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I was trying to solve this second-order, autonomous, nonlinear, rational equation for $$y:I=[0,\infty)\to\mathbb{R}$$ $$y''=-\frac{k}{y^2}\qquad y(0)=p_0\quad y'(0)=v_0$$ With $k>0$ constant, for all $x\in I$.

However even mathematica failed me;

I am kind of suprised that this is so hard to solve. Here was my attempt: $$\underbrace{y'y''}_{=\frac{1}{2}(y'\,^2)'}=k\frac{y'}{y^2}$$ Integrate both sides with respect to y $$\frac{1}{2}(y')^2=k\frac{1}{y}+c_1$$ $$y'=\pm\sqrt{2\left(k\frac{1}{y}+c_1\right)}$$ (Can we safely assume this square root is real? I think yes, except if the (infinite) 2nd derivative at y=0 lets the function explode in finite time. $c_1$ is chosen such that there definitely is a neighbourhood of 0 where a solution exists.)

This is the point where I get stuck. Firstly: how do I deal with the $\pm$, secondly, is there actually an explicit solution? An Implicit solution is given by

$$\int \frac{1}{\pm\sqrt{2\left(k\frac{1}{y}+c_1\right)}}\text{d}y=x+c_2$$

(Mathematica can integrate it, but cannot solve it); The Implicit solution is given by

$$\frac{y\sqrt{c+\frac{k}{y}}}{\sqrt2c}-\frac{k\log(k+2cy+2y\sqrt{c(c+\frac{k}{y})})}{\sqrt22c^{3/2}}$$

If not: This is a simplification of a 1-dimensional 2-body problem, where one of the masses is chosen so large that its movement can be ignored. I thought the 2-body problem was solved in general, how isnt this possible for this equation?

Is there maybe an easier way to calculate the solution? The method of denoting $u_1=y\quad u_2=y'$ and solving the system of differential equations as we learned in our analysis reading does not really help there, becuase the system is non-linear...

Thank you!

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1  
For your first integration, it should be $+ k/y+c_1$. –  1015 Feb 7 '13 at 17:56
    
@julien I did this error 5 times already today >_< –  CBenni Feb 7 '13 at 17:58
    
So, are you stuck to solve the last integral? –  Babak S. Feb 7 '13 at 18:01
2  
It is actually easier to get a parametric solution than an explicit solution. Take a look at this for the answer of a similar ODE. –  achille hui Feb 7 '13 at 18:46
3  
The parametrization $\frac{dx}{dt} = \frac{1}{\sqrt{2E}} y(t)$ is more or less specific to this problem. but the general idea is you can put whatever you want there as long as you can simplify the equation on $y'$. It is more an art rather than science. In general, if you can bring $y'$ to the square root of a low order polynomial, then there are parametric solutions in terms of well known special functions (sin/cos/sinh/cosh for quadratic, elliptic functions for cubic/quartic polynomials,...) –  achille hui Feb 7 '13 at 19:04

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