I have an expression


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


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

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.


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

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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

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!


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

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

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


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

which yields

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

it follows

$$x \geq 0$$

so solution is

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

  • $\begingroup$ 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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