I am having trouble figuring out a method for finding

khi1la2f1qv 2021-11-21 Answered
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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Expert Answer

navratitehf
Answered 2021-11-22 Author has 273 answers
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
Answered 2021-11-23 Author has 284 answers
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
\(\displaystyle{\frac{{{1}}}{{{2}}}}{u}'-{\frac{{{2}}}{{{r}}}}{u}={r}\)
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Thank you and my apologies.

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