With constant k, the solution is cube. Answer given below.

@Kitonum 

Thanks for going through my step by step statements. Thanks for finding the mistake mede by me.

Ramakrishnan V

@Kitonum 

I am interested in knowing what the formula is for because it is not clear what logic leads to such a formula.

Can you give the detail calculation with numeric numbers term by erm for M = 5 and K =4

M:=5:  K:=4:  K1:=K-2*K2:

Sum(M!/(M-K1-K2)! * K!/(K1! * K2)! * 1/M^K * 1/2^K2, K2=0..K/2);

value(%);

Step 1

k2=0

5!/(5-4-0)! * 4!/1 * 1/5^4 *1/2 ^0

value(%);

Step 2

k2=1

5!/(5-2-1)! * 4!/(2! * 1)! * 1/5^4 *1/2 ^1

Step 2

k2=2

5!/(5-1-4)! * 4!/(1 * 4)! * 1/5^4 *1/2 ^2

Is this ok?

What is the significance?

Ramakrishnan V

@J4James Thanks. I tried yr problem. Answers is x(t) =e^-t

and y(t) = e^-t - e^-1.

I shall proceed on this line. Thanks.

@Rouben Rostamian  

Thank you so much. Your points are very clear, simple and understandable. I shall do the home work and come back with more details to get help from mapleprimes.

Just for information only i give below two statements. It may or may not help you to help me.

1. The problem is to find temperature variation inside a cylinder (IC engine) varying with cam angle (piston position varies with theta(t))

2. Combustion takes place during one stroke (third stroke i.e. first half of second revolution) only out of four strokes each 180 degrees.

I shall anyway come back with mathematical parts only soon to get possible help from all of ou in maleprimes.

Ramakrishnan V

@J4James I am amazed at the support provided by J4James and Rouben. Thank you both of you. I will be coming with formulation and boundary conditions for my model. I am getting confidence that i will be able to solve. Please bear with me for some more time to come back to you. Thanks a lot once again.

i give below an example

polynomial_example.mw 

@Preben Alsholm 

Thank you for the answer. It was interesting to read and understand more ideas from your comments and suggestions. I shall tryDEtools and learn more in a few days time and hope to come back with clear ideas. Thanks once again. Cheers. Ramakrishnan V

@Preben Alsholm 

y(t) = a*exp(-t)*sin(.57736*t)+b*exp(-t)*cos(.57736*t);
`assign%`;
print(`output redirected...`); # input placeholder
assign%
a, b := 2, 1;
plot(y(t), t = 0 .. 5);
Error, (in Explore) No parameters to explore

Animation command for the above answer wld help me greatly!

Solution i.e. graph obtained after plot command. To animate i went to explore, but error indicated.

Hope the previous and this efforts on my maple worksheet wld give you atleast some light upon the answer.

Cheers and thanks for yr active help.

Ramakrishnan V

@Preben Alsholm 

dl := 3*(diff(y(t), t, t))+6*(diff(y(t), t))+4*y(t) = 0;
print(`output redirected...`); # input placeholder
/ d / d \\ / d \
3 |--- |--- y(t)|| + 6 |--- y(t)| + 4 y(t) = 0
\ dt \ dt // \ dt /
"(->)"
/1 (1/2) \ /1 (1/2) \
y(t) = _C1 exp(-t) sin|- 3 t| + _C2 exp(-t) cos|- 3 t|
\3 / \3 /

Above copy paste i do not know how distorted! Sorry.

Line by line i hv given below.

 dl := 3*(diff(y(t), t, t))+6*(diff(y(t), t))+4*y(t) = 0

 y(t) = _C1*exp(-t)*sin((1/3)*sqrt(3)*t)+_C2*exp(-t)*cos((1/3)*sqrt(3)*t)

DEplot(dl, y(t), t = 0 .. 10, [seq([y(0) = y0, (D(y))(0) = 0], y0 = -3 .. 3)]);
print(`output redirected...`); # input placeholder
DEplot(dl, y(t), t = 0 .. 10, [[y(0) = -3, D(y)(0) = 0],

[y(0) = -2, D(y)(0) = 0], [y(0) = -1, D(y)(0) = 0],

[y(0) = 0, D(y)(0) = 0], [y(0) = 1, D(y)(0) = 0],

[y(0) = 2, D(y)(0) = 0], [y(0) = 3, D(y)(0) = 0]])

Plot does not appear

The following worked and a curve from t=0 to 4 sec appeared based on comments received in maple prime.

Explore(plot(a*exp(-1.*t)*sin(.57736*t)+b*exp(-1.*t)*cos(.57736*t), t = 0 .. 4))

The following command also worked as per yr suggestion.

 plots:-animate(DEplot, [ode, y(t), t = 0 .. 5, [seq([y(0) = y0, (D(y))(0) = y1], y0 = -3 .. 3)]], y1 = -2 .. 2)

y(t) = _C1 exp(-1. t) sin(0.57736 t)

Flg is answer not understandable! but graph has come.

+ _C2 exp(-1. t) cos(0.57736 t)
-1. + 0.57736 I
-1. - 0.57736 I
2
lambda + 2. lambda + 1.333344570
/ d \ / d / d \\
1.333344570 y(t) + 2. |--- y(t)| + |--- |--- y(t)|| = 0
\ dt / \ dt \ dt //

I want the previous curve animated from 0 to 2 seconds. Like wise all curves in yr solution from t=0 to 4 sec.

can i attach mw files here?

Ramakrishnan V

@Preben Alsholm 

Thanks for clarifying range from 0 to 5.

I also need for the graph to animate for time 0 to 5 also. The development of the values of y at t=0 to y at t=5 sec for each curve.

Instead of a basic curve animating from y1=0 to 2 (-2 to 2), i want the graph to be generated from blank space say at value 3 along y coordinate at t=0 to y value at t=5.I know it must be simple by stating the sequence for t instead of y1. I am not getting the syntax correct.

Also every time shd we restart and execute?

Thanks for yr time.

Cheers. Ramakrishnan V

@Markiyan Hirnyk 

Thanks. It works. Cheers. Ramakrishnan V

@Kitonum 

Thanks. It works. Cheers. Ramakrishnan V

Thank you. It works well.

Pardon me if my question is very incorrect. For animation, the values you give are the initial conditions of differential y at time zero. Am i correct.

Can we also give command for the same 7 curves to be animated from time zweo to five?

Would be grateful for yr pecious time in helping out in this Cheers. Ramakrishnan V

I think tutor assistant is there to learn step by step answer.

Try to explore tutor assistant.