After

\def\MyText{\textbf{My Text}}
\let\MySaved\MyText


\MySaved and \MyText have the same \meaning. What I would like to have further down in the document is

% \MyText = \textbf{My Text}
\def\MyText{\textit{More Text}}
\let\MySaved{\MySaved \MyText}
% \MySaved = \textbf{My Text} \textit{More Text}


but obviously \let does not allow a group of tokens in the second argument.

Is there any way to append like this with the same specific (non-)expansion properties \let has?


Question: Solving a two point boundary problem for a piecewise system of equations
Answer: Here is now the - hopefully - correct (numeric) solution of the original problem.

To begin with, we define a time t0 such that y = 0 for 0&lt;=t&lt;=t0. 

Für x(t) we have x(t) = 1 for  0&lt;=t&lt;=t0, otherwise we have first to solve the equation

sx = DSolve[x'[t] == x[t]^(3/4) - x[t] &amp;&amp; x[0] == 1, x[t], t]


We can ignore the error Messages.

(* Out[701]= {{x[t] -&gt; E^-t (-2 + E^(t/4))^4}, {x[t] -&gt; 
   E^-t ((-1 - I) + E^(t/4))^4}, {x[t] -&gt; E^-t ((-1 + I) + E^(t/4))^4}} *)


Taking the real-valued solution:

x[t] /. sx[[1]]

(* Out[700]= E^-t (-2 + E^(t/4))^4 *)


Having caluclated that, we now define the complete function x[t,t0]

x[t_, t0_] := E^(-t + t0) (-2 + E^((t - t0)/4))^4 /; t &gt; t0

x[t_, t0_] := 1 /; 0 &lt;= t &lt;= t0


Plotting it (for t0 = 5)

Plot[{x[t, 5]}, {t, 0, 30}, PlotRange -&gt; {0, 1}]




Now inserting x[t,t0] we can NDSolve for y(t).
Here we have selected the parameter t0 by trial and error such that y[10]=2

t0 = 1.03214; 
ta = 0;
te = 12;
sn = NDSolve[{D[y[t], t] == 1 - y[t] (1/x[t, t0]^(1/4) - 1), y[t0] == 0}, 
    y[t], {t, ta, te}][[1]];
{ta, te} = Head[(List @@ sn[[1]])[[2]]][[1, 1]];
yy[t_] = y[t] /. sn;
Print["y[10] = ", yy[10]];
Plot[{2, Evaluate[x[t, t0]], yy[t]}, {t, ta, te}, PlotRange -&gt; {-1, 2.2}, 
 PlotLabel -&gt; 
  "Solution of the two-point boundary problem\nred curve = x(t), green curve \
= y(t)\nt0 = " &lt;&gt; ToString[t0] &lt;&gt; "\ty(10) = " &lt;&gt; ToString[yy[10]]]

y[10] = 2.




Of course, y(t) must be set to zero below t=t0. This is trivial and not shown in the picture.

This seems to be the only solution. For greater values of t0 the boundary value y(10) decreases, and for smaller it increases.

Regards,
Wolfgang