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I have an expression

$$y+\sqrt{y^2-1}=e^x$$

How to find $y$?


I tried it by squaring both sides, after that I also tried to solve by different substitutions like putting $y = \sec (\theta)$, but nothing worked.

Note: Mathematica may give the final answer but I want to know the intermediate steps hence I prefer solving it by hand, not with Mathematica.

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    $\begingroup$ Perhaps you may try first leave only $\sqrt{y^2-1}$ on the LHS, and then proceed with squaring... $\endgroup$ Nov 6, 2023 at 7:20
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    $\begingroup$ Use $y^2-1=(e^x-y)^2$ then expand the left side and rearrange to get a quadradic in $y$ from which you can use the quadradic formula for the solution. $\endgroup$ Nov 6, 2023 at 7:25
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    $\begingroup$ Try $y=\cosh t$ instead and solve for $t$ in terms of $x$ $\endgroup$ Nov 6, 2023 at 7:26

6 Answers 6

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Using the conjugate $y-\sqrt{y^2-1}≠0$, we have :

$$ \begin{align}\begin{cases}y+\sqrt {y^2-1}=e^x\\ e^x\left(y-\sqrt{y^2-1}\right)=1\end{cases}\end{align} $$

which is equivalent to the following system of equations :

$$ \begin{align} &\begin{cases}y+\sqrt {y^2-1}=e^x\\ y-\sqrt{y^2-1}=e^{-x}\end{cases}\\ \implies &\bbox[5px,border:2px solid #C0A000]{y=\frac {e^x+e^{-x}}{2}\thinspace,\thinspace\thinspace x≥0\thinspace .}\end{align} $$


$\rm {Explanation :}$

We observe that, for the system of equations $\begin{align}\begin{cases}y+\sqrt {y^2-1}=e^x\\ y-\sqrt{y^2-1}=e^{-x}\end{cases}\end{align}$ to be compatible, the necessary and sufficient condition is that :

$$ \begin{align}&e^x≥e^{-x}\\ \iff &x≥-x\\ \iff &x≥0\thinspace .\end{align} $$

Therefore, the original equation or equivalently the above system of equations has a solution iff, when $x≥0$ .

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    $\begingroup$ Always amazed by your clever solutions $\endgroup$
    – PotatoDude
    Nov 6, 2023 at 12:19
  • $\begingroup$ It implies $y=\frac {e^x+e^{-x}}2$ but the converse is false if $x<0.$ $\endgroup$ Nov 6, 2023 at 14:59
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    $\begingroup$ @AnneBauval I considered your criticism and added this unclear point to my answer . Thanks . $\endgroup$ Nov 7, 2023 at 21:51
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For those question which does contain both square root and square, you should consider the $\cosh$ function and $\sinh$ function with their properties. $$\cosh(x) = \dfrac{e^x + e^{-x}}{2}$$ $$\sinh(x) = \dfrac{e^x - e^{-x}}{2}$$ $$\cosh^2(x) - \sinh^2(x) = 1$$ Substitute $y = \cosh(t)$ it will give you $$\cosh(t) + \sqrt{\cosh^2(t)-1} = e^x$$ $$\cosh(t) + \sinh(t) = e^x$$ $$\dfrac{e^t + e^{-t}}{2} + \dfrac{e^t - e^{-t}}{2} = e^x$$ $$e^t = e^x \rightarrow t = x$$ $$y = \cosh(x)$$ Most of the cases, if substitution to $\sin, \cos, \tan, \sec$ not to work, usually $\cosh$ function will do the trick.

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$y+\sqrt{y^2-1}=e^x\longleftrightarrow y^2-1=e^{2x}-2ye^x+y^2\longleftrightarrow2ye^x=e^{2x}+1\longleftrightarrow y=\frac{e^{2x}+1}{2e^x}$

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    $\begingroup$ This is the same as my prior answer, except that you forgot the necessary condition for the first equivalence. $\endgroup$ Nov 6, 2023 at 14:53
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    $\begingroup$ obviously it is good you mentioned it, but they wanted to get this answer, so I didn't overwrite. BTW, @AnneBauval, do you think you could look at my profile and tell me what you think about my questions/answers(becuase a lot of my friends told me that people on this site are "toxic") $\endgroup$ Nov 6, 2023 at 16:08
  • $\begingroup$ @AnneBauval. It's just that you have a lot of reputation so I thought that you would know if I write my questions/answers(in general) good(for math stack exchange,) $\endgroup$ Nov 6, 2023 at 16:42
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I want to present a solution, albeit not as elegant as others; hopefully, you will feel you could have discovered it.

You have already tried using some trigonometric function to substitute $y$. I believe this is due to the $\sqrt{y^2 - 1}$ part in the equation, which looks like the Pythagorean theorem. So let's try something different for $y$ like $y=\cos(\theta)$. Now we have: $$\sqrt{y^2-1} = \sqrt{\cos^2(\theta) - 1} = i\sqrt{1-\cos^2(\theta)} = i \sin(\theta)$$ The left side of the equation becomes $\cos(\theta) + i\sin(\theta)$, which should remind you of some beautiful formula: $$\cos(\theta) + i\sin(\theta) = e^{i\theta}$$ Since the left sides are equal ($y+\sqrt{y^2-1} = \cos(\theta) + i\sin(\theta)$), the right sides have to be equal too: $$e^{i\theta} = e^x \Rightarrow x = i\theta \Leftrightarrow \theta = -i x$$ With this, we have a relation between $x$ and $\theta$, so we can plug it into our substitution for $y$: $$y=\cos(\theta) = \cos(-ix) = \cos(ix)$$ We could stop here, but a complex cosine function is maybe not that pretty. There also exists the hyperbolic trigonometric function so we can write $y=\cosh(x)$, but I do prefer the exponential notation here: $$y = \cos(\theta) = \frac{e^{i\theta} + e^{-i\theta}}{2} = \frac{e^{i \cdot ix} + e^{-i \cdot ix}}{2} = \frac{e^{-x} + e^{x}}{2} = \frac{e^{x} + e^{-x}}{2}$$

Further discussion There is nothing special about $y=\cos(\theta)$, and you could also choose $y=\sin(\theta)$. The resulting equation would look like $$\sin(\theta) + i\cos(\theta) = e^{ix}$$ By dividing both sides with $i$ and a bit of fiddling with the minus sign, you would end up with the same result as above, but with some extra steps!

I hope this helps someone to understand how to derive the result from the equation with a minor substitution. Continuously solving such problems using this method makes you more aware of where you could use substitution and solve equations much faster!

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You need no substitution and no trick:

$$ \begin{align}&y^2-1=(e^x-y)^2\\ \iff &-1=e^{2x}-2ye^x\\ \iff &y=\frac{e^{2x}+1}{2e^x}\end{align} $$

and the constraint for the initial squaring to be legit is $y\le e^x$.

Therefore, the solution if $x\ge0$ is $y=\cosh x$, whereas there is no solution if $x<0.$

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    $\begingroup$ Can the downvoter please explain how this answer could be improved? I did my best for it to be both rigorous and simple. $\endgroup$ Nov 7, 2023 at 22:16
  • $\begingroup$ Yet a second downvote today: what is the matter? $\endgroup$ Nov 8, 2023 at 12:56
  • $\begingroup$ I edited your post and made a little visual improvement. If you do not like my edit, please rollback . $\endgroup$ Nov 8, 2023 at 17:56
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    $\begingroup$ @AnneBauval, Your answer is concise and clear, Also I guess it was the first to point out the condition that $x\geq0$, +1 $\endgroup$ Nov 9, 2023 at 14:39
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Why didn't $y=\sec t$ work?, Let's try

$$\sec t+|\tan t|=e^x\tag 1$$

also use identity $\sec t+|\tan t|=\frac{1}{\sec t-|\tan t|}$

which gives

$$\sec t-|\tan t|=e^{-x} \tag 2$$

adding first $(1)$ and $(2)$

$$\sec t=\frac {e^x+e^{-x}}{2}$$

subtracting $(2)$ from $(1)$

$$|\tan t|=\frac {e^x-e^{-x}}{2}$$

resubstitute $\sec x$=y

$$y=\frac {e^x+e^{-x}}{2}$$

for solution to be valid

$$|\tan t|\geq 0 $$

meaning

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

which yields

$$e^{2x} \geq 1$$

it follows

$$x \geq 0$$

so solution is

$$y=\frac {e^x+e^{-x}}{2}, x \geq 0$$

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  • $\begingroup$ I would appreciate if the downvoter comments if there is something to criticize in this answer, At least leave a comment to let me know where I can Improve $\endgroup$ Nov 9, 2023 at 18:30

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