Can anyone solve X'' - 1/x^2 =0?

[bcyang]

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

Yang

 

[bcyang]

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

 

[berkeman]

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

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.

 

[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.
Thank you all in advance.

Yang

 

[dextercioby]

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

 

[bcyang]

Perhaps a little more hint:)

Thank you, Kurt.

 

[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 ?

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

 

[bcyang]

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

 

[bcyang]

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

 

[dextercioby]

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

 

[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:(

 

[dextercioby]

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

 

[bcyang]

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

 

[dextercioby]

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

 

[bcyang]

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

 

[bcyang]

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

 

[bcyang]

No, I don't know how to separate variables. Please help...

 

[dextercioby]

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

Yes, now integrate in both terms.

 

[bcyang]

Thank you, Daniel.