AHSAN

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1 years, 138 days

MaplePrimes Activity


These are replies submitted by AHSAN

@acer  actually i want to check the of effect of involved parameter like M, beta, lambda on velocity by fixing x=0 and and on domain [o,1]. The file i attached i already mention the value of M, beta and lambda. Grid lines show tje velocity.  Sir if still im failed to explain my problem then thanks for your time and apologies dor wasting your time.

@acer sir I already draw the graph for my problem which attached in my question. Now i am interested to show the  view of my same problem that i attached in my worksheet like the picture i shared you yesterday. 
note: i was confused about name that is why i mention 3 D so please forget about 3 D and legends. 

@acer sir i am waiting for your valuable guidance.

@acer  sir i want to show my graph like the picture upload. I was confuse that may be this will be drawn by using the command of 3 D graph . Sorry for inconvenience.

@acer  thanks sir, this is also ok but i am looking to draw some think this view of the graph. 
note this is not exact graph for this problem it is to show you just for an example. 

@vv i want to ahow only 3D view of my velocity profile . The value of involved parameters are given in plot section. 
how to use textplot can you please chnage in my worksheet 

I am looking for the answer of my above question.

@Kitonum thanks sir for your help,  can you please correct my worksheet attached below. The same thing you did above for simple equation, I applied for complex equation but could not get the answer. if it is possible to solve numerically the how could a obtain the expression for lambda?

help_ode_solution.mw

 

@Mariusz Iwaniuk  thanks sir, I am not well familiar with maple use. can you correct my worksheet attached below. The same thing you did above I applied for complex equation but could not get the answer

help_ode_solution.mw

 

@Mariusz Iwaniuk  thanks sir, why you put this 6*(h - 1)*(h - k)/(h^2*(-k^2 + 1)) in the last line, and what is the meaning of %

@Kitonum bundle of thanks sir by the way why we take eval(rhs(%), x = 1);? rhs stand for right-hand side or is there any other meaning and % meaning?

@Kitonum sir i am looking for how to find the value of lambda?

@Kitonum why it gives error

restart;
h := k - (k - 1)*x;
DE := diff(p(x), x) = 6/h^2 - 12*lambda/h^3;
             d                6              12 lambda    
      DE := --- p(x) = ---------------- - ----------------
             dx                       2                  3
                       (k - (k - 1) x)    (k - (k - 1) x) 
BC := p(0) = 0, p(1) = 0;
                    BC := p(0) = 0, p(1) = 0
simplify(dsolve({BC, DE}, p(x)));
Error, invalid input: simplify uses a 1st argument, s, which is missing

@Kitonum thanks sir, it is working well, i have confusion in the last step can you please explain how and why you did this

p:=unapply(eval(p(x),%), x);
h:='h':
eval(p(x),x=solve(h=k-(k-1)*x,x));
simplify(%);

and how to fine the value of lambda=k/1+k after integration

@Kitonum Thanks sir for your effort but i am looking for the solution which shows in the attached picture. first, need to put the value of h in the given first order ode and then solve obtained expression by using BCS that you b the same as p in the picture and neet to put the expression p=0 solve for lambda. but I do not obtained the exact value of p that is given papers

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