I've been staring at this for a while and looking around the internet to see if I can find a solution, but no success. I think it probably has an exact solution, since I got it from a first year college physics test. In particular, this equation describes the motion of a body falling in a gravitational field.
How to solve $y'' = -\frac{k}{y^2}$, with k > 0?
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$\begingroup$
ordinary-differential-equations
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0it makes a difference with simple harmonic motion of the form $y''=ky$ so I would not be surprised if it made a difference here – 2012-02-28
1 Answers
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Mutiply by 2y' to get
(y'^2)' = \left(\frac{-2k}{y}\right)'
Thus giving
y'^2 = C - \frac{2k}{y}
which should be solveable by taking square roots etc (the sign depending on the physical assumptions)
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0It wasn't my physics test, it was one some friends in my same year did. But never mind that, thank you for all your help. – 2012-02-29