# 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.