or, writing ${\displaystyle T}$ for the constant ${\displaystyle {\frac {aAt}{v}},}$

${\displaystyle {\begin{aligned}\phi _{0}&=-Tk_{0}{\frac {d\gamma _{0}}{dx}}\mp k_{0}\gamma _{0}{\frac {dV}{dx}},\\{\frac {c_{0}\phi _{0}}{k_{0}}}&=-Tc_{0}{\frac {d\gamma _{0}}{dx}}\mp c_{0}\gamma _{0}{\frac {dV}{dx}},\\\textstyle \sum _{0}\displaystyle {\frac {c_{0}\phi _{0}}{k_{0}}}&=-T\textstyle \sum _{0}\displaystyle {\frac {c_{0}d\gamma _{0}}{dx}}\end{aligned}}}$

[The terms ${\displaystyle \mp {\frac {dV}{dx}}k_{0}\gamma _{0}}$ disappear in the algebraic sum since ${\displaystyle \textstyle \sum \displaystyle c_{1}\gamma _{1}=\textstyle \sum \displaystyle c_{2}\gamma _{2}}$. For a similar reason]

${\displaystyle \textstyle \sum _{0}\displaystyle \mp {\frac {c_{0}\phi _{0}}{k_{0}}}=-{\frac {dV}{dx}}\textstyle \sum _{0}\displaystyle c_{0}k_{0}\gamma _{0}.}$

The first equation makes ${\displaystyle {\frac {\textstyle \sum _{0}\displaystyle c_{0}\gamma _{0}}{dx}}}$ constant throughout the tube, and since ${\displaystyle \textstyle \sum _{0}\displaystyle c_{0}k_{0}\gamma _{0}''=\textstyle \sum _{0}\displaystyle c_{0}k_{0}\gamma _{0}',}$ ${\displaystyle \textstyle \sum _{0}\displaystyle c_{0}k_{0}\gamma _{0}}$ must be constant throughout the tube. The second equation then makes ${\displaystyle {\frac {dV}{dx}}}$ constant throughout the tube. Let ${\displaystyle X=-{\frac {dV}{dx}}\cdot }$

Our original equation is

${\displaystyle \phi _{0}=-Tk_{0}{\frac {d\gamma _{0}}{dx}}\pm Xk_{0}\gamma _{0}.}$

Now with ${\displaystyle X}$ constant this is easily integrated.

${\displaystyle {\begin{aligned}{\frac {\phi _{0}}{Xk}}&=-{\frac {T}{X}}{\frac {d\gamma _{0}}{dx}}\pm \gamma _{0},\\\pm {\frac {T}{X}}{\frac {d\gamma _{0}}{dx}}&=\gamma _{0}\mp {\frac {\phi _{0}}{Xk_{0}}},\\{\frac {d\gamma _{0}}{\gamma _{0}\mp {\frac {\phi _{0}}{Xk_{0}}}}}&=\pm {\frac {X}{T}}dx,\\\log \left(\gamma _{0}\mp {\frac {\phi _{0}}{Xk_{0}}}\right)&=\pm {\frac {X}{T}}x+\log H_{0},\\\gamma _{0}\mp {\frac {\phi _{0}}{Xk_{0}}}&=H_{0}e^{\pm {\frac {X}{T}}x}.\end{aligned}}}$

To determine ${\displaystyle H_{0}}$ we have

${\displaystyle \gamma _{0}''-\gamma _{0}'=H_{0}\left(e^{\mp {\frac {X}{T}}x''}-e^{\pm {\frac {X}{T}}x'}\right).}$

If we put the origin of coordinates in the middle of the tube we have

${\displaystyle x'=-x''.}$