@dharr  i try another method but you did the same as paper did , i am looking for a methods for finding this kind of pdes i will try more and more  thanks a lot 

@janhardo i need solving not writing 

@dharr  i know is confusing becuase this topic  have a lot kind of solution of pde, Soliton=5 type, lump=3 type, breather=2 type, interaction between them also have at least 8 type each on of them the solution is change, but i am sure if for the first two one is correct and without doing substitution is give me zero it  will be  done , if you watch carfully you will see after the first solution make pde be zero in (pdetest(eq19, pde)) then when N=2 start we look for second solution  if that solution make our pde be zero then we say that our substitution is true ,  but how find that condition which make our pdes be zero i am sure is related with that part becuase if you want be sure set all of the function of (t) be one, you will see  the second solution is satisfy pdes

@dharr i know where is problem but i don't know how find the correct one watch this picture

i did this change (red part) this red part is not true is not work if do this change it work just for first pde for other is not true, but i know there is a condition with one  exchange (we look for that true exchange) and substitute it can be any exchanging it will work for all so how i find that i will take another shot in below mw file

another-c.mw

@dharr  i solved one another before he didn't make any problem  and without this issue it  give me zero, maybe i am in mistake i trust your way about the last pde which make it to zero by substituting and really it work but know about this which not give it to me zero i am in dubt about that i think  without replacing parameter and function must give me zero, so in here i will sure which i did a mistake and replacing not work is my openion 

in my pde  i did a substitution  if you watch  which is  alpha(t)=beta(t) i did that by myself becuase for finding R i need to do something like that, if i don't do this substitution is not give me zero in first pde but also this substitution must be true for second time, if not work for second so we figure out we did mistake at there at that substitution, i don't know how act with that substitution do you have any idea  about that at least one function of (t) remain in pde  which worked for both pde, there is a lot of pde which if i can find this condition  for them i will be super work

really need that changing of function also we can change them to a constant but at least one of them remain and make our two step pde be zero i am sure if untill second one be zero it will work for other type too, i hope find this i will mark in red color 

C-find.mw

@dharr in here why when i test without substittue parameter is not give me zero but when i do substitute is give me zero?

really is problem how this is possible?

problem.mw

@dharr   thanks  i will do and go to solve a lot of them  by this way, becuase by number is zero so it mean we did a true way thanks a lot 

@dharr  i try to evaluate but is failed 

p-test.mw

@C_R  thank you i didn't see it before , but is so usfull 

@dharr i don't know when i will learn this, Every time I think about what I would have done without your guidance, I can’t find the words to thank you properly. My dear Dr. David, I truly owe you a lot.

@dharr  is been while i am search for error why is not solution, i think i found but i can't apply you remove some part  in orginal mw.file which is replacing i think i marked in here you didn't used at all becuase of that is not give me a solution even B[1] and B[2] not appear in parameter which it is wrong i have paper  if you didn't undrestand that red part i mention please tell me to send you a solved problem, also this is so  different becuase have a lot of equations, tthat system you make it is exactly what i am want but must be a solution of ode becuase of that Red part which have a replacing  is not giving me solution how we can fix that ?

@dharr this is so near what i am looking but still remain, i am looking parameter, in this method i change  the series solution and try to use your code but i don't know  in one place not worked, also is so clear for me which you determine the monomial each monomial term is realated to a equation in (eqs) respectively but if be more clear it will be better like  i mention is blank place which be clear for writing equation when i do write for latex form is clear in mw.file  the place of equation, 

system1.mw

becuase the equation is to larg i can't solved by hand if you by clasical way like mine you will see that is too much for writing by seperating hand and my project is stuck becuase of this and really need a techntechnician maple work to do this  is not easy 

@dharr i am in shok , is more than something regular  i will try to do for other i hope no problem come up, but you did a (....) i don't know what to say is more than a perfect job, thank you so so much dear dctor huge cup for you....

i fixed just be remove one term in (eqw) remove the term which have k[i] it will fixed and make the pde to zero 

@dharr you are right about that, it is a function of t  i don't have idea the only way for knowing that is that pdes make zero for other function  but in here i  get  error in pdes test  i don't know how work on it , and also if we change the function to sin(t) and cos(t) and even t must be work , but i get exccet 10000000

tester2.mw