@AHSAN 

For : k=2,lambda=1.

first_ode.mw

@AHSAN 

Try:

h := k - (k - 1)*x;
lambda := k/(1 + k);
ode := diff(p(x), x) = 6/h^2 - 12*lambda/h^3;
bc := p(0) = 0, p(1) = 0;
simplify(dsolve({bc, ode}, p(x)));
is(eval(p(x), %) = 6*(h - 1)*(h - k)/(h^2*(-k^2 + 1)));

@Thomas Richard 

 When can we expect implementation for fractional ODE in Maple?.It is possible in the very near future or not at all?

@Scot Gould 

Try:

(diff(sum(x^n, n = 0 .. infinity), x) assuming (-1 < x < 1));

or:

diff(sum(x^n, n = 0 .. infinity, formal), x);

@Sradharam 

Not so easy to solve, a nonliner ode with power series solution: See

Maybe his helps  see

Series.mw

@Sradharam 

With  infinite series solution I don't know how to it.

I solved with another method by Homotopy perturbation.If you what  to, see attached file.

HOMOTOPY.mw

Nobody wants to have to type in all your code.It's impolite of you to expect an answer without providing the code to reproduce.

People here generally like users to post (complete) code as Maple code instead of just images,pdf or TeX, so they can copy-paste it(Use green arrow and upload worksheet). It makes it convenient for them and more likely you will get someone to help you.

Maple 17: release date is: March 13, 2013.

Maple 2017: release date is: May 25, 2017

During these 4 years int command it has been improved and is more powerfull.

@mwahab 

With Maple 2020.2 works fine.Maybe is version issue.

See attached file.

solution_bug_1_ver2.mw

Looks like a bug.

Submit Software Change Request.

@Nick_99

Maple can compute only 5.1268948 > t >0 because singularity.

I update the answer.

From Help pages:

The discont option is not available in the 3-D case.  In this situation, it is recommended that you divide the plotting ranges into subranges to avoid the discontinuities, generate the individual plots and then combine them using the plots[display] command.  The view option can be used to control the ranges in the final display.
 

@Carl Love

An example: 

DSolve is symbolic solver.

NDSolve is numeric solver.

What value have c parameter? Or is c(t,x) ?

For k(x)  function using FourierTransform it just: k(x)= (1/4)*sech((Pi*x)/2) .