# Lec 7 | MIT 18.03 Differential Equations, Spring 2006 This is also written in the
form, it’s the k that’s on the right hand side.
Actually, I found that source is of considerable difficulty.
And, in general, it is.
For these, the temperature concentration model,
it’s natural to have the k on the right-hand side,
and to separate out the (q)e as part of it.
Another model for which that’s true is mixing,
as I think I will show you on Monday.
On the other hand, there are some common
first-order models for which it’s not a natural way to
separate things out. Examples would be the RC
circuit, radioactive decay, stuff like that.
So, this is not a universal utility.
But I thought that that form of writing it was a sufficient
utility to make a special case, and I emphasize it very heavily
in the nodes. Let’s look at the equation.
And, this form will be good enough, the y prime.
When you solve it, let me remind you how the
solutions look, because that explains the
terminology. The solution looks like,
after you have done the integrating factor and
multiplied through, and integrated both sides,
in short, what you’re supposed to do, the solution looks like y
equals, there’s the term e to the negative k out
front times an integral which you can either make definite or
q of t times e to the kt inside dt,
it will help you to remember the opposite signs if you think
that when q is a constant, one, for example,
you want these two guys to cancel out and produce a
constant solution. That’s a good way to remember
that the signs have to be opposite.
But, I don’t encourage you to remember the formula at all.
It’s just a convenient thing for me to be able to use right
now. And then, there’s the other
term, which comes by putting up the arbitrary constant
explicitly, c e to the negative kt.
So, you could either write it this way, where this is somewhat
vague, or you could make it definite by putting a zero here
and a t there, and change the dummy variable
inside according to the way the notes tell you to do it.
Now, when you do this, and if k is positive,
that’s absolutely essential, only when that is so,
then this term, as I told you a week or so ago,
this term goes to zero because k is positive as t goes to
infinity. So, this goes to zero as t
goes, and it doesn’t matter what c is, as t goes to infinity.
This term stays some sort of function.
And so, this term is called the steady-state or long-term
solution, or it’s called both, a long-term solution.
And this, which disappears, gets smaller and smaller as
time goes on, is therefore called the
transient because it disappears at the time increases to
infinity. So, this part uses the initial
condition, uses the initial value.
Let’s call it y of zero, assuming that you
started the initial value, t, when t was equal to zero,
which is a common thing to do, although of course not
necessary. The starting value appears in
this term. This one is just some function.
Now, the general picture or the way that looks is,
the steady-state solution will be some solution like,
I don’t know, like that, let’s say.
So, that’s a steady-state solution, the SSS.
Well, what do the other guys look like?
Well, the steady-state solution has this starting point.
Other solutions can have any of these other starting points.
So, in the beginning, they won’t look like the
steady-state solution. But, we know that as time goes
on, they must approach it because this term represents the
difference between the solution and the steady-state solution.
So, this term is going to zero. And therefore,
whatever these guys do to start out with, after a while they
They must, in short, be asymptotic to it.
So, the solutions to any equation of that form will look
like this. Up here, maybe it started at
127. That’s okay.
After a while, it’s going to start approaching
that green curve. Of course, they won’t cross
each other. That’s the rock star,
and these are the groupies trying to get close to it.
Now, but something follows from that picture.
curve? All these other white solution
curves have that same property, the same property that all the
other white curves and the green curve, too, are trying to get
close to them. In other words,
there is nothing special about the green curve.
It’s just that they all want to get close to each other.
And therefore, though you can write a formula
like this, there isn’t one steady-state solution.
There are many. Now, this produces vagueness.
the usual answer is whichever one looks simplest.
Normally, the one that will look simplest is the one where c
is zero. But, if this is a peculiar
function, it might be that for some other value of c,
you get an even simpler expression.
either you integrate that, don’t use an arbitrary
constant, and use what you get, or pick the simplest.
Pick the value of c, which gives you the simplest
and that’s what usually called the steady-state solution.
Now, from that point of view, what I’m calling the input in
this input response point of view, which we are going to be
using, by the way, constantly, well,
pretty much all term long, but certainly for the next
month or so, I’m constantly going to be coming back to it.
The input is the q of t. In other words, it seems rather peculiar.
But the input is the right-hand side of the equation of the
differential equation. And the reason is because I’m
always thinking of the temperature model.
The external water bath at temperature T external,
the internal thing here, the problem is,
given this function, the external water bath
temperature is driving, so to speak,
the temperature of the inside. And therefore,
the input is the temperature of the water bath.
I don’t like the word output, although it would be the
natural thing because this temperature doesn’t look like an
output. Anyone might be willing to say,
yeah, you are inputting the value of the temperature here.
This, it’s more likely, the normal term is response.
This thing, this plus the water bath, is a little system.
And the response of the system, i.e.
the change in the internal temperature is governed by the
driving external temperature. So, the input is q of t,
and the response of the system is the solution to
the differential equation. Now, if the thing is special,
as it’s going to be for most of this period, it has that special
form, then I’m going to, I really want to call q sub e
the input. I want to call q sub e the
input, and there is no standard way of doing that,
although there’s a most common way.
So, I’m just calling it the physical input,
in other words, the temperature input,
or the concentration input. And, that will be my (q)e of t,
and by the subscript e, you will understand
that I’m writing it in that form and thinking of this model,
or concentration model, or mixing model as I will show
you on Monday. By the way, this is often
handled, I mean, how would you handle this to
get rid of a k? Well, divide through by k.
So, this equation is often, in the literature,
written this way: one over k times y prime plus y
is equal to, well, now they call it q of t,
not (q)e of t because they’ve gotten rid of this funny factor.
But I will continue to call it (q)e.
So, in other words, and this part this is just,
frankly, called the input. It doesn’t say physical or
anything. And, this is the solution,
it’s then the response, and this funny coefficient of y
prime, that’s not in standard linear
form, is it, anymore? But, it’s a standard form if
you want to do this input response analysis.
So, this is also a way of writing the equation.
I’m not going to use it because how many standard forms could
this poor little course absorb? I’ll stick to that one.
Okay, you have, then, the superposition
principle, which I don’t think I’m going to– the solution,
which solution? Well, normally it means any
solution, or in other words, the steady-state solution.
Now, notice that terminology only makes sense if k is
positive. And, in fact,
there is nothing like the picture, the picture doesn’t
look at all like this if k is negative, and therefore,
the terms would steady state, transient would be totally
inappropriate if k were negative.
So, this assumes definitely that k has to be greater than
zero. Otherwise, no.
So, I’ll call this the physical input.
And then, you have the superposition principle,
which I really can’t improve upon what’s written in the
notes, this superposition of inputs.
Whether they are physical inputs or nonphysical inputs,
if the input q of t produces the response, y of t, and q two of t produces the
response, y two of t, — then a simple calculation
with the differential equation shows you that by,
so to speak, adding, that the sum of these
two, I stated it very generally in the notes but it corresponds,
we will have as the response y1, the steady-state response y1
plus y2, and a constant times y1.
That’s an expression, essentially,
of the linear, it uses the fact that the
special form of the equation, and we will have a very
efficient and elegant way of seeing this when we study higher
order equations. For now, I will just,
the little calculation that’s done in the notes will suffice
for first-order equations. If you don’t have a complicated
equation, there’s no point in making a fuss over proofs using
it. But essentially,
it uses the fact that the equation is linear.
Or, that’s bad, so linearity of the ODE.
In other words, it’s a consequence of the fact
that the equation looks the way it does.
And, something like this would not, in any sense,
be true if the equation, for example,
Everything I’m saying this period would be total nonsense
and totally inapplicable. Now, today, what I wanted to
discuss was, what’s in the notes that I gave you today,
which is, what happens when the physical input is trigonometric?
For certain reasons, that’s the most important case
there is. It’s because of the existence
of what are called Fourier series, and there are a couple
of words about them. That’s something we will be
studying in about three weeks or so.
What’s going on, roughly, is that,
so I’m going to take the equation in the form y prime
plus ky equals k times (q)e of t,
and the input that I’m interested in is when this is a
simple one that you use on the visual that you did about two
points worth of work for handing in today, cosine omega t. So, if you like,
k here. So, the (q)e is cosine omega t.
That was the physical input. And, omega, as you know,
is, you have to be careful when you use the word frequency.
I assume you got this from physics class all last semester.
But anyway, just to remind you, there’s a whole yoga of five or
six terms that go whenever you’re talking about
trigonometric functions. Instead of giving a long
explanation, the end of the second page of the notes just
gives you a reference list of what you are expected to know
for 18.03 and physics as well, with a brief one or two line
description of what each of those means.
So, think of it as something to refer back to if you have
forgotten. But, omega is what’s called the
angular frequency or the circular frequency.
It’s somewhat misleading to call it the frequency,
although I probably will. It’s the angular frequency.
It’s, in other words, it’s the number of complete
oscillations. This cosine omega t
is going up and down right? So, a complete oscillation as
it goes down and then returns to where it started.
That’s a complete oscillation. This is only half an
oscillation because you didn’t give it a chance to get back.
Okay, so the number of complete oscillations in how much time,
well, in two pi, in the distance,
two pi on the t-axis in the interval of length two pi
because, for example, if omega is one,
cosine t takes two pi to repeat itself,
right? If omega were two,
it would repeat itself. It would make two complete
oscillations in the interval, two pi.
So, it’s what happens to the interval, two pi,
not what happens in the time interval, one,
which is the natural meaning of the word frequency.
There’s always this factor of two pi that floats around to
make all of your formulas and solutions incorrect.
Okay, now, so, what I’m out to do is,
the problem is for the physical input, (q)e cosine omega t, find the response.
In other words, solve the differential
equation. In short, for the visual that
you looked at, I think I’ve forgot the colors
now. The input was in green,
maybe, but I do remember that the response was in yellow.
I think I remember that. So, find the response,
yellow, and the input was, what color was it,
green? Blue, blue.
Light blue. Okay, so we’ve got to solve the
differential equation. Now, it’s a question of how I’m
going to solve the differential equation.
I’m going to use complex numbers throughout,
A because that’s the way it’s usually done.
B, to give you practice using complex numbers,
and I don’t think I need any other reasons.
So, I’m going to use complex numbers.
I’m going to complexify. To use complex numbers,
what you do is complexification of the problem.
So, I’m going to complexify the problem, turn it into the domain
of complex numbers. So, take the differential
equation, turn it into a differential equation involving
complex numbers, solve that, and then go back to
the real domain to get the answer, since it’s easier to
integrate exponentials. And therefore,
try to introduce, try to change the trigonometric
functions into complex exponentials,
simply because the work will be easier to do.
All right, so let’s do it. To change this differential
equation, remember, I’ve got cosine omega t here. I’m going to use the fact that
e to the i omega t, Euler’s formula,
that the real part of it is cosine omega t.
So, I’m going to view this as the real part of this complex
function. But, I will throw at the
imaginary part, too, since at one point we will
need it. Now, what is the equation,
then, that it’s going to turn into?
The complexified equation is going to be y prime plus ky
equals, and now, instead of the right hand side,
k times cosine omega t, I’ll use the whole complex
exponential, e i omega t. Now, I have a problem because y, here, in this equation,
y means the real function which solves that problem.
I therefore cannot continue to call this y because I want y to
be a real function. I have to change its name.
Since this is complex function on the right-hand side,
I will have to expect a complex solution to the differential
equation. I’m going to call that complex
solution y tilda. Now, that’s what I would also
use as the designation for the variable.
So, y tilda is the complex solution.
And, it’s going to have the form y1 plus i times y2. It’s going to be the complex
solution. And now, what I say is,
so, solve it. Find this complex solution.
So, find the program is to find y tilde, —
— that’s the complex solution. And then I say,
all you have to do is take the real part of that,
and that will answer the original problem.
Then, y1, that’s the real part of it, right?
It’s a function, you know, like this is cosine
plus sine, as it was over here, it will naturally be something
different. It will be something different,
but that part of it, the real part will solve the
original problem, the original,
real, ODE. Now, you will say,
you expect us to believe that? Well, yes, in fact.
I think we’ve got a lot to do, so since the argument for this
is given in the nodes, so, read this in the notes.
It only takes a line or two of standard work with
differentiation. So, read in the notes the
argument for that, why that’s so.
It just amounts to separating real and imaginary parts.
Okay, so let’s, now, solve this.
Since that’s our program, all we have to find is the
solution. Well, just use integrating
factors and just do it. So, the integrating factor will
be, what, e to the, I don’t want to use that
formula. So, the integrating factor will
be e to the kt is the integrating factor.
If I multiply through both sides by the integrating factor,
then the left-hand side will become y e to the kt,
the way it always does, prime, Y tilde, sorry, and the right-hand side will
be, now I’m going to start combining exponentials.
It will be k times e to the power i times omega t plus k. I’m going to write that k plus
omega t. i omega t plus k. Thank you.
i omega t plus k, or k plus i omega t. kt?
Sorry. So, it’s k times e to the i
omega t times e to the kt. So, that’s (k plus i omega) times t. Sorry. So, y tilda e to the kt
is k divided by, now I integrate this,
so it essentially reproduces itself, except you have to put
down on the bottom k plus i omega.
I’ll take the final step. What’s y tilda equals,
see, when you do it this way, then you don’t get a messy
looking formula that you substitute into and that is
scary looking. This is never scary.
Now, I’m going to do two things simultaneously.
First of all, here, if I multiply,
after I get the answer, I’m going to want to multiply
it by e to the negative kt, right,
to solve for y tilda. If I multiply this by e to the
negative kt, then that just gets rid of the k that I put in,
and left back with e to the i omega t.
So, that side is easy. All that is left is e to the i
omega t. Now, what’s interesting is this
thing out here, k plus i omega.
I’m going to take a typical step of scaling it.
And you scale it. I’m going to divide the top and
bottom by k. And, what does that produce?
One divided by one plus i times omega over k. What I’ve done is take these
two separate constants, and shown that the critical
thing is their ratio. Okay, now, what I have to do
now is take the real part. Now, there are two ways to do
this. There are two ways to do this.
Both are instructive. So, there are two methods.
I have a multiplication. The problem is,
of course, that these two things are multiplied together.
And, one of them is, essentially,
in Cartesian form, and the other is,
essentially, in polar form.
You have to make a decision. Either go polar,
it sounds like go postal, doesn’t it, or worse,
like a bear, savage, attack it savagely,
which that’s a very good, aggressive attitude to have
when doing a problem, or we can go Cartesian.
Going polar is a little faster, and I think it’s what’s done in
the nodes. The notes to do both of these.
They just do the first. On the other hand,
they give you a formula, which is the critical thing
that you will need to go Cartesian.
I hope I can do both of them if we sort of hurry along.
How do I go polar? To go polar,
what you want to do is express this thing in polar form.
Now, one of the things I didn’t emphasize enough,
probably, when I talked to you about complex numbers last time
is, so I will remind you, which saves my conscience and
doesn’t hurt yours, suppose you have alpha as a
complex number. See, this complex number is a
reciprocal. The good number is what’s down
below. Unfortunately,
it’s downstairs. You should know,
like you know the back of your hand, which nobody knows,
one over alpha. So that’s the form.
The number I’m interested in, that coefficient,
it is of the form one over alpha.
One over alpha times alpha is equal to one. And, from that,
it follows, first of all, if I take absolute values,
if the absolute value of one over alpha times the absolute
value of this is equal to one, so, this is equal to one over
the absolute value of alpha. I think you all knew that.
I’m a little less certain you knew how to take care of the
Well, the argument of the angle, in other words,
the angle of one over alpha plus, because when you multiply,
Plus, the angle associated with alpha has to be the angle
associated with one. But what’s that?
One is out here. What’s the angle of one?
Zero. Therefore, the argument,
the absolute value of this thing is want over the absolute
value. That’s easy.
And, you should know that the argument of want over alpha is
equal to minus the argument of alpha.
So, when you take reciprocal, the angle turns into its
negative. Okay, I’m going to use that
now, because my aim is to turn this into polar form.
So, let’s do that someplace, I guess here.
So, I want the polar form for one over one plus i times omega
over k. Okay, I will draw a picture. Here’s one.
Here is omega over k. Let’s call this angle phi. It’s a natural thing to call
it. It’s a right triangle,
of course. Okay, now, this is going to be
a complex number times e to an angle.
Now, what’s the angle going to be?
Well, this is a complex number, the angle for the complex
number. So, the argument of the complex
number, one plus i times omega over k is how much? Well, there’s the complex
number one plus i over one plus i times omega over k. Its angle is phi.
So, the argument of this is phi, and therefore,
the argument of its reciprocal is negative phi.
So, it’s e to the minus i phi. And, what’s A? A is one over the absolute
value of that complex number. Well, the absolute value of
this complex number is one plus omega over k squared. So, the A is going to be one
over that, the square root of one plus omega over k,
the quantity squared, times e to the minus i phi. See, I did that. That’s a critical step.
You must turn that coefficient. If you want to go polar,
you must turn is that coefficient, write that
coefficient in the polar form. And for that,
you need these basic facts about, draw the complex number,
draw its angle, and so on and so forth.
And now, what’s there for the solution?
Once you’ve done that, the work is over.
What’s the complex solution? The complex solution is this.
I’ve just found the polar form for this.
So, I multiply it by e to the i omega t,
which means these things add. So, it’s equal to A,
this A, times e to the i omega t minus i times phi. Or, in other words,
the coefficient is one over, this is a real number,
now, square root of one plus omega over k squared. And, this is e to the,
see if I get it right, now.
And finally, now, what’s the answer to our
real problem? y1: the real answer.
I mean: the really real answer. What is it?
Well, this is a real number. So, I simply reproduce that as
the coefficient out front. And for the other part,
I want the real part of that. But you can write that down
instantly. So, let’s recopy the
coefficient. And then, I want just the real
part of this. Well, this is e to the i times
some crazy angle. So, the real part is the cosine
of that crazy angle. So, it’s the cosine of omega t
minus phi. And, if somebody says, yeah, well, okay,
I got the omega k, I know what that is.
That came from the problem, the driving frequency,
driving angular frequency. That was omega,
and k, I guess, k was the conductivity,
the thing which told you how quickly the heat that penetrated
the walls of the little inner chamber.
So, that’s okay, but what’s this phi?
Well, the best way to get phi is just to draw that picture,
but if you want a formula for phi, phi will be,
well, I guess from the picture, it’s the arc tangent of omega,
k, divided by k, over one,
which I don’t have to put in.
So, it’s this number, phi, in reference to this
function. See, if the phi weren’t there,
this would be cosine omega t, and we all know what that looks like.
The phi is called the phase lag or phase delay,
something like that, the phase lag of the function.
What does it represent? It represents,
let me draw you a picture. Let’s draw the picture like
this. Here’s cosine omega t. Now, regular cosine would look
sort of like that. But, I will indicate that the
angular frequency is not one by making my cosine squinchy up a
little too much. Everybody can tell that that’s
the cosine on a limp axis, something for Salvador Dali,
okay. So, there’s cosine of
something. So, what was it?
Blue? I don’t have blue.
Yes, I have blue. Okay, so now you will know what
I’m talking about because this looks just like the screen on
your computer when you put in the visual for this.
So, this is cosine omega t. Now, how will cosine omega t minus phi look? Well, it’ll be moved over.
Let’s, for example, suppose phi were pi over two.
Now, where’s pi over two on the picture?
Well, what I do is cosine omega t minus this.
I move it over by one, so that this point becomes that
one, and it looks like, the site will look like this.
In other words, I shove it over by,
so this is the point where omega t equals pi over two. It’s not the value of t.
It’s not the value of t. It’s the value of omega t. And, when I do that,
then the blue curve has been shoved over one quarter of its
total cycle, and that turns it, of course, into the sine curve,
which I hope I can draw. So, this goes up to there,
and then, it’s got to get back through.
Let me stop there while I’m ahead.
So, this is sine omega t, the yellow thing, but that’s also,
in another life, cosine of omega t minus pi over
two. The main thing is you don’t
subtract, the pi over two is not being subtracted from the t.
It’s being subtracted from the whole expression,
and this whole expression represents an angle,
which tells you where you are in the travel,
a long cosine to this. What this quantity gets to be
two pi, you’re back where you started.
That’s not the distance on the t axis.
It’s the angle through which you go through.
In other words, does number describes where you
are on the cosine cycle. It doesn’t tell you,
it’s not aiming at telling you exactly where you are on the t
axis. The response function looks
like one over the square root of one plus omega over k squared
times cosine omega t minus phi. And, I asked you on the problem
set, if k goes up, in other words,
if the conductivity rises, if heat can get more rapidly
from the outside to the inside, for example,
how does that affect the amplitude?
This is the amplitude, A, and the phase lag.
In other words, how does this affect the
response? And now, you can see.
If k goes up, this fraction is becoming
smaller. That means the denominator is
becoming smaller, and therefore,
the amplitude is going up. What’s happening to the phase
lag? Well, the phase lag looks like
this: phi one omega over k. If k is going up, then the size of this side is
going down, and the angle is going down.
Now, that part is intuitive. I would have expected everybody
to get that. It’s the heat gets in quickly,
more quickly, then the amplitude will match
more quickly. This will rise,
and get fairly close to one, in fact, and there should be
very little lag in the way the response follows input.
But how about the other one? Okay, I give you two minutes.
The other one, you will figure out yourself.

### 31 thoughts on “Lec 7 | MIT 18.03 Differential Equations, Spring 2006”

• February 29, 2008 at 7:07 pm

congratulations!!!!

• February 29, 2008 at 7:07 pm

congratulations!!!!

• March 28, 2009 at 11:27 pm

The application(s) is(are) trivial. The theory is where the difficulty lies. Though the application(s) are historically where the theory/theories come from.

• September 15, 2009 at 2:03 pm

MIT isn't (!) any better than any german university!

• December 2, 2009 at 11:50 pm

Germans, always think that their better than anyone else.

• June 10, 2010 at 4:02 pm

@cesfigas Been half a year in Germany, have a german exchange student at home right now and I can tell that what you're saying is absolutely… right. They're such stuck ups! I mean they consider seing the flaws of other people as the highest virtue… yeah, like it's such a hard thing to do.

• January 13, 2012 at 8:22 am

I Really Like The Video From Your First-order Linear with Constant Coefficients: Behavior of Solutions, Use of Complex Methods

• April 7, 2012 at 9:14 pm

are the notes available

• April 20, 2012 at 8:32 am

it is found at the website in the info box above

• June 28, 2012 at 4:54 pm

Home > Courses > Mathematics > Differential Equations > Unit I: First Order Differential Equations > Constant Coefficients

• July 20, 2012 at 7:59 am

thats not correct!!
not every german thinks that way…

• October 16, 2012 at 5:27 am

why is the constant of integration ignored when evaluating y~ (y tilde/complexified equation)? wouldn't the constant be a real part of y~ and so form a part of y (ie. Re{y~}) as the solution to the ODE?

• October 20, 2012 at 12:09 am

Most professors start writing things up early so jump right into things. You'll notice that at the beginning, everyone's talking (just like anyone would before class), but when he begins to speak, they stop talking for the remainder of class.

• April 12, 2013 at 1:12 am

Great lecture but the last bit was a tad bit confusing 🙁

• June 2, 2013 at 4:48 am

Phi would not be given in a real application problem, but k and w will be. Therefore, it's better to have 2 given constants instead of one hidden phi. By the way, the variable here is t.

• January 18, 2014 at 3:31 pm

The quality of this video is so bad I can't read everything on the black board

• May 29, 2014 at 3:38 am

24:30 gee, that's embarassing

• March 19, 2015 at 12:03 am

I get the impression that this video started a few minutes in to the lecture…

• October 14, 2015 at 9:51 am

hey, im a law student interested in mathematics especially theoretical mathematics. i have not found a suitable program in my country and therefore Im taking your lectures. my concern is whether all the lecturers from the first to the last is available and to what extent have you edited to remove parts of it (should i worry about whats not there in essence.)
thanks

• May 5, 2016 at 2:06 pm

Did anyone else get (k^2*cos(w*t) + k*w*sin(w*t))/(k^2 + w^2) + Ce^(-k*t) while solving it independently?

• May 7, 2016 at 2:08 am

Question, would mathematics alone allow you to understand in the future Einstein's general relativity equations and if so how long would it take to you reach that and if so in terms of mathematics courses, for the record currently a math and econ freshman in uk.

• August 11, 2016 at 12:14 am

Great lecture in the series. A small matter is that the constant of integration seems to be left out at 25:18 which will carry the e^-kt for a transient component to the solution. The model may not call for it, but should perhaps be noted for a general case. (I'll die happy if I can learn a fraction of this man's skills)

• June 7, 2017 at 7:21 pm

At 25:13 where is the constant c at the equation?

• November 20, 2018 at 10:42 pm

Why does the steady state solution entail a positive K?

• December 31, 2018 at 12:38 am

So this is where QE and QT came from.

• January 14, 2019 at 1:45 am

video quality so poor. . .not visible

• February 4, 2019 at 1:01 pm

At 37:38: phi = pi/2 but that's only possible if arctan(w/k) = pi/2, meaning w is infinitely large or k = 0, which does not make any physical sense. Can anyone answer?

• March 5, 2019 at 7:24 am

OMG!!!
SS!(S)

• April 21, 2019 at 3:31 pm

MY GOD this method seems way more complicated than how Gilbert Strang solved it in his lecture. Perhaps he's using complex analysis so students get used to it?
Using undetermined coefficients makes it much easier to solve "response to oscillatory/complex input"…

• June 9, 2019 at 1:21 pm
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