Homework Help: First Order ODE

1. Oct 8, 2007

robbondo

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

$$x^{2} y y\prime = (y^{2} - 1)^{\frac{3}{2}}$$

2. Relevant equations

3. The attempt at a solution

I know I have to use separation of variables because it isn't linear.

so I get

$$\frac{ydy}{(y^{2} - 1)^{\frac{3}{2}}} = \frac{1}{dxx^{2}}$$

Now I'm kinda stuck at how to integrate this. I think I'm supposed to use partial fraction expansion, but since there's the 3/2 exponent, I'm confused as to how to go about doing that. Tips???

2. Oct 8, 2007

dashkin111

Check the right hand side of your equation to make sure the dx is in the right spot. Try substitution on the left

3. Oct 8, 2007

robbondo

Thanks, that helped. So now I substituted take the integral of both sides and I get

$$x = (y^{2}-1)^{\frac{1}{2}} + c$$

This isn't matching up with the correct answer though. I took the integral and got -1/x on the left and 1 / - (y^2 - 1)^1/2 + c on the right. So I just took out the negative signs and changed to the reciprocal on the both sides, whatcha think?

4. Oct 8, 2007

rock.freak667

Well that would be correct if the constant wasn't there but because it is...you just can't simply invert both sides

if you brought $$\frac{1}{\sqrt{y^2-1}} +c$$ to the same base and then invert...that would be correct

5. Oct 8, 2007

robbondo

ahh I see. I got the correct answer the only thing that confuses me is that in the book the answer says $$y \equiv \pm 1$$... Actually I still don't really know what "equivalent" represents in this class.

6. Oct 8, 2007

rock.freak667

well normally for the function to be continuous i thought it would be y not equal to +/- 1

7. Oct 8, 2007

Dick

In addition to the solutions you get by integrating, the equation has solutions where y' is identically equal to zero. These are them.

8. Oct 8, 2007

robbondo

So if if it didn't say " all solutions " I could just say that y=1, but what's the point of having the weird three line equal sign?

9. Oct 8, 2007

Dick

It just means y(x)=1 for all x. I.e. y is IDENTICALLY equal to one. Sure, if you don't need all solutions, then you could just pick an easy one.