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visit MIT OpenCourseWare at ocw.mit.edu. HERBERT GROSS: Hi, in our lesson
last time, we showed that, to find the general solution of
a linear differential equation, we first looked for the general
solution of the homogeneous or reduced equation--
that is the equation where the right-hand side was 0-- and then found a
particular solution of the original equation. And then the sum of
these two solutions would be the general solution. Consequently, we agreed
that we would now separate our study
into two pots, one to find a general solution
of the reduced or homogeneous equation and the other to
find a particular solution of the given equation. In today's lesson, we
are going to tackle the problem of finding
the general solution of the homogeneous equation,
but subject to a very special case, special in the
sense that it's easy to handle, but also special
in the sense that many practical applications live
up to this particular model. What we're going to tackle today
is solving the linear equation L of y equals 0 subject
to the condition that we have constant
coefficients. In other words, we
are going to deal with y double prime plus
2a y prime plus by equals 0 where a and b are constants. The only reason I
put the 2 in here is to simplify some
algebra later on. After all, if a is an
arbitrary constant, so is 2a. We'll talk about that in
more detail where I need it. The thing is that,
to solve an equation of this particular type
where a and b are constants, we essentially try for
a solution in the form y equals e to rx, recalling that,
when we differentiate this, we get re to the rx. When we differentiate
it a second time, we get r squared e to the rx. Consequently, substituting
this into here, we get that e to the rx
is a constant factor. And we're then left
with a polynomial in r equal to r squared
plus 2ar plus b. In other words, mechanically,
notice that the power of r replaces the given
derivative. r squared replaces y double prime. r to the first power
replaces y prime. And this is the r to
the 0 term replaces the 0 derivative term,
namely the zeroth derivative is y itself. At any rate, since e
to the rx cannot be 0, to find the values of r
that satisfy this equation, then what we have to do
is set this polynomial, in this case a
quadratic equation. And those, of course, if I
had order greater than 2, the same theory would hold. I have simply, as usual,
picked second-order equations for illustrative purposes. The exercises will take
care of higher orders. At any rate, the values
of r are determined by the equation r squared
plus 2ar plus b equals 0. Now you see, if I use
the quadratic formula and solve this
equation for r, I find that r is minus a plus
or minus the square root of a squared minus b. That's the reason I used
2a instead of a here. You see, if I had used
a here without the 2, then the quadratic
formula would have said r is equal to minus a
plus or minus the square root of a squared minus 4b over 2a. And, with blackboard
space at a premium, I hate to have to keep writing
in a 2a as a denominator. And, every time
I do that, I lose a line of space on my board. So, for convenience sake,
I chose the 2 in here so I could write r this way. Now notice that,
in general, this determines two values of r, one
with the positive square root, one with the
negative square root. Let me call r1 the
root that I get with the positive sign, r2 the
root with the negative sign. And the point that
I want to mention is that the nature of the
magnitude of a squared minus b determines what the
values of r look like. For example, if a squared
minus b is positive, then r1 and r2 are
unequal real roots. If a2 minus b-- if a squared minus b
is 0, then r1 and r2 are both equal to negative a. And, if a squared
minus b is negative, r1 and r2 are
complex conjugates. Now the idea is simply this. What we're going
to do now is I'm going to summarize for you
what these results are. I'm going to use the
cookbook technique first of illustrating all
three cases for you, giving you examples
of each, and then I will conclude today's
lesson simply by showing you why the recipes work. And, with a little bit of
luck, for the first time in our course, we may
actually have a short lecture. But let's see what
actually goes on here. Let's not be too presumptuous. The three cases are this. Either a squared
is greater than b-- case 1, a squared
is greater than b-- so r1 is unequal to r2,
and they're both real. And, in this case,
the general solution of the homogeneous
equation, y double prime plus 2a y prime
plus by equals 0, is simply a linear
combination of e to the r1 x and e to the r2 x, namely,
c1 e to the r1 x plus c2 e to the r2 x. For the second case, when a
squared equals b, in that case, r1 equals r2. And, because they're equal, let
me just call both of them r1. Notice that, if I
came back up here, this would really not be a
general solution in this case because, with r1 equal to r2, e
to the r1 x is a common factor. I could then, well, amalgamate
the two arbitrary constants here and write what? That y sub G, the
general solution, is just c1 plus c2
times e to the r1 x. And, if c1 and c2 are
arbitrary constants, then their sum is simply
a arbitrary constant. It turns out that, in
this particular case, another solution of this
equation, meaning what? Another one other
than e to the r1 x, which is not a constant
multiple of e to the r1 x, is xe to the r1 x. And we'll verify that
later in the lecture. For now, just take
this for granted. And, in that case, it turns
out that the general solution is given by the linear
combination of e to the r1 x and xe to the r1
x, c1 e to the r1 x plus c2 xe to the r1 x. In the third case, we have
that a squared is less than b. That means that r1 and r2
are complex conjugates, which means that r has to form a real
number plus or minus i times a real number. Calling the real part alpha
and the imaginary part beta, all we're saying is, if a
squared minus b is negative, alpha is playing the
role of negative a here. And beta is just the negative
of what's under the radical sign here. In other words, if a
squared minus b is negative, b minus a squared is positive. In other words, the beta
part is just the square root of b minus a squared, you see? And, in this case, it turns
out that the general solution is a linear combination,
a very interesting one. Well, let me factor out
the e to the alpha x. What it says is
first take e raised to the alpha x
power, where alpha is the real part
of our root r, OK? And then multiply that by a
linear combination of sine beta x and cosine beta x where beta
is the imaginary part of r. By the way, without
going into this because the textbook
does it very eloquently, and we'll have enough
exercises on this to hammer this home later, notice
the intrinsically different geometric nature of interpreting
these various solutions. For example, in this
particular case, just speaking very
quickly, here we see that the general solution
is the sum of two exponentials. Coming down to here, notice
that, if alpha happened to be 0, in other words,
if r were purely imaginary, notice that the
solution would be what we call oscillatory motion. And, if alpha is not 0, notice
that e to the alpha x c1 cosine beta x plus c2 sine beta x is
an oscillation that is either blown up if alpha is positive
or shrunk if alpha is negative. In other words, the
oscillatory part stays here, but, if alpha happens
to be negative, say, as x gets very large,
e to the alpha x goes to 0. That pulls this whole
expression down to 0. And that's the
case where you get what's called damped
oscillatory motion. In other words, you
get oscillation, but the amplitude
keeps shrinking. You see, physically, different
things start to happen here. And this is a very
interesting subtlety that just small changes
change the entire solution or the meaning of a solution. And maybe the best
way to illustrate that is to show examples
of all three of these cases where the examples
almost look alike. Well, let me show what I mean. For example, we
already saw last time that, if y double prime minus
4y prime plus 3 y equals 0, then the general solution was c1
e to the x plus c2 e to the 3x. That's precisely this
case here because, if we look at the equation
that determines r, it's r squared minus
4r plus 3 equals 0. That says that r1 is 1. r2 is 3. The roots are 1 and 3. So two solutions are e
to the x, e to the 3x. r1 and r2 are real and unequal. That's exactly the
type that we were talking about back here when
we were talking about case one. As far as case two
is concerned, notice that y double prime
minus 4y prime plus 4y equals 0 leads
to the equation r squared minus 4r
plus 4 equals 0. That factors into r
minus 2 squared is 0. And that says that r equals
2 is a multiple root. To illustrate case
two, again, the proof coming shortly, but not
now, what we're saying is that, in this case,
e to the 2x and xe to the 2x will be non-constant
multiples of one another, the solutions, and
that, consequently, the general solution will
be c1 e to the 2x plus c2 x e to the 2x. The third case that I'll
pick is y double prime minus 4y prime plus 5y equals 0, which
leads to the equation r squared minus 4r plus 5 equals 0. And, solving by the
quadratic formula for r, this leads to r
equals 4 plus or minus the square root of
16 minus 20 over 2. The square root of 16 minus 20--
that's the square root of minus 4-- is 2i. 4 plus or minus 2i/2 just
leads to 2 plus or minus i. Since alpha is the real part
of the root, alpha will be 2. Since beta is the magnitude of
the imaginary part of the root, beta will be 1. In this case, alpha is 2. Beta is 1. And, therefore, the
general solution is what? e to the 2x times c1
cosine x plus c2 sine x. And, by the way, before
I leave this example, here's what I meant by
saying how subtle this was. Notice that all three
of these equations are identical, except
for the coefficient of y. In the first equation,
the coefficient of y was 3, in the second equation,
4, and, in the third equation, 5. Other than that, they all
looked alike, the equations. And yet, in one case, we got
the two real and unequal roots. In the second case, we got
the real but equal roots and, in the third case the
non-real complex conjugates, all right? That's all there
is to this lesson, except for giving you
some insight as to why these recipes hold. From this point on,
the rest is drill that there's not
much more we can do, other than to have you look
to see what's happening and work on the
exercises, but I thought that it might help if I give
you a few little supplementary notes as to why these
things work out. And they work out easily enough. So I've decided to do
the supplementary notes as part of the lecture. Namely, first of
all, notice that, in the case where r1 is unequal
to r2, and they're both real, that e to the r1 x and e to the
r2 x are both real solutions. And, more importantly, since one
cannot be a constant multiple of the other, their linear
combination must be the general solution. That's what we saw last time. Remember, the only
time we were in trouble is if one was a constant
multiple of the other. Why can't e to the r1
x and e to the r2 x be constant multiples
of one another? The answer quite simply is this. Assume that they were. Divide both sides
by e to the r2 x. We then get e to the r1
minus r2 x is a constant. But look at this is e
raised to a variable power. The only way that e
raised to a variable power can be a constant is if
the multiplier of the x is itself 0 so that the
x doesn't appear here. In other words, in
that case, notice that r1 minus r2 must be 0. And that's the same as
saying that r1 equals r2. In other words, the only
way that e to the r1 x can be a constant
multiple of e to the r2 x is if r1 equals r2, but
we're given in case one that r1 is not equal r2. That's why the solution works. As far as case two
is concerned, notice that what we have
to establish is it's obvious that xe to the r1 x is
not a constant multiple of e to the r1 x. After all, their ratio is
x, which is not a constant. The hard thing to
do is to justify how we can find out
that xe to the r1 x is a solution in that case. And the answer is
look at that case occurs when a squared equals b. If a squared equals
b, notice that y double prime
plus 2ay prime plus by becomes replaced
by y double prime plus 2ay prime plus a squared y. Now we very cleverly
invoke the trick of writing 2a y prime as
ay prime plus ay prime. See, we take this and
write it like this. Now we factor out an a
from this expression here. We have a times y prime plus ay. Notice that these two terms
are just the derivative of y prime plus ay. In other words, if I let
u equal y prime plus ay, then this being u means
that, with the prime on this, this is u prime. This just becomes u
prime plus au equals 0. That's a first-order
differential equation in u in which the variables
are separable. See, in other words,
this becomes this. From this, I can very
quickly find that u is equal to ce to the minus ax. I didn't want c. I wanted-- I didn't want u. I wanted y. I remember that, by
definition of my substitution, u is y prime plus ay. Consequently, this
leads to y prime plus ay is ce to the minus ax. This is a first-order linear
differential equation. We learnt how to solve
that in a previous unit. Leaving the details
again to you, it now follows that a particular
solution of this equation will be y sub p is
xe to the minus ax. Recalling that, in this
case, r1 is minus a, this is xe to the r1 x. This together with e to
the r1 x form your two linearly independent solutions. And, consequently, that's
why the general solution in this case is c1 e to the
r1 x plus c2 xe to the r1 x. The third case where we've
got the imaginary roots or the non-real roots hinges
on a very interesting property of linearity plus
complex functions. If u and v are real,
notice that L of u plus iv, by linearity, is L
of u plus iL of v because, after all,
i is a constant, be it imaginary or otherwise. Consequently, by linearity,
since L of u plus iv is L of u plus iL of v, the
only way that L of u plus iv can equal 0 is if L of
u plus iL of v equals 0. And the only way a
complex number can equal 0 is if both its real and
imaginary parts are 0. In other words, then,
if L of u plus iv is 0, it means that L of u and L
of v must themselves be 0. See, the real part must be 0. The imaginary part must be 0. Now how do we apply
this to case three? Notice that, in case
three, we solved for r and got that r was equal to
alpha plus or minus i beta. Let me pick the positive root. And we said our solution was
e to the alpha plus i beta x. Now the idea is this
looks quite imaginary, and it this in this form. But what we've just
seen is that, since this is a solution of this
homogeneous equation, the real and the imaginary part
separately must be solutions. Consequently, if I write this in
the traditional u plus iv form, then u and v separately
will also be solutions. Well, let's see how I write
e to the alpha plus i beta x in traditional form. First of all, this is e to the
alpha x times e to the i beta x by the rules of exponents. We recall from our lecture
on complex functions that e to the i beta x is cosine
beta x plus i sine beta x. Consequently, e to
the alpha plus i beta x is just e to the alpha x
times this expression here. Or, separating it into
the form u plus iv, e to the alpha plus i beta x is
e to the alpha x cosine beta x, which is a real number because
alpha and beta are real, plus i times e to the
alpha x sine beta x. And e to the alpha x sine
beta x is also a real number because alpha and beta are real. In other words, u,
the alpha x cosine beta x, that's the real
part of this complex number. v is e to the alpha
x sine beta x, which is the imaginary part of this. Since we've just seen that the
real and the imaginary parts of this function must
satisfy this equation also, that tells us that e to
the alpha x cosine beta x and e to the alpha x sine beta
x also satisfy this equation. And that, by the way, is exactly
what we saw in case three, only we factored out
the e to the alpha x. Notice, by the way, that
these two equations cannot be constant multiples of one
another because, if we divide, say, the top by the bottom,
the e to the alpha x cancel. Sine beta x over cosine
beta x is tan beta x. And, for beta not
equal to 0, tan beta x can't be a constant so that
these two solutions do indeed generate the general solution. Now that's all there is to
this particular problem. The rest, as I say, is
left to the exercises. What we're going to do next time
is tackle the companion part of this problem. And that is, how do you
find the particular solution or a linear
differential equation with constant coefficients where
the right-hand side is not 0, in other words, the
non-homogeneous case? But more about that next time. So, until next time, goodbye. Funding for the
publication of this video was provided by the Gabriella
and Paul Rosenbaum Foundation. Help OCW continue to provide
free and open access to MIT courses by making a donation
at ocw.mit.edu/donate.
