# Tough ODE problem

1. Sep 19, 2010

### justaboy

1. The problem statement, all variables and given/known data

Solve the ODE:

$$y'(x) = - \frac{y(x)}{\sqrt{a^2-y(x)^2}}$$

3. The attempt at a solution

To be honest I'm having trouble even classifying this ODE. 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. I know I haven't shown much work, but if you could nudge me in the right direction I would be extremely grateful. Thanks.

Last edited: Sep 19, 2010
2. Sep 19, 2010

### snipez90

Isn't this just http://en.wikipedia.org/wiki/Method_of_separation_of_variables" [Broken]?

Last edited by a moderator: May 4, 2017
3. Sep 19, 2010

### justaboy

If you separate it, you get
$$\frac{\sqrt{a^2-y^2}}{y}dy=-dx$$
but I still have the same problem... how do I integrate the left side?

Last edited by a moderator: May 4, 2017
4. Sep 19, 2010

### Staff: Mentor

I would start with a trig substitution, say y = a sin(u).

5. Sep 19, 2010

### justaboy

Okay... after many trig subs, I found

$$a\log(a^3 y)+\sqrt{a^2-y^2}-a\log(2a\sqrt{a^2-y^2}+a) = -x$$

How do I go about finding a y(x)?

6. Sep 19, 2010

### Staff: Mentor

That looks fine, but you forgot to add the constant of integration.
I don't believe that you can solve this equation for y as an explicit function of x. That's OK - just leave it in the form you have.

7. Sep 19, 2010

### justaboy

The question asks to solve for a curve y=y(x) explicitly.

8. Sep 19, 2010

### Staff: Mentor

I don't see any way of solving for y, so I don't have any advice for getting y = f(x), but you could write x = g(y). That would be pretty simple.

9. Sep 20, 2010

### justaboy

Does anyone see a way of solving for y=f(x)?

My teacher is quite positive that it can be done. Apparently this is an old geometry problem that Fermat, Descartes, Bernoulli and Leibniz worked on in the late 17th century... the curve y(x) should be such that the length of the segment of the tangent
to the curve between the curve and the x-axis is equal to a constant "a" at all points on the curve.

Last edited: Sep 20, 2010
10. Sep 20, 2010

### D H

Staff Emeritus
Why don't you just make the substitution like your teacher suggested?

Hint: Differentiate both sides of $z^2=a^2-y^2$ with respect to $x$.

11. Sep 20, 2010

### D H

Staff Emeritus
On the right hand side you need to differentiate z with respect to x, not y. Forget about dz/dy. You want an expression that involves dz/dx. On the left hand side, you already know what dy/dx is. Use that knowledge.

12. Sep 20, 2010

### justaboy

Thanks very much. I end up with this:

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

Which has turned out to be difficult to solve.

13. Sep 20, 2010

### D H

Staff Emeritus
Tough, but doable. You should get x as a function of z, and from that, x as a function of y. Inverting this to get y as a function of x: Hmmmm.

14. Sep 20, 2010

### diggy

Maybe he didn't mean for the z in the substitution to be squared.