@Markiyan Hirnyk Please kindly show me an example of a solution set returned by solve, or a similar command, which contains a trivial equation and yet represents only a finite number of solutions.

Here's how you can quickly check if any solution set (or list) S contains a trivial equation of the form z = z. Do

evalb(`*`(map(lhs-rhs, S)[]) = 0);

The result will be true if and only if S contains a trivial equation.

Here's how you can quickly check if any solution set (or list) S contains a trivial equation of the form z = z. Do

evalb(`*`(map(lhs-rhs, S)[]) = 0);

The result will be true if and only if S contains a trivial equation.

@fixedeggs But dsolve itself is evaluating the derivatives by using the right sides of dsys8. There is no "built-in interpolant" for this particular step. 

But if you really want to do it your way, you just need to add one equation and one initial condition to the system for each derivative that you want to evaluate. For example,

Sys:= {diff(y(t),t) = y(t)}: IC:= {y(0)=1}:
Sys1:= {diff(y(t), t) = w(t)}: IC1:= {w(0)=1}:
Sol:= dsolve(`union`(Sys,Sys1,IC,IC1), {y(t), w(t)}, numeric);
          proc(x_rkf45_dae)  ...  end;
Sol(1);
          [t = 1., w(t) = 2.7182818283577044,  y(t) = 2.7182818283577044]

@fixedeggs But dsolve itself is evaluating the derivatives by using the right sides of dsys8. There is no "built-in interpolant" for this particular step. 

But if you really want to do it your way, you just need to add one equation and one initial condition to the system for each derivative that you want to evaluate. For example,

Sys:= {diff(y(t),t) = y(t)}: IC:= {y(0)=1}:
Sys1:= {diff(y(t), t) = w(t)}: IC1:= {w(0)=1}:
Sol:= dsolve(`union`(Sys,Sys1,IC,IC1), {y(t), w(t)}, numeric);
          proc(x_rkf45_dae)  ...  end;
Sol(1);
          [t = 1., w(t) = 2.7182818283577044,  y(t) = 2.7182818283577044]

@Preben Alsholm This is a new usage of "so" as an introduction rather than as a continuation, which seems quite prevalent among academics. There is even a weekly radio show Le Show, hosted by Harry Shearer, that has a segment each week (about 2-3 minutes) devoted to playing clips of interviewees introducing their comments with "so". (Recorded episodes available at harryshearer.com/le-show/)

@LEETZ 

The boundary conditions from your original question will work just fine.

S := pdsolve({pde, bc});

@LEETZ 

The boundary conditions from your original question will work just fine.

S := pdsolve({pde, bc});

@digerdiga 

Sorry, I misunderstood what you meant by "parametrically", but I understand now. My fault: You didn't use the word incorrectly. You're right: The display method will not work to create the parametric plot of one dependent variable versus another.

@digerdiga 

Sorry, I misunderstood what you meant by "parametrically", but I understand now. My fault: You didn't use the word incorrectly. You're right: The display method will not work to create the parametric plot of one dependent variable versus another.

@digerdiga 

Any plots can be merged with plots:-display as long as they have the same dimensionality, i.e., all 2D or all 3D.

@digerdiga 

Any plots can be merged with plots:-display as long as they have the same dimensionality, i.e., all 2D or all 3D.

@pawarabhijit7 You wrote:

My all values must be less than one because i have normalised all the terms

Does that mean that they are all less than one in absolute value? That's a stricter requirement than just being less than one.

And when you say that they are normalised, do you mean that the sum of the squares is one? If so, then that's an additional equation. Or do you mean the weaker condition that each has been divided by its own absolute value?

I didn't say that we needed three more equations; I said that we would need three more equations if we wanted to get a unique solution (or, indeed, any finite number of solutions). Do you have reason to suspect a finite number of solutions?

@pawarabhijit7 You wrote:

My all values must be less than one because i have normalised all the terms

Does that mean that they are all less than one in absolute value? That's a stricter requirement than just being less than one.

And when you say that they are normalised, do you mean that the sum of the squares is one? If so, then that's an additional equation. Or do you mean the weaker condition that each has been divided by its own absolute value?

I didn't say that we needed three more equations; I said that we would need three more equations if we wanted to get a unique solution (or, indeed, any finite number of solutions). Do you have reason to suspect a finite number of solutions?

Adri, The theorem that you cite is about systems of ODEs. The Asker is asking about systems of algebraic equations.