# Solving Equation with Second Derivative

1. Jan 4, 2012

### Swimmingly!

1. The problem statement, all variables and given/known data
Solve the following equation for x(y). (use no differential functions)

$image=http://latex.codecogs.com/gif.latex?-x^{-2}=d^{2}x/dy^{2}&hash=0b6db8781e593b1854b7258796e704bf$

x(0)' and x(0) are known.

2. Relevant equations

$image=http://latex.codecogs.com/gif.latex?-x^{-2}=d^{2}x/dy^{2}&hash=0b6db8781e593b1854b7258796e704bf$

3. The attempt at a solution
I'm a bit unsure as to what to do next but I can easily make a messy formula up to approximate the result.

$image=http://latex.codecogs.com/gif.latex?\left\{\begin{matrix}%20x(0)=k_{1}%20\\%20x(0)%27=k_{2}%20\\%20x(y)%27%27=-x^{-2}%20\\%20x(y+E)\approx%20x(y)+x(y)%27\times%20E+x(y)''\times%20E^{2}/2%20\\%20x(y+E)%27\approx%20x(y)%27+%20x(y)%27%27\times%20E%20\end{matrix}\right.&hash=ade0b5b60f18ff4b7daf16c3aac1ab3f$

k1 and k2 are a constant value.
The bigger E is, the bigger the error. For an infinite recursive use of this formula with an infinitesimal E the right result would be achieved. I don't know what it converges to though.

Any help for this problem or/and similar problems would be great. Thank you.

Last edited: Jan 4, 2012
2. Jan 4, 2012

### CompuChip

What does "use no differential functions" mean?

[strike]Why don't you just integrate the whole thing twice, or separate variables?[/strike]

It was a bit more complex than I thought - never mind the second line there[/edit]

Last edited: Jan 4, 2012
3. Jan 4, 2012

### quZz

Hi,
the equation does not contain explicit dependency on y, so try multiplying both sides on dx/dy. You'll get a sort of "conservation of energy" for the problem:
(dx/dy)^2/2 + 1/x = const
const depends on the initial conditions. This way you are left with solving 1st order diff equation

4. Jan 4, 2012

### Swimmingly!

CompuChip:
It means that I want something with which I can calculate values directly. I do not want Integrals or Derivatives in my final formula and I think this is very much possible.

quZz:
I'm not sure what you mean by explicit but for a given k1, k2 and y there's a single well defined value of x.
Also I'm sorry I'm not very good at handling differentials, I don't even have much practice with calculus. I don't understand how you got that from multiplying the first equation by dx/dy. (assuming that was what you did?)

Last edited: Jan 4, 2012
5. Jan 4, 2012

### quZz

so you just need an answer, right?

6. Jan 4, 2012

### Swimmingly!

I've had my time for solving it alone for fun. But it doesn't look so simple and I am extremely curious as for the answer.

I'm trying to find an answer but with the proof too. Otherwise I'll just forget it because I won't understand it.
If you could do that, it'd be really nice. Thanks.

7. Jan 4, 2012

### quZz

try wolframalpha.com...
http://www.wolframalpha.com/input/?_=1325716343605&i=d^2x%2fdy^2%3d-1%2fx^2&fp=1&incTime=true

8. Jan 4, 2012

### HallsofIvy

Staff Emeritus
$$\frac{d^2x}{dy^2}= x^{-2}$$
As quZz said, y does not appear explictly so you can use "quadrature". let v= dx/dy. Then
$$\frac{d^2x}{dy^2}= \frac{dv}{dy}= \frac{dv}{dx}\frac{dx}{dy}= v\frac{dv}{dx}= x^{-2}$$

That is now a separable first order equation:
$$v dv= x^{-2}dx$$
$$\frac{1}{2}v^2= -x^{-1}+ C$$

$$v^2= 2(C- x^{-1})$$
$$v= \frac{dx}{dy}= 2\sqrt{C- x^{-1}}$$
which is also a separable first order equation:
$$\frac{dx}{\sqrt{C- x^{-1}}}= 2dy$$

9. Jan 4, 2012

### Swimmingly!

Thanks!
I didn't know how to write second derivatives in Wolfram Alpha and I thought it wouldn't give a proof for such a complex problem.

Anyway thanks a lot. Problem solved.

Edit: There's a slight mistake on the your result by the way, v^2=2*(C-x^-1). It's times 2, not times "2 squared". It was much easier to follow then wolfram though. Thanks.

Last edited: Jan 4, 2012