# I am having trouble figuring out a method for finding

I am having trouble figuring out a method for finding a solution to
$$\displaystyle{r}+{\left({\frac{{{2}}}{{{r}}}}\right)}{\left({r}'\right)}^{{2}}-{r}{''}={0}$$
I have tried substitution of $$\displaystyle{w}={r}'$$ to obtain
$$\displaystyle{r}+{\left({\frac{{{2}}}{{{r}}}}\right)}{w}^{{2}}={\frac{{{d}{w}}}{{{d}{r}}}}{w}$$
But then I am not sure how to proceed from there.
Any suggestions are welcome. Tha

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navratitehf
What if we let PSKu(t)=1/r(t)? Then
$$\displaystyle{r}'=-{\frac{{{1}}}{{{u}^{{2}}}}}{u}'$$ and $$\displaystyle{r}{''}={\frac{{{2}}}{{{u}^{{3}}}}}{\left({u}'\right)}^{{2}}-{\frac{{{1}}}{{{u}^{{2}}}}}{u}{''}$$
Hence
$$\displaystyle{r}+{\frac{{{2}}}{{{r}}}}{\left({r}'\right)}^{{2}}-{r}{''}={\frac{{{1}}}{{{u}}}}+{2}{u}{\left(-{\frac{{{1}}}{{{u}^{{2}}}}}{u}'\right)}^{{2}}-{\frac{{{2}}}{{{u}^{{3}}}}}{\left({u}'\right)}^{{2}}+{\frac{{{1}}}{{{u}^{{2}}}}}{u}{''}$$
$$\displaystyle={\frac{{{u}+{u}{''}}}{{{u}^{{2}}}}}$$
Can you take it from here?
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Donald Valley
A side note - you were so close following your approach.
$$\displaystyle{r}+{\frac{{{2}}}{{{r}}}}{w}^{{2}}={w}'{w}={\frac{{{1}}}{{{2}}}}{\frac{{{d}}}{{{d}{r}}}}{w}^{{2}}$$
sitting $$\displaystyle{w}^{{2}}={u}$$ then you get
$$\displaystyle{\frac{{{1}}}{{{2}}}}{u}'-{\frac{{{2}}}{{{r}}}}{u}={r}$$