@glcrawfo You have many, many typos. First answer my question about RC and LC. Once you do that I can finish the problem, and I'll post a worksheet done in plaintext Maple Input. 2D Input is crap, which is not your fault, because Maple stupidly made it the default input mode.

@glcrawfo Why have you changed the ODE so fhat the coefficient of the 2nd derivative is L*C when it used to be R and that of the 1st derivative is R*C when it used to be L? Which version is correct?

It will be much easier for me to explain the rest of your numerous typos to you if you switch your entry style to Maple Input. It's currently 2D Input. You could simply open the worksheet that I attached to a previous Answer. It's written in Maple Input (all the input code is reddish-brown bold monospaced characters). Let me know if you can do that. 

@glcrawfo Are you sure that there's supposed to be an x(t) on the right side on the ODE? That's not what you had at first, and it's not the standard way that ODEs are written, (although it's still mathenatically valid). So, it seems strange.

@glcrawfo Okay, that's better; I'm begining to understand. Should "(-1) and n" actually be "(-1) to the n^th power", which is denoted in Maple (-1)^n?

I told you before that your piecewise function makes no sense, and you haven't changed it! It doesn't even have balanced parentheses! I understand the rest of the problem fine. A piecewise function (for this context) has alternating inequalities and algebraic expressions. You just have a single algebraic expression inside piecewise(...); you have no inequalities.

Do not post this as a new Question! If you do it again, I'll delete it. Edit this version instead.

@Oliveira Are you saying that when you solved this in Maple, you only got the middle piece of the piecewise, as shown by the plaintext result in your Question?

Please re-enter the problem. The piecewise part makes no sense to me. I don't think that you've accurately transcribed what your professor wrote.

@glcrawfo See the Answer that I just posted below.

I deleted the Question where you asked essentially the same question as this but tried to put the whole problem in the title box. Please keep titles somewhat general and word-based and put specifics like actual equations in the main body box.

I am not familiar with the term "component graph". Can you define it? I can certainly show you how to plot x(t) over 0 <= t <= 5, or any other interval.

@mmcdara Yes, the error you got is a version issue. You'll need to change TRX1 from

(n::posint)-> add(GG(TRS(X), TRS(Y)), 1..n)

to

proc(n::posint) local k; add(GG(TRS(X), TRS(Y)), k= 1..n) end proc

@mmcdara The binary-infix-operator precedence of = is higher than that of ->, so 

':-Mean'= ()-> 1/Pi 

is equivalent to

(':-Mean'= ())-> 1/Pi,

which is nonsense. Hence, Acer's parentheses. See ?operators,precedence.

@syhue Ah, you're using Maple's 2-D Input (which is total garbage, for reasons such as you're about to see). In that case, use

C &\^ d mod n

The backslash will not appear as you type it. It's an escape character that tells it that the ^ is attached to the & and it's not meant to perform its usual 2-D Input function, which is superscripting. If you switch to 1-D input (aka Maple Input), which is what I always use, then just enter what I said before

C &^ d mod n

In other words, don't use backslash in this context in 1-D input.

To avoid the risk of numeric overflow, you should do this for all modular exponentiation (unless the exponent is -1). So, do

C:= M &^ e mod n

also, even though e is fairly small in this case.

@acer Is it possible to use Statistics:-Specialize to achieve the goal that I stated in the first paragraph of my Reply above? That is, the creation of custom parametrized Distributions that can be used exactly like the stock parametrized distributions? If so, then that's the way to do it.

I converted your Post to a Question.

It's extremely difficult to for us to re-execute your code in its current format. It's also difficult to follow references to equation numbers, such as the all-important equation 19. Please upload the worksheet using the green up arrow on the toolbar.

There's a mysterious symbol uminus0 or `&uminus0;` that begins to appear in your output as of MatrixInverse(XprimedX). I think that this has something to do with your trouble. Symbols with names like that are special mathematical symbols that come from the palettes and are often inadvertently pasted into code. There's no way to tell whether this is actually the cause of your trouble without close inspection of the actual worksheet.

Great Question; vote up. I don't have an Answer, but this is something that I've thought about before: How can I create my own parametrized custom distribution MyDist such that it can be invoked MyDist(a,b) in any situation where one could invoke, say, Normal(mu, sigma)? That is your goal, correct? If so, I think that you're well on your way to a method. I think that the key is constructing the module that is the 3rd attribute (see setattribute at ?attributes).

Here are five very small steps to help you on your way:

1. I see that you're using with, apparently simply to list the exports of the 3rd attribute. It seems to work, but I think that it's risky because with modifies the global state. You could replace with(A) with exports(A).

2. In all cases where you've defined a PDF, you've used both z and t, but you need to reduce that to a single name.

3. You've used ParentNames where you should've used ParentName.

4. No doubt you've read the examples at ?Statistics,Distribution, but they're worth re-reading carefully a few more times.

5. Avoid compound conditions in piecewise whenever possible: piecewise(a <= t and t < b, 1/(b-a), 0) can be and should be replaced by piecewise(t < a, 0, t < b, 1/(b-a), 0). I find that the latter form causes less problems, for example with symbolic integration. The only benefit that I know of the former form is that its prettyprinted display is more akin to standard mathematical typesetting. Maple's definition of piecewise guarantees that the leftmost condition that evaluates to true is the one that's used, which is what makes the two forms mathematically equivalent (for all real t). There's no need to construct a piecewise such that only one condition can be true because it's only the leftmost one that's true that matters.