@Kitonum Sir, I have finished the plotting for the ratio of two expressions but am unable to extract expressions for k greater and equal to zero.

Extract_required_solution.mw

@Kitonum sir, Actually, According my problem requiermnets the integration is required of expression "Pre" for x=0..1 and need to find a symbolic expression in the of form of k. 
Besides this, need to plot Expression of Pre_max versus k by taking Pre_max along vertical axis and k along horizontal axis.

Also, need to plot the ratio of Pre / Pre_max versus x for different vakue of k such as 2.14,3,20,4,30.

@Kitonum , Thanks, sir, for your response. Actually, my question has two parts: first, we need to find the maximum expression at the point (critical), where the function is maximum.

and

In the second part, we need to find a symbolic expression. In this part, we obtained 4 answers... [2] When I integrate the given expression, it gives 4 cases. How do we separate the required expression for the k condition?..

I have uploaded the file with more explanation; can you please have a look at it?

Maximum_expression_New_1.mw

@mmcdara Thanks a lot sir

@mmcdara Thanks; when I fix the range for j=1...2 I got the error about Offset=-1/4 being unexpected, so I removed this command, code was run and produced the results, but some curves vanished, and even the values were not aligned on the bars

Error_Matrix_index_out_of_range.mw

@mmcdara thanks sir for time and help. By the way how to fix the error  "Matrix index out of range
"Error_Matrix_index_out_of_range.mw

@mmcdara thanks alot for response and affort. In my work I do not want ignore the negative values as you define new data by using adata := abs~(data)......my opinion is to change only display of colum graphs inside the graphs is look like the view you produced either it is on positive or negative sides. 

@mmcdara Thanks, sir, Is it possible to change the bar chats, Means look like rectangles for the example display in the below picture? Besdie, there is a box with a white line in graphs, it possible to remove them? Also, I tried by adding more option but have error, please correct sir.

another_try.mw

look like the graphs given below

@Carl Love pp is notation and stand for pressure profile.  Below is the picture that I take from a paper of this.

In Mathematica file (pdf) written a simplified form after putting u=diff(psi,y) and its simplified form is gievn in equation 62 and its BCs are also given Mathematica file.

firstly Need to solve for  eq 62 for psi against y and then use the solution of psi to get the solution for u against and similarly use the solution of psi to plot diff(p,x) against x and integrate the solution of diff(p,x) and find solution for p against x .  In mathematica file pp stand for solution of p. I am not understanding mathematics code and how to transform that in Maple?

@acer , sir, thanks for the correction. Is it possible to create one more column for the answer check? Means residual

@mmcdara Sir, can you help with my problem even though I lost the file you uploaded

Help_Numeric_solution.mw

@mmcdara Dear Sir, unfortunately, my laptop got damaged, and could not access the website. Managed a new laptop, found all accounts, logged in to the website, and saw your responses on my question again. I even do not see your reply due to my laptop issue.  I am sorry for the inconvenience. Could you please upload your response(Maple worksheet) to my problem? Sir waiting for your response.

@mmcdara is it possible to check that our obtained solution is correct or not? means compression of solution with any other method available in Maple? and why did you make Hypothesis := p(y) = a+b*y for diff(p(x),x)? even in original problem,we have .diff(p(x), x).but you pleased diff(p(y), y)how could I plot for diff(p(x),x), rather than p(y)?

@mmcdara thanks for your help. The results look like this

1. for u(y) against y for different values of k like 0.1,0.3,0.5,0.7,0.9 ( we show only odd values not all )

2. and the results for diff(p(x), x) against x expected to be look like as 

Note: you plot diff(p(y), y), which wrong

3. Finally the results for T(y) against y look like as