# Solving Equation with Second Derivative

by Swimmingly!
Tags: differential, equation, second derivative
 P: 45 1. The problem statement, all variables and given/known data Solve the following equation for x(y). (use no differential functions) $-x^{-2}=d^{2}x/dy^{2}$ x(0)' and x(0) are known. 2. Relevant equations $-x^{-2}=d^{2}x/dy^{2}$ 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. $\left\{\begin{matrix} x(0)=k_{1} \\ x(0)'=k_{2} \\ x(y)''=-x^{-2} \\ x(y+E)\approx x(y)+x(y)'\times E+x(y)''\times E^{2}/2 \\ x(y+E)'\approx x(y)'+ x(y)''\times E \end{matrix}\right.$ 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.
 Sci Advisor HW Helper P: 4,300 What does "use no differential functions" mean? Why don't you just integrate the whole thing twice, or separate variables? It was a bit more complex than I thought - never mind the second line there[/edit]
 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
 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?)
 P: 125 so you just need an answer, right?
P: 45
 Quote by quZz 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.

 Math Emeritus Sci Advisor Thanks PF Gold P: 39,682 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: $$\frac{dx}{\sqrt{C- x^{-1}}}= 2dy$$