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I'm trying to figure out how we go from:

$$v \frac{\mathrm dv}{\mathrm dr} = \frac{c_s^2}{r^2v}\frac{\mathrm d}{\mathrm dr}\left(r^2v\right) - \frac{GM_{\odot}}{r^2}$$

to

$$\frac{1}{v}\frac{\mathrm dv}{\mathrm dr}\left(c_s^2-v^2\right) = \frac{2c_s^2}{r}-\frac{GM_{\odot}}{r}$$

where $c_s^2$ is isothermal sound speed.

How would I take this further and solve the following:

I'm stuck on how to go from:

$$\left(v^2-c_s^2\right)\frac{1}{v}\frac{\mathrm dv}{\mathrm dr} = 2\frac{c_s^2}{r^2}(r-r_c) $$

to:

$$\left(\frac{v}{c_S}\right)^2-\ln\left(\frac{v}{c_s}\right)^2= 4 \ln\left(\frac{r}{r_c}\right)+4\frac{r_c}{r}+C$$

I've tried integrating it as you would a regular ODE but get stuck on the answer of:

$$\frac{v^2}{2}-c_s^2\ln(v) = 2c_s^2\left[\ln(r)+\frac{r_c}{r}\right]+C$$

Apologies if this is a duplicate, I made some errors to the previous question and felt I should add extra detail. This problem is taking the previously derived expression and solving it as a separable ODE

My attempt:

Integrate both sides with respect to $r$ to get $$ \int\frac{v(r)^{2}-c_{s}^{2}}{v(r)}\frac{dv}{dr}(r)dr=2c_{s}^{2}\int\frac{r-r_{c}}{r^{2}}dr. $$ Perform the substitution $u\equiv v(r)$ to get $$ \int\frac{u^{2}-c_{s}^{2}}{u}du=2c_{s}^{2}\int\frac{r-r_{c}}{r^{2}}dr. $$ Therefore, $$ \left.\frac{u^{2}}{2}-c_{s}^{2}\log u\right|_{v(a)}^{v(b)}=\left.2c_{s}^{2}\left(\frac{r_{c}}{r}+\log r\right)\right|_{a}^{b}. $$ Now, multiply both sides by $2/c_{s}^{2}$ to get $$ \left.\frac{u^{2}}{c_{s}^{2}}-2\log u\right|_{v(a)}^{v(b)}=4\left.\left(\frac{r_{c}}{r}+\log r\right)\right|_{a}^{b}. $$ Since $c_{s}$ and $r_{c}$ are constants, this is equivalent to $$ \left.\frac{u^{2}}{c_{s}^{2}}-2\log u-2\log c_{s}^{2}\right|_{v(a)}^{v(b)}=4\left.\left(\frac{r_{c}}{r}+\log r-\log r_{c}\right)\right|_{a}^{b}. $$ Simplifying, $$ \left.\frac{u^{2}}{c_{s}^{2}}-\log\left(\frac{u}{c_{s}^{2}}\right)^{2}\right|_{v(a)}^{v(b)}=4\left.\left(\frac{r_{c}}{r}+\log\left(\frac{r}{r_{c}}\right)\right)\right|_{a}^{b}. $$

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  • $\begingroup$ Have you considered the chain rule? $\endgroup$ – Kyle Kanos Nov 16 at 18:28
  • $\begingroup$ It's not a derivative. It's the parker solar model? $\endgroup$ – H98 Nov 16 at 18:28
  • $\begingroup$ Read the article DYNAMICS OF THE INTERPLANETARY GAS AND MAGNETIC FIELDS* by E. N. Parker $\endgroup$ – Alex Trounev Nov 16 at 18:29
  • $\begingroup$ Just take the derivative of the first term on the right and rearrange, is really as simple as that $\endgroup$ – Kyle Kanos Nov 16 at 18:29
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    $\begingroup$ Why is the last term changing from $\frac{GM_{\odot}}{r^2} \rightarrow \frac{GM_{\odot}}{r}$? Otherwise it's a derivative chain rule and a rearrangement. $\endgroup$ – Cinaed Simson Nov 16 at 19:58
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This is a straight-forward application of the product rule $$ \frac{\mathrm d}{\mathrm dr}\left(r^2v\right)=\frac{\mathrm dr^2}{\mathrm dr}\cdot v+r^2\cdot\frac{\mathrm dv}{\mathrm dr} $$ After a quick simplification of the above, it's just a matter of rearranging the variables to obtain the second equation.

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  • $\begingroup$ I'm getting that but now how to rearrange $\frac{1}{v}\frac{\mathrm dv}{\mathrm dr}\left(c_s^2-v^2\right)$ $\endgroup$ – H98 Nov 16 at 18:45
  • $\begingroup$ By adding and subtracting, then factoring $\endgroup$ – Kyle Kanos Nov 16 at 19:07

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