# X'' - 1/x^2 =0

1. Sep 10, 2007

### bcyang

Can anybody solve this problem?
I don't think this is something that we see in an elementary diff eq textbook.

Yang

2. Sep 10, 2007

### bcyang

Sorry. I should have put this into "diff eq."

3. Sep 10, 2007

### Staff: Mentor

I moved it to Homework Help, Calculus. Homework and coursework questions (even if not for homework, as long as they are homework-like) need to be posted in the Homework Help forums, and you need to show us your work before we can offer tutorial help.

So how would you approach solving this type of equation?

Welcome to the PF, BTW.

4. Sep 10, 2007

### bcyang

Thanks. I was solving (which I thought is) a trivial diff eq.
where there is some repeling force from an origin to an electron, say, at x.
So, the diff eq. really boils down to

x'' - 1/x^2 = 0

which I wonder if it even has a closed form solution.

Yang

5. Sep 10, 2007

### dextercioby

HINT: Multiply the equation by x' and then try to write the whole LHS as a total derivative.

6. Sep 10, 2007

### bcyang

Perhaps a little more hint:)

Thank you, Kurt.

7. Sep 10, 2007

### dextercioby

I'm not Kurt. My name is Daniel. Kurt's a part of the name appearing in the signature.

How about writing it like that ?

$$x' x'' -\frac{x'}{x^{2}}=0$$

8. Sep 10, 2007

### bcyang

Sorry, Daniel. I guess it's been too long since I left college. My brain is all rusty:(

9. Sep 10, 2007

### bcyang

Perhaps $$\frac{d}{dt} (x')^2 = 2 x' x''$$ helps?

10. Sep 10, 2007

### dextercioby

Yes, it does. Also $\frac{d}{dt} \frac{1}{x'}=-\frac{x'}{x^{2}}$.

11. Sep 10, 2007

### bcyang

Thanks a lot. So, I have
$$\frac{1}{2}(x')^2 - \frac{1}{x} = C$$ where C is a constant.
I feel so hopeless. I still don't see how I can solve this:(

12. Sep 10, 2007

### dextercioby

Well, there's a plus before the 1/x. Check it out. Separate the derivative and then separate variables.

13. Sep 10, 2007

### bcyang

Thank you again. This is not how I should approach it, isn't it?
$$x' = 2\sqrt{C-1/x}$$

14. Sep 10, 2007

### dextercioby

Okay, now, do you know how to separate variables ?

15. Sep 10, 2007

### bcyang

Oh, I see. Isn't it
$$\frac{dx}{2\sqrt{C-1/x}} = dt$$

16. Sep 10, 2007

### bcyang

I guess no (though I learned it once upon a time). Sorry.

17. Sep 10, 2007

### bcyang

18. Sep 10, 2007

### dextercioby

Yes, now integrate in both terms.

19. Sep 10, 2007

### bcyang

Thank you, Daniel.