I stumbled upon this differential equation I would like to solve.

$y'=13\frac{\sqrt{y^2-1}}{\sqrt{x^2-1}}$ for $y(0)= 0$

I searched everywhere and tried all of the online calculators for differential equations. But I couldn't find anything that would work for this equation.

I got to the part where I need to integrate this:


When I used the integral calculator to find the solution to these and put them in and solve for

$x=0$ and $y=0$

I always get a solution that the constant equals -1


and when I plug it in I get


which doesn't make any sense. Although I think my problem is that I can't figure out how to simplify the equation to the right format to solve for easier integrals. If anyone could explain me how to do that that would be enough and I could continue by myself.

Thank you.

  • $\begingroup$ How are you dealing with $\sqrt{-1}$ issues here? $\endgroup$ – Ian Feb 25 at 21:16
  • $\begingroup$ I haven't come across $\sqrt{-1}$ $\endgroup$ – S. Kopecký Feb 25 at 21:17
  • $\begingroup$ At your initial point, the square roots involve $\sqrt{-1}$. What are you doing about that? $\endgroup$ – Ian Feb 25 at 21:18
  • 1
    $\begingroup$ The thing is, that's built into how the problem is defined, the problem makes no sense until you've decided how to interpret that. (For example, what even is $y'$ at the initial point? Is it $+13$? If so, then you probably actually want to look at $y'=13 \frac{\sqrt{1-y^2}}{\sqrt{1-x^2}}$, in which case you can easily solve to get $\arcsin(y)=13\arcsin(x)+C$ and for your initial point you have $C=0$.) $\endgroup$ – Ian Feb 25 at 21:20
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    $\begingroup$ Note that really the only other interpretation that makes any sense is that the two $\sqrt{-1}$'s have the opposite sign, in which case $y=\sin(-13 \arcsin(x))$. $\endgroup$ – Ian Feb 25 at 22:00

Well, we have:


Finding the integrals gives:


Now, using $\text{y}(0)=0$ we get:

$$\text{arccosh}\left(0\right)=\text{n}\cdot\text{arccosh}\left(0\right)+\text{C}\space\Longleftrightarrow\space\text{C}=-\frac{\pi i}{2}\cdot\left(\text{n}-1\right)\tag3$$


$$\text{arccosh}\left(\text{y}\left(x\right)\right)=\text{n}\cdot\text{arccosh}\left(x\right)-\frac{\pi i}{2}\cdot\left(\text{n}-1\right)\tag4$$


$$\dfrac{\mathrm dy}{\mathrm dx}=13\sqrt{\dfrac{y^2-1}{x^2-1}} \\ \int\dfrac{\mathrm dy}{\sqrt{y^2-1}}=13\int\dfrac{\mathrm dx}{\sqrt{x^2-1}}\\ \cosh^{-1}y =13\cosh^{-1}x+C$$

$$y(0)=0 \iff \dfrac{i\pi}{2}=\dfrac{13i \pi }{2}+C\implies C=-6i\pi \\ \boxed{\cosh^{-1}y=13\cosh^{-1}x-6i\pi}$$

You can solve $\int \mathrm dx/\sqrt{x^2-1}$ using a lot many ways. I present three of them here:

  1. Know that the derivative of $\cosh^{-1}x$ is $1/\sqrt{x^2-1}$. $$\dfrac{\mathrm d}{\mathrm dx}\cosh^{-1}x=\dfrac{1}{\sqrt{x^2-1}}$$
  2. Make a Trig Substitution. $\sec\theta$ should work and pave your way through to the anti-derivative $$\begin{bmatrix}x \\ \mathrm dx\end{bmatrix}=\begin{bmatrix}\sec\theta \\ 2\sec^2\theta \tan\theta \mathrm d\theta\end{bmatrix} \\ \int \dfrac{\mathrm dx}{\sqrt{x^2-1}}=\int\dfrac{\sec\theta\tan\theta\mathrm d\theta}{\tan\theta}=\ln\mid\sec\theta+\tan\theta\mid +C\\ \int\dfrac{\mathrm dx}{\sqrt{x^2-1}}=\ln\mid x +\sqrt{x^2-1}\mid+C$$

  3. Make the Euler Substitution of the $3^{\text{rd}}$ kind namely let $\sqrt{x^2-1}=(x-1)t$. $$\sqrt{x^2-1}=(x-1)t\implies x=-\dfrac{1+t^2}{1-t^2}\iff \mathrm dx= \dfrac{-4t}{(1-t^2)^2}\mathrm dt$$


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