In Problem Session 7, Problem 3 has a sign error in the forcing function. The problem statement says the forcing function is Sin[t] - UnitStep[[t - 2Pi]Sin[t] with a minus sign before UnitStep.

However, the Wolfram Language statement has this equation:

eqn=y''[t]+9y[t]\[Equal]Sin[t]+UnitStep[t-2Pi]Sin[t];   (* Note the plus sign before UnitStep *)

The minus sign reappears when you plot the function with its forcing function, and the plot shows that the forcing function terminates after one cycle. Therefore, the plus sign in the equation is wrong.

ALSO: The discussion says the solution oscillates indefinitely because there is no damping term. However, my solution is in fact dampened entirely by the (negative) forcing function. It's a beautiful example with the negative sign.

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.

In Examples 3, 4, and 5 of Lesson 26 (Convolution Theorem), the analysis stops at the same point, showing that the solution has the form

Integrate[ someFunction[ t-r ]*phi[r], {r,0,t}]

where phi[r] represents the UnitStep function. Since the Wolfram Language apparently does not understand the meaning of phi, it does not resolve the integral . HOWEVER, it would help the student if you would point out in all these cases that the UnitStep function (phi) simply equals 1 in the region of integration {r, 0, t}, and therefore the integral reduces to Integrate[ someFunction [t-r], {r,0,t} ]. I'm sure it would also be useful to remind the student that L[UnitStep] is 1/s .

In the Exercises for Lesson 24 (Discontinuous Forcing Functions), Exercise 1 solves a different equation than posed at the top of the problem. At the top, the equation is

2*y''[t] + y'[t] +2*y[t] == UnitStep[t-2]   (* Note the sign of the y'[t] term. *)

The equation you actually solve for is

eqn = 2*y''[t] -y'[t] + 2*y[t] == UnitStep[t-2]      (* Note the sign of the y'[t] term. *)

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

Hello Jason,

I'll inquire about this with Luke.

Thank you,

Christine Owens
Wolfram U Project Manager
Wolfram U

@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.

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.

Thanks, I do recognize your point about y being the same as y(t) where t is the independent variable. I was approaching that aspect incorrectly.

I will say, going back to the other issue about exactness, I did provide an answer to the question on the quiz regarding whether all autonomous equations are exact that was nuanced as it took into account the different forms in which the equation could be expressed which determines whether the equation is exact and it marked it as incorrect. So for that I will need the teacher's input because I don't know if the form of the equation is being deemed as relevant - although I am not sure how it wouldn't be.

Hello,

Thank you for bringing this to our attention. Can you please confirm your name? I'll follow up with you via email once this has been reviewed and updated.

Best,
Christine Owens
Wolfram U Project Manager
Wolfram U

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

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

Sorry to submit so many questions -- but not really. They come up because the course is so rewarding and has so much potential. Thanks for the challenge!

In Lesson 19 (Series Solutions near a Singular Point), Example 2, I'm able to duplicate everything up until the solution to the recursion relations (notebook attached). When I try to use RSolve[ ], I get expressions involving Pochhammer[...], which I've never heard of before. Wolfram documentation for Pochhammer indicates it should reduce to ratios of factorials if it is given integer arguments, but the solutions I come up with all involve a fraction.

QUESTION: When RSolve[ ] gives us Pochhammer, is there anything we can do to simplify it? Or is there something in particular we should look for to fix it?

Attachments:

Lesson_19_Exampl...nb

Thank you very much for all of your comments, Richard. I really do appreciate it. We've put a lot of work into this course and we want to make it the best course we can. Your comments have helped us a lot with identifying some of the errors or improvements that we've overlooked.

Hello Jason,

No, the quizzes are not timed. There is a deadline to pass all quizzes in the framework to earn the Completion certification for Daily Study Group: Introduction to Differential Equations, which is May 8.

Otherwise, you can earn Completion within the course framework.

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

The voiceover for Lesson 18 (Euler Equations) includes an important sentence that is NOT in the written notes: "The equation can be simplified by assuming the solution is some power of x." [At time stamp 2:09 or so.] The written notes then say "Euler observed that by utilizing the power rule (x^r}' == r*x^(r-1), the differential equation could be transformed into a simpler form ..."

SUGGESTION: Include the voiceover assumption that y[x]==x^r in the written notes. It's implicit in the Wolfram Language transformation ( /.y->Function[x,x^r] ) but has no justification if one is just reading the notes.

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.

In Problem Set 4, Problem 7 says to solve an initial value problem with this differential equation:

y''[x] - y[x] == Exp[2]*x^2   (* Note the constant exponent 2 for Exp[] *)

However, the solution of the problem implicitly assumes g[x] equals Exp[x]*x^2 , where Exp[ ] has an argument of x instead of 2. I had no trouble solving the problem as stated with Exp[2], but perhaps that's a typo in the problem?

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 have a similar problem with the free-response question on Quiz 4. I'll do as you suggest.

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

Hello Richard,

I've created a ticket per your inquiry for review.

Christine Owens
Wolfram U Project Manager
Wolfram U

I believe I figured out my original problem: The solution to Problem 5 refers to the Method of Undetermined Coefficients even when it is applying Variation of Parameters. Please check: Above the equation

u1=-Integrate[y2[x]Sin[x]/Wronskian[{y1[x],y2[x]},x],x]+C[1] 

do you mean to say "Method of Variation of Parameters" instead of "Method of Undetermined Coefficients"?

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.

Thank you very much for your comments, Richard. Sorry that I couldn't give a better reply during the session today, but posting these comments here helps us keep track of everything that needs to be improved in the course.

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]

Thank you very much. It does look like the solution for those two exercises have been switched. We'll have to fix that. Thank you for pointing that out. We really appreciate it.

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.

In Section 12 of the course, the symbol Lambda was used for the characteristic equation. Can you explain why that particular symbol was used there? It seems that choice was no accident, but I see no clues to why that was chosen. TY.

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.

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.

No, I'm NOT looking for a link to the recording. Cassidy has been sending those out each day. What I'm looking for is a link to the presentation ABOUT TO BE PRESENTED -- in other words, the live class session. For example, I just received the reminder for today's course session. It has a link [Join Daily Study Group] and immediately below that another link headed by "Join using your custom link:"

Both links take me to to a page describing the webinar, with a graphic that sometimes contains an arrowhead that lets me into the course and sometimes does not. Yesterday it did not. Today it does not. I have dialed into the session (Dial-In Number 176487# and Passcode 8751#) and am now listening to some very nice hold music ...

I'm trying to generate nice legends like the ones in the Autonomous Equations section of Lesson 6:

C[1] = -2  C[1] = -1  C[1] = 0  C[1] = 1  C[1] = 2 

I tried Table["C[1] = "j, {j, -2, 2}] but this generated

-2 C[1] =
-C[1] = 
0
C[1] =
2 C[1] =

In other words, Mathematica did math instead of lexical substitution!
QUESTION: Is there an analog of the C language sprintf function by which we can generate formatted expressions?

Something seems odd in the "Integrating Factor Method" section of Lesson 5 (Linear 1st-Order Eqns). The input

D[Exp[f[x]],x] 

generates Exp[f[x]] f'[x] on my system, as expected. However, in the lesson notebook it generates output with an exponent that seems to be a detailed list of possibilities. Pasting it into this box gives an unholy mess, but the first set of possibilities seems to be

1 if 0 <= x < 1 and
0 otherwise

and the second set of possibilities looks like

0 if x is not equal to zero or one;
Indeterminate otherwise

My guess is that the function f is carrying a definition over from a previous cell. SUGGESTION: clear the contents of f before calling D[Exp[f[x]],x]

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?

Can r be i?

Thank you very much for your comments, Richard. Yes, the main focus of this course is the learn the math and theory behind differential equations while using the Wolfram Language simply as a tool to do some of the work. I appreciate your comment about how we should show more of the intermediate steps and explicitly show what is being integrated before doing the integration. We will take this into account when making future improvements to this course.