{1+ \sinh^2 2u}}{2 \sinh 2u}##

##x=\dfrac{-1±\sqrt{1+ \sinh^2 2u}}{\sinh 2u}##

##x=\dfrac{-1+\sqrt{1+ \sinh^2 2u}}{\sinh 2u}## or ##x=\dfrac{-1-\sqrt{1+ \sinh^2 2u}}{\sinh 2u}##

##x=\dfrac{-1+1+2\sinh^{2} u}{2\sinh u \cosh u}##

##x=\dfrac{2\sinh^{2} u}{2\sinh u \cosh u}##

##x=\dfrac{\sinh u}{\cosh u }##

 

[fresh_42]

Firstly, we have the possibility ##x=0 \Longleftrightarrow u=0.## Next, we assume ##x,u\neq 0.## Then

\begin{align*}
0&=x^2+\dfrac{2}{\sinh 2u}x-1\\[6pt]
x&=-\dfrac{1}{\sinh 2u} \pm \sqrt{\dfrac{1}{\sinh^2 2u}+1}=-\dfrac{1}{\sinh 2u} \pm \dfrac{\sqrt{1+\sinh^22u}}{\sinh 2u}\\[6pt]
x&=\dfrac{1}{\sinh 2u}\left(-1\pm \cosh 2u\right)=\dfrac{1}{2\sinh u\cosh u}\left(-1\pm \left(1+2\sinh^2 u\right)\right)\\[6pt]
x_1&=\dfrac{\sinh u}{\cosh u}\, , \,x_2=-\dfrac{1+\sinh^2u}{\sinh u\cosh u}=-\dfrac{\cosh u}{\sinh u}\\[6pt]
x_1&=\tanh u=1-\dfrac{2}{e^{2u}+1}\, , \,x_2=-\coth u=-1-\dfrac{2}{e^{2u}-1}
\end{align*}

This is the first half with all you left out. Now, you have to find ##u_1=u_1(x_1)## and ##u_2=u_2(x_2).##

 

[Mark44]

My working;
##x=\dfrac{-2±\sqrt{4+4 \sinh^2 2u}}{2 \sinh 2u}##
##x=\dfrac{-2±2\sqrt{1+ \sinh^2 2u}}{2 \sinh 2u}##
##x=\dfrac{-1±\sqrt{1+ \sinh^2 2u}}{\sinh 2u}##

Sort of a style issue: you don't need to start each equation with "x = ...". Instead, you can have a continued equality.
E.g.:
##x=\dfrac{-2±\sqrt{4+4 \sinh^2 2u}}{2 \sinh 2u} = \dfrac{-2±2\sqrt{1+ \sinh^2 2u}}{2 \sinh 2u}##
##=\dfrac{-1±\sqrt{1+ \sinh^2 2u}}{\sinh 2u}##

##x=\dfrac{-1+\sqrt{1+ \sinh^2 2u}}{\sinh 2u}## or ##x=\dfrac{-1-\sqrt{1+ \sinh^2 2u}}{\sinh 2u}##

No need for the line above. The last expression I wrote gives exactly the same values of x.

##x=\dfrac{-1+1+2\sinh^{2} u}{2\sinh u \cosh u}##

You lost me here -- how did you get from the +/- values of x to this line?

##x=\dfrac{2\sinh^{2} u}{2\sinh u \cosh u}##

##x=\dfrac{\sinh u}{\cosh u }##

The expression on the right can be written as ##\tanh u##.

 

[chwala]

Sort of a style issue: you don't need to start each equation with "x = ...". Instead, you can have a continued equality.
E.g.:
##x=\dfrac{-2±\sqrt{4+4 \sinh^2 2u}}{2 \sinh 2u} = \dfrac{-2±2\sqrt{1+ \sinh^2 2u}}{2 \sinh 2u}##
##=\dfrac{-1±\sqrt{1+ \sinh^2 2u}}{\sinh 2u}##
No need for the line above. The last expression I wrote gives exactly the same values of x.
You lost me here -- how did you get from the +/- values of x to this line?

The expression on the right can be written as ##\tanh u##.

Noted @Mark44 ... I will get back on lost part later in the day...cheers mate

 

[chwala]

Sort of a style issue: you don't need to start each equation with "x = ...". Instead, you can have a continued equality.
E.g.:
##x=\dfrac{-2±\sqrt{4+4 \sinh^2 2u}}{2 \sinh 2u} = \dfrac{-2±2\sqrt{1+ \sinh^2 2u}}{2 \sinh 2u}##
##=\dfrac{-1±\sqrt{1+ \sinh^2 2u}}{\sinh 2u}##
No need for the line above. The last expression I wrote gives exactly the same values of x.
You lost me here -- how did you get from the +/- values of x to this line?

The expression on the right can be written as ##\tanh u##.

by...

##x=\dfrac{-1±\sqrt{1+ \sinh^2 2u}}{\sinh 2u}=\dfrac{-1±\sqrt{\cosh^2 2u}}{\sinh 2u}=\dfrac{-1±\cosh 2u}{\sinh 2u}=\dfrac{-1+1+2\sinh^{2} u}{2\sinh u \cosh u}## ...

 

[fresh_42]

by...

##x=\dfrac{-1±\sqrt{1+ \sinh^2 2u}}{\sinh 2u}=\dfrac{-1±\sqrt{\cosh^2 2u}}{\sinh 2u}=\dfrac{-1±\cosh 2u}{\sinh 2u}=\dfrac{-1+1+2\sinh^{2} u}{2\sinh u \cosh u}## ...

If you write ##x=a\pm b## then this is an abbreviation of ##x\in \{a-b,a+b\}.##

If you split it, then you get ##x=a+b## or ##x=a-b.##

What you cannot do is ##x=a\pm b \neq a +b.##

You should really read what I answer to you. See my post #2. Why do I even read your questions ...

 

[chwala]

If you write ##x=a\pm b## then this is an abbreviation of ##x\in \{a-b,a+b\}.##

If you split it, then you get ##x=a+b## or ##x=a-b.##

What you cannot do is ##x=a\pm b \neq a +b.##

You should really read what I answer to you. See my post #2. Why do I even read your questions ...

aaargh i get you @fresh_42 ...should have double checked... i will amend my response in post ##5## as follows ...just in following with your working,

##x=\dfrac{-1±\sqrt{1+ \sinh^2 2u}}{\sinh 2u}=\dfrac{-1±\sqrt{\cosh^2 2u}}{\sinh 2u}=\dfrac{-1±\cosh 2u}{\sinh 2u}=\dfrac{-1±(1+2\sinh^{2} u)}{2\sinh u \cosh u}##

Now, we have two possibilities;

##x_1=\dfrac{-1+1+2\sinh^{2} u}{2\sinh u \cosh u}=\dfrac{2\sinh^{2} u}{2\sinh u \cosh u}=\dfrac{\sinh u}{\cosh u}##

i can also see different variations on this by you and also by @Mark44 to be specific;

##\dfrac{\sinh u}{\cosh u}=\tanh u##.

Also,

##x_2=\dfrac{-1-1-2\sinh^{2} u}{2\sinh u \cosh u}=\dfrac{-2-2\sinh^{2} u}{2\sinh u \cosh u}=-\dfrac{2(1+\sinh^{2} u)}{2\sinh u \cosh u}=-\dfrac{1+\sinh^{2} u}{\sinh u \cosh u}##

##=-\dfrac{\cosh^{2} u}{\sinh u \cosh u}=-\dfrac{\cosh u}{\sinh u }##

Thanks @fresh_42 , i actually did not go through earlier posts in detail because of other work related duties...regards,

 

[Mark44]

Also,
##x_2= <snip>=-\dfrac{\cosh u}{\sinh u }##

##= -\coth u##.

 

[SammyS]

aaargh i get you @fresh_42 ...should have double checked... i will amend my response in post ##5## as follows ...just in following with your working,
. . .

Thanks @fresh_42 , i actually did not go through earlier posts in detail because of other work related duties...regards,

The last line @fresh_42 wrote in Post #6, which you just responded to.

You should really read what I answer to you. See my post #2. Why do I even read your questions ...

@fresh_42 has some great ending lines. From Post #2 :

This is the first half with all you left out. Now, you have to find ##u_1=u_1(x_1)## and ##u_2=u_2(x_2).##

That "first half" he refers to is your amendment (Post #7).

What you left out is to find ##u## in terms of ##x## .

 

[chwala]

The last line @fresh_42 wrote in Post #6, which you just responded to.

@fresh_42 has some great ending lines. From Post #2 :

That "first half" he refers to is your amendment (Post #7).

What you left out is to find ##u## in terms of ##x## .

Noted...let me look at this later...cheers mate.

 

[chwala]

Just by quick working; i am having

From,

##x^2\sinh 2u+2x-\sinh2u=0##

i will get,

##2x=\sinh 2u - x^2\sinh2u=\sinh 2u(1-x^2)##

therefore,

##\dfrac{2x}{1-x^2}=\sinh 2u##

##2u=\sin^{-1} \left[\dfrac{2x}{1-x^2}\right]####u=\dfrac{1}{2}\sin^{-1} \left[\dfrac{2x}{1-x^2}\right]=\dfrac{1}{2}\sin^{-1} \left[\dfrac{1}{1-x}-\dfrac{1}{1+x}\right]##

 

[Mark44]

Just by quick working; i am having

From,

##x^2\sinh 2u+2x-\sinh2u=0##

i will get,

##2x=\sinh 2u - x^2\sinh2u=\sinh 2u(1-x^2)##

therefore,

##\dfrac{2x}{1-x^2}=\sinh 2u##

The next line below doesn't follow from the line above. With as much LaTeX as you usually put in, it would be a good idea to proofread your work before posting it. The icon in the menu bar that looks like a sheet of paper with a magnifying glass is the Preview button.

##2u=\sin^{-1} \left[\dfrac{2x}{1-x^2}\right]####u=\dfrac{1}{2}\sin^{-1} \left[\dfrac{2x}{1-x^2}\right]=\dfrac{1}{2}\sin^{-1} \left[\dfrac{1}{1-x}-\dfrac{1}{1+x}\right]##

 

[chwala]

The next line below doesn't follow from the line above. With as much LaTeX as you usually put in, it would be a good idea to proofread your work before posting it. The icon in the menu bar that looks like a sheet of paper with a magnifying glass is the Preview button.

Let me amend that now...

 

[chwala]

Amendment to post ##11##

From,

##x^2\sinh 2u+2x-\sinh2u=0##

i will get,

##2x=\sinh 2u - x^2\sinh2u=\sinh 2u(1-x^2)##

therefore,

##\dfrac{2x}{1-x^2}=\sinh 2u##

##2u=\sinh^{-1} \left[\dfrac{2x}{1-x^2}\right]####u=\dfrac{1}{2}\sinh^{-1} \left[\dfrac{2x}{1-x^2}\right]=\dfrac{1}{2}\sinh^{-1} \left[\dfrac{1}{1-x}-\dfrac{1}{1+x}\right]##