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Solving Equation with Second Derivative 
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#1
Jan412, 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. 


#2
Jan412, 03:21 PM

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What does "use no differential functions" mean?
[edit]It was a bit more complex than I thought  never mind the second line there[/edit] 


#3
Jan412, 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 


#4
Jan412, 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?) 


#5
Jan412, 04:12 PM

P: 125

so you just need an answer, right?



#6
Jan412, 04:31 PM

P: 45

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. 


#7
Jan412, 04:33 PM

P: 125

try wolframalpha.com...
http://www.wolframalpha.com/input/?_=1325716343605&i=d^2x%2fdy^2%3d1%2fx^2&fp=1&incTime=true 


#8
Jan412, 04:39 PM

Math
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PF Gold
P: 39,497

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] 


#9
Jan412, 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*(Cx^1). It's times 2, not times "2 squared". It was much easier to follow then wolfram though. Thanks. 


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