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

  • Thread starter bcyang
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In summary, the conversation is about solving a differential equation involving an electron and an origin. The equation is simplified to x'' - 1/x^2 = 0 and it is questioned if it has a closed form solution. Hints are given to use the total derivative and separate variables in order to solve the equation. The conversation ends with the integration of both terms to solve the equation.
  • #1
bcyang
13
0
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
 
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  • #2
Sorry. I should have put this into "diff eq."
 
  • #3
bcyang said:
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.
 
  • #4
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
 
  • #5
HINT: Multiply the equation by x' and then try to write the whole LHS as a total derivative.
 
  • #6
Perhaps a little more hint:)

Thank you, Kurt.
 
  • #7
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 ?

[tex]x' x'' -\frac{x'}{x^{2}}=0 [/tex]
 
  • #8
Sorry, Daniel. I guess it's been too long since I left college. My brain is all rusty:(
 
  • #9
Perhaps [tex]\frac{d}{dt} (x')^2 = 2 x' x'' [/tex] helps?
 
  • #10
Yes, it does. Also [itex]\frac{d}{dt} \frac{1}{x'}=-\frac{x'}{x^{2}} [/itex].
 
  • #11
Thanks a lot. So, I have
[tex] \frac{1}{2}(x')^2 - \frac{1}{x} = C [/tex] where C is a constant.
I feel so hopeless. I still don't see how I can solve this:(
 
  • #12
Well, there's a plus before the 1/x. Check it out. Separate the derivative and then separate variables.
 
  • #13
Thank you again. This is not how I should approach it, isn't it?
[tex] x' = 2\sqrt{C-1/x} [/tex]
 
  • #14
Okay, now, do you know how to separate variables ?
 
  • #15
Oh, I see. Isn't it
[tex] \frac{dx}{2\sqrt{C-1/x}} = dt [/tex]
 
  • #16
I guess no (though I learned it once upon a time). Sorry.
 
  • #17
No, I don't know how to separate variables. Please help...
 
  • #18
bcyang said:
Oh, I see. Isn't it
[tex] \frac{dx}{2\sqrt{C-1/x}} = dt [/tex]

Yes, now integrate in both terms.
 
  • #19
Thank you, Daniel.
 

1. What is the value of x that solves the equation 1/x^2 = 0?

The answer is that there is no real value of x that solves this equation. When x approaches infinity or negative infinity, the value of 1/x^2 approaches 0, but it never equals 0.

2. Can any number be substituted for x to make the equation 1/x^2 = 0 true?

No, there is no real number that can be substituted for x to make this equation true. As mentioned before, the value of 1/x^2 never equals 0.

3. Is this equation solvable using basic algebra?

No, this equation cannot be solved using basic algebraic methods. It involves a division by 0, which is undefined in mathematics.

4. What is the significance of this equation in mathematics?

This equation is significant because it is an example of an indeterminate form in calculus. It helps illustrate the concept of limits and how certain values may approach a limit but never reach it.

5. Can this equation be solved using advanced mathematical techniques?

Yes, this equation can be solved using advanced mathematical techniques such as complex analysis or calculus of variations. However, the solution will still involve the concept of a limit and will not result in a single real value for x.

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