# Math Help - Tough ODE problem

1. ## Tough ODE problem

Solve the ODE:

$\frac{dy}{dx}=-\frac{y}{\sqrt{a^2-y^2}}$

3. The attempt at a solution

My teacher hinted that the substitution $z^2=a^2-y^2$ could be helpful, but once I make the substitution, I can't seem to take the next step.

$z^2=a^2-y^2$

Differentiating WRT x, we have:

$2z\frac{dz}{dx}=-2y\frac{dy}{dx}$

Substituting back in we have:

$\frac{dz}{dx}=\frac{a^2-z^2}{z^2}$

But I can't take it from here... much less obtain an equation for y(x). Any thoughts?

2. Originally Posted by justaboy
Solve the ODE:

$\frac{dy}{dx}=-\frac{-y}{\sqrt{a^2-y^2}}$

3. The attempt at a solution

My teacher hinted that the substitution $z^2=a^2-y^2$ could be helpful, but once I make the substitution, I can't seem to take the next step.

$z^2=a^2-y^2$

Integrating WRT x, we have:

Substituting back into the equation, we have:

$\frac{dz}{dx}=\frac{a^2-z^2}{z^2}$

But I can't take it from here... much less obtain an equation for y(x). Any thoughts?
$\displaystyle \frac{dx}{dy} = \frac{- \sqrt{a^2 - y^2}}{y} \Rightarrow x = - \int \frac{\sqrt{a^2 - y^2}}{y} \, dy$.

Now use the substitution $u = a^2 - y^2$ etc.

3. Originally Posted by mr fantastic
$\displaystyle \frac{dx}{dy} = \frac{\sqrt{a^2 - y^2}}{y} \Rightarrow x = \int \frac{\sqrt{a^2 - y^2}}{y} \, dy$.

Now use the substitution $u = a^2 - y^2$ etc.
Ok, making the sub $z^2=a^2-y^2$ and plugging that into $x = -\int \frac{\sqrt{a^2 - y^2}}{y} \, dy$ I get

$x = \int \frac{z}{a^2 - z^2} \, dz$. Sort of the same as I ended up with in the OP, only the inverse. How do I take this from here?

4. Originally Posted by justaboy
Ok, making the sub $z^2=a^2-y^2$ and plugging that into $x = -\int \frac{\sqrt{a^2 - y^2}}{y} \, dy$ I get

$x = \int \frac{z}{\sqrt{a^2 - z^2}} \, dz$. Sort of the same as I ended up with in the OP, only inversed. How do I take this from here?

5. You're the boss. If $u=a^2-y^2$ then

$x = \int \frac{du}{2u\sqrt{a^2-u^2}}$

6. Originally Posted by justaboy
You're the boss. If $u=a^2-y^2$ then

$x = \int \frac{du}{2u\sqrt{a^2-u^2}}$
No, you don't get that. Nevertheless, what you do get is not directly helpful. So forget my initial suggestion. The given hint of $u^2 = a^2 - y^2$ is better.

You get $\displaystyle \int \frac{u^2}{a^2 - u^2} \, du = - \int \frac{u^2}{u^2 - a^2} \, du = - \int 1 + \frac{a^2}{u^2 - a^2} \, du$.

Partial fraction decomposition should be used to integrate the second term.

7. Originally Posted by mr fantastic

You get $\displaystyle \int \frac{u^2}{a^2 - u^2} \, du = - \int \frac{u^2}{u^2 - a^2} \, du = - \int 1 + \frac{a^2}{u^2 - a^2} \, du$.
I had accidently added a negative sign to the initial equation in my first post... anyways, with that in mind, I get:

$\int \frac{u^2}{a^2-u^2} \, du$

so $x=\frac{1}{2}(-a\log(u-a)+a\log(a+u)-2u)+c$

8. Originally Posted by justaboy
I had accidently added a negative sign to the initial equation in my first post... anyways, with that in mind, I get:

$\int \frac{u^2}{a^2-u^2} \, du$

so $x=\frac{1}{2}(-a\log(u-a)+a\log(a+u)-2u)+c$
That looks OK: http://www.wolframalpha.com/input/?i...-+y%5E2%5D%2Fy

(Note that the form given by Wolfram is not the simplest)

9. Wolfram also returns this solution:

$x = a*arctanh(\frac{\sqrt{a^2-y^2}}{a})-\sqrt{a^2-y^2}+c$

In either case is it possible to solve for y(x)?

Thanks for you time.

10. Originally Posted by justaboy
Wolfram also returns this solution:

$x = a*arctanh(\frac{\sqrt{a^2-y^2}}{a})-\sqrt{a^2-y^2}+c$

In either case is it possible to solve for y(x)?

Thanks for you time.
No.

11. I just want to point out that there are integrals of the following type
$\displaystyle\int x^m(a+bx^n)^p dx$
that can be solved only in three cases:

1. $p\in\mathbb{Z}$
2. $\frac{m+1}{n}\in\mathbb{Z}$ substitute $a+bx^n=z^s$ where $s$ is the denominator of the fraction $p$
3. $\frac{m+1}{n}+p\in\mathbb{Z}$ substitute $ax^{-n}+b=z^s$ where again $s$ is the denominator of the fraction $p$

The integral here is one such integral $\displaystyle\int\frac{\sqrt{a^2-y^2}}{y}dy=\int y^{-1}(a^2-y^2)^{\frac{1}{2}}dy$ in which $m=-1$, $n=2$ and $p=\frac{1}{2}$. Notice that $\frac{m+1}{n}=0\in\mathbb{Z}$ hence it can be solved using the substitution $a^2-y^2=z^2.$

Of course, your teacher gave you that as a hint, and you and the rest of the crew here solved the problem. I think it is worth noting that this integral is nothing special.

12. I don't think my prof would have assigned an impossible question to me though... there's no way to find y(x)?

13. Originally Posted by justaboy
I don't think my prof would have assigned an impossible question to me though... there's no way to find y(x)?
There is nothing wrong with an implicit solution like the one that has been found. I suggest you ask your professor what his/her expectation is.