Derivation of $\frac{du}{dt}$ from The Chemical Basis of Morphogenesis by A. Turing I am currently reading The Chemical Basis of Morphogenesis
 by A. Turing. On page $56$ of his article, Turing uses the substitution
\begin{align}
\xi&=b(u+v), \\
\eta&=(p-a')u+(p'-a')v,
\end{align}
to transform
\begin{align}
\frac{d\xi}{dt}&=a'\xi+b\eta+R_1(t),\\
\frac{d\eta}{dt}&=c\xi+d'\eta+R_2(t),
\end{align}
into $$\frac{du}{dt}=pu+\frac{p'-a'}{(p'-p)b}R_1(t)-\frac{R_2(t)}{p'-p}+\xi\frac{d}{dt}\left(\frac{p'-a'}{(p'-p)b}\right)-\eta\frac{d}{dt}\left(\frac{1}{p'-p}\right).$$ Here, $p$ and $p'$ are the (real) roots of $(p-a')(p-d')=bc$. I am seeking advice/hints on how I can derive this final expression.
 A: Assuming $p \neq  {p'}$ and $b \neq  0$, we have
\begin{equation}
\renewcommand{\arraystretch}{1.5}  \left\{\begin{array}{rcl}{\xi}&=&b \left(u+v\right)\\
{\eta}&=&\left(p-{a'}\right) u+\left({p'}-{a'}\right) v
\end{array}\right. \Longleftrightarrow  \left\{\begin{array}{rcl}u&=&\displaystyle  \frac{{p'}-{a'}}{\left({p'}-p\right) b} {\xi}-\frac{1}{{p'}-p} {\eta}\\
v&=&\displaystyle -\frac{p-{a'}}{\left({p'}-p\right) b} {\xi}+\frac{1}{{p'}-p} {\eta}
\end{array}\right.
\end{equation}
Hence
\begin{equation}
{u'} = \frac{d u}{d t} = \frac{d}{d t} \left(\frac{{p'}-{a'}}{\left({p'}-p\right) b}\right) {\xi}-\frac{d}{d t} \left(\frac{1}{{p'}-p}\right) {\eta}+\frac{{p'}-{a'}}{\left({p'}-p\right) b} \frac{d {\xi}}{d t}-\frac{1}{{p'}-p} \frac{d {\eta}}{d t}
\end{equation}
Hence
\begin{equation}
{u'}-p u = \frac{d}{d t} \left(\frac{{p'}-{a'}}{\left({p'}-p\right) b}\right) {\xi}-\frac{d}{d t} \left(\frac{1}{{p'}-p}\right) {\eta}+\frac{{p'}-{a'}}{\left({p'}-p\right) b} \left(\frac{d {\xi}}{d t}-p {\xi}\right)-\frac{1}{{p'}-p} \left(\frac{d {\eta}}{d t}-p {\eta}\right)
\end{equation}
As $p$ and $p'$ are the two roots of the equation $x^2-(a'+d')x + (a' d' - b c) = 0$, it follows that $p + p'=a'+d'$ and $p p'=a' d' - b c$ by Vieta's formulas, hence we have
\begin{equation}
\renewcommand{\arraystretch}{1.5}  \begin{array}{rcccl}\displaystyle  \frac{d {\xi}}{d t}-p {\xi}&=&\displaystyle  \left({a'}-p\right) {\xi}+b {\eta}+{R}_{1}&=&\left({p'}-{d'}\right) {\xi}+b {\eta}+{R}_{1}\\
\displaystyle  \frac{d {\eta}}{d t}-p {\eta}&=&\displaystyle  c {\xi}+\left({d'}-p\right) {\eta}+{R}_{2}&=&c {\xi}+\left({p'}-{a'}\right) {\eta} + R_2
\end{array}
\end{equation}
Turing's formula clearly follows from these relations because $(p'-a')(p'-d')-b c = 0$ and
\begin{equation}
\renewcommand{\arraystretch}{2}  \begin{array}{rcl}\displaystyle  \frac{{p'}-{a'}}{\left({p'}-p\right) b} \left(\frac{d {\xi}}{d t}-p {\xi}\right)-\frac{1}{{p'}-p} \left(\frac{d {\eta}}{d t}-p {\eta}\right)&=&\displaystyle  \frac{\left({p'}-{a'}\right) \left({p'}-{d'}\right)-b c}{\left({p'}-p\right) b} {\xi}\\
&&\displaystyle  \qquad +\frac{\left({p'}-{a'}\right) b-\left({p'}-{a'}\right) b}{\left({p'}-p\right) b} {\eta}\\
&&\displaystyle  \qquad +\frac{{p'}-{a'}}{\left({p'}-p\right) b} {R}_{1}-\frac{1}{{p'}-p} {R}_{2}\\
&=&\displaystyle  0+0+\frac{{p'}-{a'}}{\left({p'}-p\right) b} {R}_{1}+\frac{1}{{p'}-p} {R}_{2}
\end{array}
\end{equation}
