Editing SSC/Dynamics/IsothermalSimilaritySolution (section)

==Combined==
Now, before returning to the first ODE, let's write <math>~\Delta U</math> in terms of <math>~\ell</math> and, hereafter, use <math>~\ell</math> as the unknown instead of <math>~U_2</math>.
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\Delta U \equiv U_2 - U_1</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(2\bar{U} - U_1) - U_1</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2\biggl[\bar{U} - U_1 \biggr]</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2\biggl[\frac{(\ell - 1)}{\zeta} - U_1 \biggr]</math>
  </td>
</tr>
</table>
</div>

Hence, the first ODE gives,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~
\ell^2 \Rho_2  
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2\biggl\{ 2\ell +\biggl[ \ell^2 - \zeta^2 \biggr]\frac{\Delta U}{\Delta\zeta} \biggr\} - \ell^2 P_1 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~
\Delta\zeta \ell^2 \Rho_2  
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(4\ell - \ell^2 P_1)\Delta\zeta  + 2( \ell^2 - \zeta^2 ) \Delta U  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(4\ell - \ell^2 P_1)\Delta\zeta  + 4( \ell^2 - \zeta^2 ) \biggl[\frac{(\ell - 1)}{\zeta} - U_1 \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~
(2 \zeta \ell^2 \Delta\zeta )\Rho_2  
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2(4\ell - \ell^2 P_1) \zeta\Delta\zeta  + 8( \ell^2 - \zeta^2 ) (\ell - 1 - \zeta U_1 ) \, . 
</math>
  </td>
</tr>
</table>
</div>

Finally, using this to replace <math>~\Rho_2</math> in the second ODE expression gives,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~
2(4\ell - \ell^2 P_1) \zeta\Delta\zeta  + 8( \ell^2 - \zeta^2 ) (\ell - 1 - \zeta U_1 ) 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\ell [ (4 - 2\Rho_1 \ell )\zeta \Delta\zeta - 4( \ell^2 - \zeta^2) ] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
\pm \ell \biggl\{ 
[ (4 - 2\Rho_1 \ell )\zeta \Delta\zeta - 4( \ell^2 - \zeta^2) ]^2 
+ 4\ell \zeta \Delta\zeta[ ( 4\Rho_1 - \Rho_1^2 \ell )\zeta \Delta\zeta + 4( \ell^2 - \zeta^2) P_1 ] 
\biggr\}^{1 / 2} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~
\frac{8}{\ell} ( \ell^2 - \zeta^2 ) (\ell - 1 - \zeta U_1 ) 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-4 \zeta \Delta\zeta   - 4( \ell^2 - \zeta^2)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
\pm \biggl\{ 
\biggl[ (4 - 2\Rho_1 \ell )\zeta \Delta\zeta - 4( \ell^2 - \zeta^2) \biggr]^2 
+ 4\ell \zeta \Delta\zeta \biggl[ ( 4\Rho_1 - \Rho_1^2 \ell )\zeta \Delta\zeta + 4( \ell^2 - \zeta^2) P_1 \biggr] 
\biggr\}^{1 / 2} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~
\frac{1}{\ell}  ( \ell^2 - \zeta^2 )\biggl[ 2  (\ell - 1 - \zeta U_1 ) + \ell \biggr] + \zeta \Delta\zeta
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\pm \frac{1}{4} \biggl\{ 
\biggl[ (4 - 2\Rho_1 \ell )\zeta \Delta\zeta - 4( \ell^2 - \zeta^2) \biggr]^2 
+ 4\ell \zeta \Delta\zeta \biggl[ ( 4\Rho_1 - \Rho_1^2 \ell )\zeta \Delta\zeta + 4( \ell^2 - \zeta^2) P_1 \biggr] 
\biggr\}^{1 / 2} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~
( \ell^2 - \zeta^2 )\biggl[ 3\ell - 2 - 2\zeta U_1 \biggr] + \ell \zeta \Delta\zeta
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\pm \frac{\ell}{4} \biggl\{ 
\biggl[ (4 - 2\Rho_1 \ell )\zeta \Delta\zeta - 4( \ell^2 - \zeta^2) \biggr]^2 
+ 4\ell \zeta \Delta\zeta \biggl[ ( 4\Rho_1 - \Rho_1^2 \ell )\zeta \Delta\zeta + 4( \ell^2 - \zeta^2) P_1 \biggr] 
\biggr\}^{1 / 2} \, .
</math>
  </td>
</tr>
</table>
</div>
This doesn't look particularly useful because, after squaring both sides, it is a sixth-order polynomial in <math>~\ell</math>, which generally has no analytic solution.