It's always annoying when the question you ask is not answered nor commented. 
Does this mean that nobody has any answers, or that the question isn't even worth spending time answering because it's completely irrelevant?
I do not consider your question lacks interest.

I'm afraid there is no solution to your problem, even in the far simpler 2D case.

Here are a few ideas ( a white elephant?) to rotate a text
Aligned.mw


This is bery basic stuff that must be enhanced to give acceptable results: increase the size of the text in t:=... (for instance font=[Helvetica, 50] for a better definition), adjust the threhold in function Tf , adjust the gridsize in densityplot, ...).

Hope this will help

Download sum.mw

Can you please keep this as a comment and avoid converting it into an answer, there is no reason to do so.

After your dsolve command type this to see that sys still contains an unknown quantity named e for which no ic/bc condition is given, so the error you get:

indets(sys1, name) minus {eta}
                              {e}

Cure: fix the value of e before executing dsolve or do

dsolve(eval(sys, e=some_numeric_value), numeric)

For instance

sol1 := dsolve(eval(sys1, e = 1), numeric);
           sol1 := proc(x_bvp)  ...  end;

the answer is NO.
But you can easily use Eclipse, GitHub or any other versioning tool provided you use mpl files and not mw files.

If you have several mw files it will take some time to convert them all by hand (open the mw file ans export it as mpl) and I don't know if these actions can be done programmatically.
Once the mpl files are created I advice you to use some tool which enables syntaxic coloring.
For my part, I use GitHub with Notepad++ but I know a coleague of mine uses Eclipse.

A hint: search "versioning" on this site, if I'm not mistaken I asked a similar question years ago (maybe from my professional account sand15).

Can you please keep this as a comment and avoid converting it into an answer, there is no reason to do so.

Download Amoeba.mw

A few workarounds colors.mw
 

Download degrees.mw

which works for polynomial expressions of whose all functions depend on the same indeterminate.

For instance

expr = x(t)*diff(z(t), t)^2,
expr = diff(x(t), t$2)+x(t)*diff(z(t), t$3),
...

but not

expr = x(t)*z(u)^2,
expr = x(t)*diff(cos(u(t)), t),
...

Download An_alternative.mw

Could this help you Fireworks?

Another exemple of trajectory fading is provided in Jonglage_4.mw.
Here is an animation of a ball bouncing within a triangular shape: at the current time T the trajectory is plotted between T-dt and T, a colorscheme whose first color is equal to the background color mimics a fadding trajectory.

Based on Thomas' example  you would obaiin this


Thomas_with_fading.mw

It is not because the display of den1 contains terms involving t that den1 INDEED depends on t.
This is a "side effect" of simplify.

Functions_of_t := print~(select(has, indets(simplify(eval(den1, para), size), function), t)):
                        exp(-I exp(I t))

                            exp(I t)

                        exp(I exp(I t))

               /1   /   (1/2)  (1/2)             \\
            exp|- I \x 3      4      - 4 exp(I t)/|
               \4                                 /

               /1   /   (1/2)  (1/2)             \\
            exp|- I \x 3      4      + 4 exp(I t)/|
               \4                                 /


 Functions_of_t := select(has, indets(expand(eval(den1, para)), function), t);
                               {}

numericden_mmcdara.mw

It is easy to verify that G(x, y) = G(y, x).
Then the minimum is to be search along the line y=x.

# Verify that G(x, y)
simplify(G(x, y)-G(y, x))
                               0

# Define a univariate quantity T
T := expand(G(x, x));
s := [fsolve(diff(T, x))];
                  s := [0., .8073218015, 1.015827227]

# Which one is a maximizer/minimizer ?
 eval~(diff(T, x$2), x=~s);
                 [-5.600000000, 30.91991289, -88.23416528]

# then s[2] is the minimizer and the minimum is equal tools

G(s[2], s[2])
                          -0.397001070

# With minimize
minimize(T, x=0..1, location=true);
      -0.3970010694, {[{x = 0.8073218015}, -0.3970010694]}

(In Statistics a model is said to be linear if it writes as a linear combination of its parameters, not of its regressors [here lambda])

Here obj simply writes C__10*U + C__10*V where both U and V are known functions of lambda.
Then your problem is linear wrt C__10 and C__10. LINEAR.mw 

Linearization is by far the simplest way to solve the problem (you can even find the solution by hand).

The first one consists in "rescaling the abscissas" of a classical ErrorPlot; the second one (which I prefer for it is more versatile) is to create "from scratch" your own error plot procedure.

ErrorPlot.mw

Example of what the second approach can produce:


As a rule I prefer  to define my own visualization procedure (of statistical data) instead of using those provided by the Statistics package.
This doesn't mean these latter are not "good" but only that they are not versatile enough to fit my requirements

Let us consider now this much simpler problem

(a*x+b*y)*diff(u(x, y), x) + (c*x+d*y)*diff(u(x, y), y) = 0

For this problem to have a solution you must have a=-d, which gives

                         1    2           1    2
               u(x, y) = - b y  + a x y - - c x 
                         2                2     

See compatibility.mw

So your problem could have a non trivial solution (aka a constant one) only for some values of values of alpha, b, c, d.
Here is an example of a non trivial solution for the case alpha=1, c=0, d=0 compatibility_2.mw

Another point of view                                                        
Your equation writes F(x, y, z, w) _° gradient(u(x, y, z, w) where F is field of directions in space [x, y, z, w] and _° denotes the dot product.
This direction field defines a family of lines such that the tangent to each of them is confounded with the direction of the field at the contact point.
Defining x(s), ..., w(s) as the cosine of this family gives:

                 d                              
                --- x(s) = alpha (b w(s) + y(s))
                 ds                             
                 d                              
                --- y(s) = -b w(s) + x(s) + z(s)
                 ds                             
                        d                
                       --- z(s) = -c y(s)
                        ds               
                    d                        
                   --- w(s) = d (y(s) - x(s))
                    ds                       

Integrating this system gives the expression of the characteristic lines.
Then u is an hypersurface made of these characteristic lines.

Look here  (characteristic system can be solved numerically but a numerical solution is provided as an illustration):

Download Characteristic_lines.mw

@Aixleft aimbot 

Here is a more readable code where some errors have been corrected.

Nevertheless it remains 4 expressions which "come from nowhere" and which I wasn't capable to check if the are correct or not: N11, N22, N33, N44 (it seems that you rewote yourself the expressions of N1, N2, N3, N4... Am I right?)

If it is so here are the expressions you are looking for:

                   2    1   /   (1/2)    \  (1/2)  3    / 4\
     tr :=    2 - v  + ---- \3 5      + 5/ 5      v  + O\v /
                  1200                                 

                       1   /   (1/2)    \  (1/2)  3    / 4\
     det :=       1 - ---- \7 5      + 5/ 5      v  + O\v /
                      1200                                 

What_I_understand.mw