# particular result of a differential equation

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:

$$\int{\frac{1}{\sqrt{y^2-1}}dy}=13\int{\frac{1}{\sqrt{x^2-1}}dx}$$

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

$$C=-1$$

and when I plug it in I get

$$y=0$$

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.

• How are you dealing with $\sqrt{-1}$ issues here? – Ian Feb 25 at 21:16
• I haven't come across $\sqrt{-1}$ – S. Kopecký Feb 25 at 21:17
• At your initial point, the square roots involve $\sqrt{-1}$. What are you doing about that? – Ian Feb 25 at 21:18
• 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 $C=0$.) – Ian Feb 25 at 21:20
• 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 $y=\sin(-13 \arcsin(x))$. – Ian Feb 25 at 22:00

Well, we have:

$$\text{y}'\left(x\right)=\text{n}\cdot\frac{\sqrt{\text{y}^2\left(x\right)-1}}{\sqrt{x^2-1}}\space\Longleftrightarrow\space\int\frac{\text{y}'\left(x\right)}{\sqrt{\text{y}^2\left(x\right)-1}}\space\text{d}x=\int\frac{\text{n}}{\sqrt{x^2-1}}\space\text{d}x\tag1$$

Finding the integrals gives:

$$\text{arccosh}\left(\text{y}\left(x\right)\right)=\text{n}\cdot\text{arccosh}\left(x\right)+\text{C}\tag2$$

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

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

So:

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

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

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

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 $$\sqrt{x^2-1}=(x-1)t$$. $$\sqrt{x^2-1}=(x-1)t\implies x=-\dfrac{1+t^2}{1-t^2}\iff \mathrm dx= \dfrac{-4t}{(1-t^2)^2}\mathrm dt$$