0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
0
$\begingroup$

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:

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

$\endgroup$
0
$\begingroup$

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


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


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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.