@SophiaAraoz 

I edited my answer.

You say: "Does the speed/norm usually need a value for t?"  of course Not.

It's only a variable name "V(10)", I could write: "V-10" or "V10" .I just wanted that variable name is at the 10th  sec.(Speed as a vector (at the 10th sec))

Sorry for my English, is bad. Google translator have only IQ 50.

@assma

Mydata := Matrix([seq([seq(w(x, y), y = 0 .. 4)], x = 0 .. 4)]);

#Export to file:

Export("data.csv", Mydata, base = homedir);

#Import from file to Maple:

Import("data.csv", base = homedir);

Home dir you can find:

kernelopts(homedir);

For more see:

help("Export");

Have fun ! 

@Preben Alsholm 

Maybe another time. I have a lot reputation points :P

Regards Mariusz.

Maybe this helps:https://www.mapleprimes.com/questions/148773-Dsolve-Numeric-Able-To-Solve-A-Threepoint-BVP

@golnaz 

Your code contains syntax errors, please correct it.

Please post actual code rather than an image of code.

@one pound 

Because Maple does not know the answer so don't be surprised if some things don't work.

Mathematica does not know either.

Well, can be expressed with a few sums:

convert(ln(a-x), FormalPowerSeries, x, n)*convert(1/ln(a+x), Sum, x, include = powers, dummy = j);
expand(%);

#(Sum((-1)^j*(-1+ln(a+x))^j, j = 0 .. infinity))*ln(a)-(Sum((-1)^j*(-1+ln(a+x))^j, j = 0 .. infinity))*x*(Sum(x^n/(a^n*(n+1)), n = 0 .. infinity))/a

Possible bug is in Windows 10(corrupted by updates). Reinstal Windows.

@one pound 

Because at the first  Maple compute integral and second convert to SUM.This integral has no closed form and that's why it can not be calculated.

@one pound 

Something like that:

convert(int(ln(a+x)/a, x), FormalPowerSeries, x)

#Sum((-1/a)^(k+2)*x^(k+2)/((k+2)*(k+1)), k = 0 .. infinity)

Maybe I'm wrong, but maybe it should be:

 c(x, 0) = (c[1]+c[2])*(1/2)

for c1=c2=1 Can Solve. (On Maple 2017.3)

 restart;
 c1 := 1; c2 := 1;
 pde := diff(c(x, t), t) = d*(diff(c(x, t), x$2));
 ibs := c(x, 0) = (c1+c2)*(1/2), c(-infinity, t) = c1, c(infinity, t) = c2;
 pdsolve([pde, ibs], HINT = boundedseries);

Gives: c(x, t) = 1

Difficult to guess what is going on. I can't even see that example, let alone replicate it for diagnosing.

Try with this code:

pdsolve(YoursPDE, HINT = strip)

Yes, you are absolutely right.

@vv 

Maybe this helps :https://files.eric.ed.gov/fulltext/EJ936611.pdf