Solving Equation with Second Derivative

[Swimmingly!]

Homework Statement

Solve the following equation for x(y). (use no differential functions)

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

Homework Equations

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.

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.

 

[CompuChip]

What does "use no differential functions" mean?

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

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

 

[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

 

[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?)

 

[quZz]

so you just need an answer, right?

 

[Swimmingly!]

so you just need an answer, right?

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.

So, yes please.
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.

 

[quZz]

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

 

[HallsofIvy]

Your equation is
$$\frac{d^2x}{dy^2}= x^{-2}$$
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:
$$\frac{dx}{\sqrt{C- x^{-1}}}= 2dy$$

 

[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.