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[WSG77] Daily Study Group: Differential Equations (begins April 13)

Posted 21 days ago

Please join us in a study group devoted to differential equations that begins Monday, April 13. This study group will meet daily, Monday to Friday, over the next two weeks. We will share the excellent lesson videos from the Wolfram U course "Introduction to Differential Equations." The study group sessions will include time for exercises, discussion and Q&A. This study group will help you achieve the course completion certificate for the "Introduction to Differential Equations" course after you complete the course quizzes.

REGISTER HERE

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POSTED BY: Luke Titus
55 Replies

Basically, you need to convert everything to a string then use <> to combine the strings you want. You can use quotes to convert the C[1] expressions to a string, but you need to use the ToString function to convert the iterator in the Table to a string. For example:

In[94]:= Table["C[1] = " <> ToString[i], {i, -2, 2}]
Out[94]= {"C[1] = -2", "C[1] = -1", "C[1] = 0", "C[1] = 1", "C[1] = 2"}
POSTED BY: Luke Titus

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.

POSTED BY: Luke Titus

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

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.

POSTED BY: Luke Titus
Posted 2 days ago

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"?

POSTED BY: Updating Name

Hello Richard,

An update has been made per your feedback. Please delete the related saved file from your cloud storage to see the updated version. https://www.wolframcloud.com/browse#Home/Copied%20Files

Thank you,
Christine Owens
Wolfram U Project Manager
Wolfram U

POSTED BY: Christine Owens

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.

POSTED BY: Luke Titus

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]
POSTED BY: Luke Titus
POSTED BY: Luke Titus

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.

POSTED BY: Luke Titus

Someone from the WolframU team will reach out to you shortly.

POSTED BY: Luke Titus

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.

POSTED BY: Luke Titus
POSTED BY: Luke Titus

It does seem like the symbol f has some other definition attached to it. If you clear f before evaluating the derivative, it should give the expected result. For example:

Clear[f]
D[Exp[f[x]], x]
POSTED BY: Luke Titus

You can get an imaginary r(t) function, for example, by using an initial condition on r[0] that is complex. However, we are mostly focusing on the real-valued functions in this course.

POSTED BY: Luke Titus

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.

POSTED BY: Luke Titus

Sure, but most (maybe all?) introductory diff. eq. courses restrict their attention to real-valued solutions. [This is a practical convention, since the course usually follows first-year, real-valued calculus.]

POSTED BY: Michael Rogers

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.

POSTED BY: Luke Titus

Thank you very much. It does look like it should be Exp[x] rather than Exp[2]. We'll get that fixed.

POSTED BY: Luke Titus
POSTED BY: Christine Owens
Posted 2 days ago
POSTED BY: Updating Name
Posted 2 days ago

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!

POSTED BY: Rich Albert

Thanks for pointing that out, Richard. We'll get that fixed.

POSTED BY: Luke Titus

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

POSTED BY: Jamie Peterson

COMMENT on END-OF-COURSE SURVEY: You ask us to indicate all the Daily Study Groups we'd be interested in, but the radio buttons only allow a single choice. SUGGESTION: Change the reply function to checkboxes.

Hello Richard,

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

Christine Owens
Wolfram U Project Manager
Wolfram U

POSTED BY: Christine Owens

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

POSTED BY: Luke Titus

To form an augmented matrix from matrix A and vector b, it's important to realize that the augmented matrix uses b as a COLUMN vector. If b is a single-level list, that amounts to a row vector (in spite of what some Mathematica documentation says), so we must transpose it. Transpose[ ] works with matrices (multi-level lists) but not with row vectors (single-level lists). Assume A is {{1,2,3},{4,5,6}} and b is {x,y}. To transpose b we must first enclose it in curly braces to form a two-level list. We then direct the Join[ ...] function to act at level 2 of the lists, instead of level 1.

Augmented matrix:: Join[A,Transpose[ { b } ], 2]

Output: {{1, 2, 3, x}, {4, 5, 6, y}}

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.

POSTED BY: Luke Titus

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

POSTED BY: Christine Owens

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

enter image description hereenter image description here

POSTED BY: Jaime Jiménez

THANK YOU. I really appreciate your taking the trouble to address my difficulty. MY REAL PROBLEM may simply be impatience. I just now tried the e-mailed link again, and this time the screen says "The webinar will begin momentarily ..." Apparently it doesn't work if you try it too soon.

Posted 12 days ago
POSTED BY: Phil Earnhardt

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}]
Posted 13 days ago

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.

POSTED BY: Phil Earnhardt

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.

POSTED BY: Luke Titus

You can get different sized arrows using VectorScaling and VectorSizes. For example:

VectorPlot[{y, -x}, {x, -3, 3}, {y, -3, 3}, VectorScaling -> Automatic, VectorSizes -> {0, 1}]

The advantage of using all vectors with a unit length and specifying the magnitudes with colors is to make the plot less cluttered when vector lengths differ a lot in the plot. When using the default unit lengths for the vectors, you can get a legend that shows the length that corresponds to a particular color using PlotLegends

VectorPlot[{y, -x}, {x, -3, 3}, {y, -3, 3}, PlotLegends -> Automatic]
POSTED BY: Luke Titus
Posted 15 days ago
POSTED BY: Phil Earnhardt
Posted 16 days ago
POSTED BY: Tingting Zhao

This is just a reminder that the Differential Equations Study Group begins tomorrow (Monday, April 13).

The Study Group will offer an excellent opportunity to get certified in Differential Equations with guidance from our popular instructor, Luke Titus.

I look forward to seeing you all!

POSTED BY: Devendra Kapadia
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