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.

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

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

In those examples the phi(r) function is meant to represent any function of r, not just the unit step function.

Hello Richard,

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

Christine Owens

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.

Question for Luke on the definition of autonomous equations. In my notes I have that you presented an example y'(t) = 5 + y(t). However this is still depending on "t", so it is not purely just depending on the dependent variable there. If you are defining an autonomous equation as y' = f(y) versus y' = f(x), that will change whether an autonomous equation is necessarily exact or not.

Hi, I'm not sure how Luke Titus would answer, but since "autonomous" is a fairly standardized concept across mathematics, I'll share my take. Let's say the independent variable is $t$, as in your first example $y'(t) = 5 + y(t)$. Then for a first-order ODE, if the equation can be expressed in terms of $y(t)$ and $y'(t)$ with no other appearance of $t$, then the ODE is autonomous. For higher-order equations, the ODE may contain $y''(t)$, $y^{(3)}(t)$, etc., by no other instances of t.

If we change the independent variable to $x$, then $y'=f(y)$ is autonomous, but $y'=f(x)$ is not (unless $f(x)$ is a constant).

Another characterization involves rearranging the ODE in the form $F(t,y,y')=0$, or $y'(t)-5-y(t)=0$ for the original example. We then consider $F(t,y_0,y_1) $, or $y_1-5-y_0$ for the example. If no $t$ appears in this formula, then the ODE is autonomous.

Finally, if one wants to push abstraction a little further, we can characterize "autonomous" as follows: If there exists a function $G(y_0,y_1)$ such that $F(t,y,y')=G(y,y')$ for all $t$, then $F$ is autonomous. The function $G$ shows that there is no dependence on $t$ other than the dependence on $y(t)$ and $y'(t)$. This is an abstraction of a common way of defining autonomous: If we write $y$ for $y(t)$ and $y'$ for $y'(t)$ in the ODE, then the ODE is autonomous if the differential equation does not depend on $t$; that is, in the example, there is no $t$ in $y' = 5+y$.

@Jason Gross, Just to be clear, I'm saying both of these are autonomous:

• y'(t) = 5 + y(t)
• y' = f(y)

And this one is not autonomous because the independent variable x appears in f(x) and f is not y:

• y' = f(x)

The first one, y'(t) = 5 + y(t), is the same as y' = 5 + y, which has no t. The fact that the solution y(t) and its derivative depend on t is not a factor in the definition of autonomous.

[Update: Edited to fix sign errors.]
As for exact, I'm not sure what definition was used, but I'm sure it's equivalent to the following. Put the autonomous differential equation in differential form: $${dy \over dx} = {M(x,y) \over N(x,y)} \Longrightarrow M(x,y)\;dx + N(x,y)\; dy = 0\;.$$ Then the equation is exact if ${\partial M \over \partial y} = {\partial N \over \partial x}$. For the autonomous $y'=f(y)$, we have $${dy \over dx} = f(y) \Longrightarrow -dx + {1 \over f(y)}\; dy = 0\;.$$ So $M$ is the constant $-1$ and $N$ is $1/f(y)$. Thus ${\partial M \over \partial y}$ and ${\partial N \over \partial x}$ are both zero. So the autonomous equation *in this form is exact. In fact, all separable equations can be made exact with the right integrating factor, and an autonomous equation is separable: $${dy \over dx} = f(y) \Longrightarrow {dy \over f(y)} = dx \;. $$ On the other hand, we also have $${dy \over dx} = f(y) \Longrightarrow -f(y)\;dx + {1}\; dy = 0\;.$$ In this form $M=-f(y)$ and $N=1$. Thus ${\partial M \over \partial y}=-f'(y)$ and ${\partial N \over \partial x}=0$. In this form, the equation is not exact unless $f'(y)=0$, that is, unless $f$ is constant. It does, however, have an integrating factor ( $1/f(y)$, for instance).

Oops, you're right about the minus sign. I was careless. [Now edited to fix the error.]

I understood that the basis for your question was that y is being set as a function of t and tried to address that. Apparently, I haven't succeeded yet. Sorry about that, but I'm also unsure how to make clear that the dependence of the solution function on t is different than the dependence of the differential equation on t.

Note that every ODE in which the independent variable is t has solutions that are functions of t. If that disqualified an ODE from being autonomous, then no ODE would be autonomous, and the term would serve no purpose. It must be the case that a different interpretation was intended.

In the equation $y'=5+y$, I think of $y'$ and $y$ as variables connected by the equation. The variables $y'$ and $y$ represent real numbers. We can plug in real numbers for them, and the equation will be satisfied or not. We can also plug in expressions for them, and the equation will be satisfied or not. As an equation there is no variable $t$. The equation becomes a differential equation by following: A solution to the differential is a function $g(t)$ such that if we plug in $y=g(t)$ and $y=g'(t)$, the equation is satisfied. That is, we have to plug the derivative of the function for $y$ into $y'$.

In treating $y'=5+y$ as a differential equation, $t$ does not appear in the equation; it appears in the solution. If we write the equation as $y'(t)=5+y(t)$, this suggests we're plugging in some solution or other. Then, to my mind, it is no longer properly a differential equation; it is the original equation with a function of $t$ plugged into it. Plugging in a function of $t$ does not make the original equation depend on $t$.

I believe most mathematicians, myself included, would think saying that $y'=5+y$ and $y'(t)=5+y(t)$ are different equations is a technical quibble. But if the presence of $t$ as an argument of the variables $y$ and $y'$ is the basis for saying the differential equation is not autonomous, then I have to resort to a technical objection.

I put the same argument in functional terms before, so I'm unsure it will convince you. Here's a rephrasing. Suppose we have a function or formula $F(a,b,c)$, such as $F(a,b,c)=c-5-b$. There is no $a$ in the example formula. A differential equation is what you get when you plug in $t,y,y'$ and set equal to zero: $F(t,y,y')=0$. In our example, we obtain $y'-5-y=0$ which is equivalent to $y'=5+y$. There was no $a$ in $F(a,b,c)$, and there is no explicit $t$ in $F(t,y,y')$. Whenever one of the variables is missing in the formula for $F$, there is something special about the differential equation. When $t$ is missing, we say the equation is autonomous.

The meaning of autonomous is "self-law" or "self-ruling." The future trajectory of $y$ is completely determined by the value of $y$: $y'=f(y)$ determines how $y$ changes in the future. How $y$ changes at a given time $t$ does not depend on the value of $t$, only on the value of itself. That may not convince that I'm right about the definition. But that is the concept of autonomous that the definition is supposed to define.

Hope that helps.

There's a typo in the setup of Problem 7 in Problem Session 6. The initial conditions in the text intro are

y(0) = 1
y'(0) = 1

However, the Mathematica code for the transform is

leqn=LaplaceTransform[y''[t]+4y'[t]+4y[t]\[Equal]Sin[t], t,s]/.{y[0]->0,y'[0]->1} 

Note that y(0) is given different values in the two places (one in the text, zero in the code).

Thank you. My name is Richard Daehler-Wilking.

Richard Daehler-Wilking daehlerr@bellsouth.net

My result for Exercise 4 in Lesson 19 (series solutions near a singular point) seems to be more interesting than yours.

eqn = xy''[x] + 2y'[x] + y[x] == 0; soln = DSolveValue[eqn, y[x], x]; series = Collect[Series[soln, {x, 0, 5}], {C[1], C[2]}]; series1 = series /. {C[1] -> 1, C[2] -> 0};

In the attached notebook, you can see that DSolveValue[ ] returns a result in terms of BesselJ[1, 2, Sqrt[x]]/Sqrt[x]]. The series approximation after collecting terms for C[1] and C[2] does NOT include a singularity for C[1], and the plot for {C[1]->1, C[2]->0} is well behaved over [0,1]. Have I made a mistake somewhere?

Attachments:

Lesson_19_Exerci...nb

Hi Richard. It looks like you did everything right. The C[1] solution appears to have a singularity at x=0, but the Bessel function has a leading order term of Sqrt[x] in its series expansion.

In[58]:= Series[BesselJ[1, 2 Sqrt[x]], {x, 0, 4}] // Normal
Out[58]= Sqrt[x] - x^(3/2)/2 + x^(5/2)/12 - x^(7/2)/144

This leading order term in the Bessel function will cancel the Sqrt[x] in the denominator which will lead to a finite expression at x=0.

In[59]:=Series[BesselJ[1, 2*Sqrt[x]]/Sqrt[x], {x, 0, 4}]//Normal
Out[59]= 1 - x/2 + x^2/12 - x^3/144 + x^4/2880

Thank you Richard. I've noted this for Luke to review.

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)!)  *)

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)!

Just a practical question here - the quizzes are not timed are they? Thanks.

In Lesson 19 (Series Solutions near a Singular Point), Example 1 glosses over a subtle, difficult, and perhaps important point. For the equation 4xy''[x] + 2*y'[x] + y[x] == 0, one recursion relation is derived leading to the sequence of coefficients {a[0], -a[0]/2!, +a[0]/4!, -a[0]/6!, ...} and another relation is derived leading to {a[0], -a[0]/3!, +a[0]/5!, -a[0]/7!, ...}. Both these sequences appear in my old CRC handbook table of trigonometric series. The first is for Cos[x] and the second is for Sin[x]. But Lesson 19 goes on to say that the sequences we've just derived are for Cos[Sqrt[x]] and for Sin[Sqrt[x]] . Where did the Sqrt[..] come from?

The answer is implicit when you show the result of Series[Cost[Sqrt[x]], {x, 0, 3}]. There, we discover that the powers of x in the Cos series are {1, x, x^2, x^3, ...}, while the powers in my trust old CRC handbook are {1, x^2, x^4, x^6, ...} (each term the square of CRC's list). For Sin, CRC shows {x, x^3, x^5, x^7, ...} instead of {x^(1/2), x^(3/2), x^(5/2), x^(7/2), ...} as shown by Mathematica for Sin[Sqrt[x]].

QUESTION: How are we to know the powers of x that go with each term in the series expansion? I could probably answer this question 55 years ago in Calc 210, but it eludes me now.

The series expansion for Cos(x) is

In[33]:= Series[Cos[x], {x, 0, 6}] // Normal
Out[33]= 1 - x^2/2 + x^4/24 - x^6/720

When you make the substitution that x -> Sqrt[x], the power of each term in the series is cut in half, so you get the series from the notebook.

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

That's a good point. Thanks for the suggestion, Richard. We'll update the notebooks to reflect that.

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.

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

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.

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.

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.

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

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

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 think there's a significant typo in the solution to Problem 11 of Problem Set 2. The problem is to plot the first four iterations of Picard's method for this initial value problem:

y'[t] == (t^2 + y[t]^2 )^2      and      y[0] == 0

The published solution starts out with

y1 = Integrate[0 + s^2, {s, 0, t}]

This omits the outer exponent of the original equation. Shouldn't it be

y1 = Integrate[ ( 0 + s^2) ^2, {s, 0, t}]

In the Basic Example of Lesson 13 (Method of Undetermined Coefficients) , DSolveValue returns a solution in terms of Cos[x] and Sin[x]. QUESTION: Is it possible to specify which basis functions DSolveValue uses? That is, can we ask DSolveValue to return an answer in terms of

Exp[x] and Exp[-x]?

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

I must have missed something. Lesson 8 is on Picard's Theorem. The first two exercises for Lesson 8 are indeed connected to Picard's Theorem, but exercises 8.3 through 8.5 seem unrelated to it. For example, Exercise 8.3 is:

Transform the following initial value problem to the equivalent problem with the initial value at the origin: 2x - y[x] + y'[x](2*y[x] - x) == 0 and y[1] == 5

The solution doesn't make sense to me, and Mathematica says the solution's transformed equation does NOT correspond to the original. The original solution is entirely real; the transformed solution is complex. Perhaps I made a typo, but I triple-checked it and also used copy-and-paste.

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}]

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

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.

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

I see. You should be getting an email after each lesson with a link to the recording. I'll talk to Cassidy today to see if we can resend those links to you.

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

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.

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"}

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]

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.

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]

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.]

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.)

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!