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Solving Equation with Second Derivative

by Swimmingly!
Tags: differential, equation, second derivative
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Swimmingly!
#1
Jan4-12, 03:15 PM
P: 45
1. The problem statement, all variables and given/known data
Solve the following equation for x(y). (use no differential functions)



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


2. Relevant equations




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.



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.
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CompuChip
#2
Jan4-12, 03:21 PM
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What does "use no differential functions" mean?

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

[edit]It was a bit more complex than I thought - never mind the second line there[/edit]
quZz
#3
Jan4-12, 03:24 PM
P: 125
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!
#4
Jan4-12, 03:49 PM
P: 45
Solving Equation with Second Derivative

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
#5
Jan4-12, 04:12 PM
P: 125
so you just need an answer, right?
Swimmingly!
#6
Jan4-12, 04:31 PM
P: 45
Quote Quote by quZz View Post
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
#7
Jan4-12, 04:33 PM
P: 125
try wolframalpha.com...
http://www.wolframalpha.com/input/?_=1325716343605&i=d^2x%2fdy^2%3d-1%2fx^2&fp=1&incTime=true
HallsofIvy
#8
Jan4-12, 04:39 PM
Math
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Thanks
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P: 39,340
Your equation is
[tex]\frac{d^2x}{dy^2}= x^{-2}[/tex]
As quZz said, y does not appear explictly so you can use "quadrature". let v= dx/dy. Then
[tex]\frac{d^2x}{dy^2}= \frac{dv}{dy}= \frac{dv}{dx}\frac{dx}{dy}= v\frac{dv}{dx}= x^{-2}[/tex]

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

[tex]v^2= 2(C- x^{-1})[/tex]
[tex]v= \frac{dx}{dy}= 2\sqrt{C- x^{-1}}[/tex]
which is also a separable first order equation:
[tex]\frac{dx}{\sqrt{C- x^{-1}}}= 2dy[/tex]
Swimmingly!
#9
Jan4-12, 05:11 PM
P: 45
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.


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