FunctionExpand[] and FullSimplify[] will convert Pochhammer[] to Gamma[]. Since Gamma[1+n] equals n !, FullSimplify[] will go further if the complexity function tells FullSimplify[] that the factorial expression is less complex than the gamma expression and the argument is an integer. Note that "less complex" is a numerical measurement and not the same a "more familiar to the user."

FullSimplify[
 -(((-1)^n 2^(1 - n) C[1])/(Pochhammer[3/2, -1 + n] Pochhammer[2, -1 + n])), 
 n \[Element] Integers && n >= 0]
(*  -(((-2)^n C[1])/(2 n)!)  *)

For the following, one needs a complexity function that favors factorials to get a factorial:

FullSimplify[
 -(((-1)^n 2^(1 - n) C[1])/(Pochhammer[2, -1 + n] Pochhammer[5/2, -1 + n])), 
 n \[Element] Integers && n >= 0]  
(*  -((3 (-2)^n C[1])/Gamma[2 + 2 n])  *)

FullSimplify[
 -(((-1)^n 2^(1 - n) C[1])/(Pochhammer[2, -1 + n] Pochhammer[5/2, -1 + n])), 
 n \[Element] Integers && n >= 0, 
 ComplexityFunction -> 
     (LeafCount[#] - 3 Count[#, _Factorial, Infinity] &)]
(*  -((3 (-2)^n C[1])/(1 + 2 n)!)  *)

Lambda is chosen just by convention. It's the symbol a lot of differential equation textbooks use. There isn't a meaning behind why Lambda is chosen, it's just the symbol that has commonly been chosen in the past so we decided to use the same symbol just to be consistent with other references.

Thank you. That definitely shouldn't be like that. I think x was holding a definition from a previous example that wasn't cleared. We'll get that fixed. Thank you so much for catching these errors.

The difference the sign of y'[t] makes is fascinating. With a minus sign, the solution diverges (swings wider and wider away from zero). With a plus sign, it converges to a value of 1/2.

Thank you. My name is Richard Daehler-Wilking.

Richard Daehler-Wilking daehlerr@bellsouth.net

Hello Richard,

I'll open a ticket including your report and notebook for Luke to review. I'll follow up with you via email.

Thank you,

Christine Owens
Wolfram U Project Manager
Wolfram U

You're welcome! I'll suggest this, too, which may sometimes be helpful:

aN2 = aN2 /. Gamma[u_] :> (u - 1)!

It will convert any Gamma[…] expression to a factorial, so you don't have to know ahead of time the exact form of the argument. For instance:

{Gamma[2 + 2 n], Gamma[n]}  /. Gamma[u_] :> (u - 1)!
(*  {(1 + 2 n)!, (-1 + n)!}  *)

P.S. (added in edit). You could also use the Pochhammer function identity:

RSolve[…] /. Pochhammer[a_, b_] :> (a + b - 1)!/(a - 1)!

In Lesson 17 about series solutions near an ordinary point, the first graph shows a sequence of plots converging to a limit. The last graph shows a similar collection of plots, and points out their convergence in the interval (-1, 1). I was confused at first because I expected a note about divergence.

SUGGESTION: For the last graph, also point out that as the order increases the plots move AWAY from the full solution (dotted line) rather than toward it as they did in the first graph. This illustrates divergence outside the region of convergence.

Problems 11 through 14 of Problem Set 4 entail physical equations with actual physical units. The use of physical units such as {kilograms, meters, seconds} was mentioned in the class, so I tried using pounds, inches, and seconds, (converting pounds to mass by dividing by gravitational acceleration). Mathematica allowed me to enter pretty-looking equations and initial conditions, but then did not seem to know what to do with them when I tried DSolveValue. Presumably I'm doing something wrong, but the sample problems don't help here.

SUGGESTION: Demonstrate use of physical units in one or two problem set solutions.

Hi Rich. Please send your solution to wolfram-u@wolfram.com They will forward your solution along to me and I will get back to you over email.

Hi Luke,

I decided to go ahead and work the quizzes for the certification, and on the second quiz regarding whether all autonomous eqtns are exact (the last question), my answer was marked as incorrect, whereas I am confident that it was correct. Is there some way that it can be reviewed by a person instead of AI so that proper credit can be given?

Thanks!

I"m trying to work Problem 5 in Problem Session 4:

Find a general solution for the following equation using Variation of Parameters: y''(x)-6y'(x)+9y(x)=sin(x)

Compare with the particular solution obtained by the Method of Undetermined Coefficients. ==> The solution goes into great detail presenting the method of undetermined coefficients. It then finishes with "It can be observed that this is equal to the particular solution derived from Variation of Parameters." However, I can't see where variation of parameters was used.

In both Lesson 24 (Discontinuous Forcing Functions) and Lesson 25 (Impulse Functions), plots are shown of the complementary solution (forcing function is zero) and the particular solution (forcing function/impulse function applied). It would be interesting to see a plot of the DIFFERENCE between results with/without forcing function.

In Lesson 24, there's not much to see. However, the difference plot of (sln2 - sln3) in Lesson 25 makes it easier to see the effects of the applied impulse functions, particular the first negative hit at t==2Pi. It's harder to see the effect of the positive hit at t==3Pi because it simply adds to the already-positive motion of the curve. SUGGESTION: Create a Manipulate box to allow the student to tinker with the timing and magnitude of the second hit. That should provide a nice feeling for coherent and destructive effects.

Yes, you are absolutely right. It should be Integrate[ ( 0 + s^2) ^2, {s, 0, t}]. Thank you for pointing this out. We'll get it fixed.

That can't be controlled from within DSolveValue, but you can use the TrigToExp function to do that conversion. For example:

In[1]:= TrigToExp[DSolveValue[y''[x] + y[x] == x, y[x], x]]
Out[1]= x + 1/2 E^(-I x) C[1] + 1/2 E^(I x) C[1] + 1/2 I E^(-I x) C[2] - 1/2 I E^(I x) C[2]

Picard's Theorem works best when the initial condition when the dependent variable equals zero when the independent variable equals zero. Exercise 3 shows how to transform a differential equation where y[0] is not equal to zero into an equivalent problem where the initial condition at zero equals zero.

Thank you for pointing that out. The Integrate code shouldn't be there. In place of the Integrate code should be

m = 2 x - y;
n = 2 y - x;

We'll get that fixed. Thanks again for pointing that out.

Here is a link to the course: https://www.wolfram.com/wolfram-u/courses/mathematics/introduction-to-differential-equations/

Michael, I have reservations about that first equation being autonomous because y is being set as a function of t which is the independent variable. For example population growth as a function of population would be clearly autonomous but population growth as a function of population set as a function of time has a somewhat different relationship wouldn't it? In his definition, Luke says that the independent variable cannot be in the equation when you have the autonomous property.

I agree with your other calculations on exactness for the other equations except that when setting the equation on the left hand side to 0, I have negative signs for the "N" terms, -1/f(y) in the first , and -1 for the second.

You're absolutely right. Thank you very much for pointing this out. We'll get that fixed.

@Michael Rogers, But so basically what you have written confirms my conclusion that he made an error in saying that first equation I listed is autonomous, just as f'(y) = f(x) is also not autonomous, unless it is a constant which is not a general condition and he didn't say that. His definition will falsely lead someone to infer that all autonomous equations are exact.

There is something odd about Lesson 27 (Review of Matrices) as it appears in the course framework. In the section "Matrix Functions", the code snippet

A={{x^2, x},{3 + x}, Exp[x]}};
A // MatrixForm

is followed by (apparently output) matrix

{{{r^2, s^2}, {r, s}}, {{r + 3, s + 3}, {Exp[r], Exp[s]}}}  (* Completely different matrix *)

When I open the downloaded lesson in Mathematica, I see the input as shown, but the output is the matrix corresponding to the input with the variable x, not the more complicated matrix with r and s as I see in the course framework..

In Problem Session 7, the discussion of Problem 2 says "Because of the lack of a damping term, the solution takes an unending oscillatory path:"

In fact, the equation for this problem (y''[t] + 2*y'[t] + y[t] == (a few step functions) ) DOES have a damping term and the plotted solution tapers off quickly.

Hello Richard,

Thank you for this feedback. I'll create a ticket per your report for Luke to review.

Christine Owens

Hello Jason,

I'll inquire about this with Luke.

Thank you,

Christine Owens
Wolfram U Project Manager
Wolfram U

Ahh OK! I thought that might've been you :-) Thank you for confirming.

Cool! Thank you. SUGGESTION: Add this to the discussion of Example 4 in Lesson 19. It seems to be a nice example of an apparent singularity that isn't "essential"; that is, it goes away when you look at it right.

Hello Richard,

Thank you for this feedback. I'll create a ticket per your report for Luke to review.

Christine Owens Wolfram U Project Manager Wolfram U

Hello Richard,

Thank you for this feedback. I'll create a ticket per your report for Luke to review.

Christine Owens Wolfram U Project Manager Wolfram U

Thanks, Richard. That's a good suggestion. We can include units in a few of those problems to show how to include units in a calculation.

Hello Richard,

I've added your follow-up to our original ticket per your report above for "problem 5, problem session 4".

Thank you,

Christine

Oops! Thanks for sharing this error in the survey question. We will fix it.

Thank you, Richard. It looks like that is doing the right thing.

Hello Richard,

Thank you for your feedback. I've created a ticket per your report for Luke and our team to review. As soon as I have more information, I'll let you know.

Best,
Christine Owens
Wolfram U Project Manager
Wolfram U

I have notice between the Exercise 11-1 and Exrcise 12-1. They are changed.

Richard, it sounds like you have an issue logging into the BigMarker servers. The link Cassidy sends out every day contains your BigMarker credentials: they're embedded in the long list of characters in that URL. Cassidy should be able to reset and re-send to you. An email to wolfram-u@wolfram.com should get the ball rolling.

As a temporary workaround, I bet you could "register" with another e-mail address and use that to access the session today. Wolfram U would prefer to fix your permanent account (since that's how they track users for certifications they earn), but that can get you running today.

Good luck. This course is hard enough for me with my BigMarker login working smoothly. :)

Thanks, but that link goes to the course framework with the recording of the course. I'm looking for a link to the daily lectures as they are given. Sometimes the link in the daily reminder e-mail works, but at least once or twice it did not seem to work for me.

In Lesson 7 / Exercises, the solution to Exercise 1 seems to be a non sequitur. The equation is

2*x - y[x] + y'[x]*(2*y[x] - x) == 0

The solution says "The functions N(x,y(x)) and M(x,y(x)) can be identified from the differential equation:"

y1=Integrate[(-s+0),{s,0,x}]

REQUEST: Can you please post a URL for the daily study group sessions? Everything I try just takes me to the registration page. It has dial-in information that doesn't work (no Dial-In ID Number), and I can't find a link to the session that has surely started by now.

The issue here is that the form of y[x] in the definition of soln2 isn't in the right form to make the replacement. This variable has the definition:

soln2={{y[x]->(b+E^(-a x+Subscript[\[ConstantC], 1]))/a}}

When making the substitution into the equation, you first need to extract just the functional form for the solution rather than the entire rule. For example:

In[24]:= soln2[[1, 1, 2]]
Out[24]= (b + E^(-a x + C[1]))/a

This form can be substituted into the equation to verify it is true.

In[25]:= (y'[x] + a*y[x] == b) /. y -> Function[x, Evaluate[soln2[[1, 1, 2]]]]
Out[25]= True

For your second question. There isn't an easy way to automatically redefine the constant term in the exponential as another constant. If the equation was in a simpler form, you could use a replacement rule such as:

In[26]:= a*Exp[C[1]] /. Exp[C[1]] -> C[2]
Out[26]= a C[2]

However, since the -a*x term is also in the exponential, more complicated pattern matching that wouldn't be general for all situations would be required pick out just the Exp[C[1]] term. In these cases it's normally easier to just redefine the arbitrary constants manually.

In Lesson 5 (Linear 1st Order Eqns), the example for constant coefficients shows a manual reworking of the solution to bring the arbitrary constant C out of the exponential function, then uses that manually reworked expression as a replacement for y when checking the solution:

y->Function[x,(b+\[ConstantC] Exp[-a x])/a]

When I tried simply replacing y with the generated solution

[In]    soln2 = Solve[soln, y[x], Reals]

    (y'[x] + a*y[x] == b) /. y -> Function[x, Evaluate[soln2]] // Simplify

[Out]    {{y[x] -> (b + E^(-a x + C[1]))/a}}

the result would not simplify to True. It doesn't perform the indicated replacements:

[In]    (y'[x] + a*y[x] == b) /. y -> Function[x, Evaluate[soln2]] // Simplify

[Out]    
{{a (y[x] -> (b + E^(-a x + C[1]))/a) + (Derivative[1][y][x] -> -E^(-a x + C[1]))}} == b

QUESTION #1: Why won't it simplify?

QUESTION #2: Is there a way to have Mathematica do the work, rather than having to rework the solution manually?

When first studying differential equations, I seem to recall we were drawing magnitudes with the vectors in the direction field. Or maybe my memory is not accurage. I mostly recall they were a pain to draw manually -- much work. :( Is there any advantage to drawing magnitudes with the vector fields, or has everyone agreed that having unit magnitude vectors is the way to go?

Does VectorPlot[] allow one to draw vector fields with something other than a unit magnitude? Have you ever encountered an application where that was useful?

Regarding Lesson 3 and the associated Problem 5: At first I was baffled why the instructor did so much work to come up with a solution that he then verified using DSolveValue[...]. Why not simply use DSolveValue[...] to begin with, as was done in the recent course on Partial Differential Equations. Then I realized that this instructor is actually teaching us math, not simply Wolfram Language functions!
SUGGESTION (in terms of Problem 5 of Problem Session 1, but can be applied to the population model in Lesson 3 as well): After the text "Move dx to the right-hand side and use Integrate on the left-hand side", show the result of moving dx to the right-hand side BEFORE integrating:
(1 / {f(x) + 1)) df(x) = 1 * dx
NOW point out that we have separated variables and can integrate each side separately. (The instructor said this, but we didn't have this form of the equation to look at, so I missed his point.)