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DAVID SHIROKOFF: Hi, everyone.

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So today I'd like to take
a look at linear equations,

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linear first-order
differential equations.

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And specifically, we're going
to look at just exactly, what

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is a linear equation?

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So for example, which of
these equations are linear?

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And then second of
all, we're going

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to take a look at superposition.

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And we ask to consider
the differential equation

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y dot plus y squared equals
some function q of t.

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And just to demonstrate that
this equation does not satisfy

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the superposition principle.

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So I'll let you take a
look at this problem,

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and try and work it
out for yourself.

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And I'll be back in a minute.

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Hi, everyone.

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Welcome back.

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All right.

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So we're asked to figure
out which of these equations

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are linear.

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So just to recap,
what does it mean

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for an equation to be linear?

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What does it mean for
a first-order equation

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to be linear?

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Well, it means that we can
write it in the general form,

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so if it's a
function of x, dy/dx

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plus some general function of x.

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And this could be anything.

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But the point is that
it's multiplied just by y.

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And then it can equal
some arbitrary function

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q of x on the right-hand side.

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So every first-order
linear equation

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can be rewritten in this form.

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So let's tackle part A. All
right, y dot plus k*y equals--

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or y dot equals k*y.

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We can just rewrite it as
y dot equals negative k*y.

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Note that just some constant
k is a function of time.

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It's a constant
function of time,

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but it can just be thought
of as a function of time.

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Same with the
right-hand side, it's 0.

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And so this equation
has the same form

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as our general first-order
linear equation.

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So this is a linear equation.

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Let's take a look at
question 2, part 2.

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We have y dot plus y squared t.

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I'll bring the y to
the left-hand side.

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If we do that, we get this.

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And just by comparing this to
our generic first-order linear

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ODE, we note that
it's not of this form.

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So specifically, this is
the bad term in question.

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We see that it's
y squared times t.

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Notice how we're not allowed
to have any term that's y

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squared in our first-order
linear equation.

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Now I will say this
just as an aside.

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This equation's also known
as a Bernoulli equation.

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And as an aside, if you were
to make a substitution u equals

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1 over y, and rewrote this
equation in terms of u

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as the independent
variable, you'd

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find out that this equation
would be linear in terms of u.

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But as it's written now in
terms of y, it's not linear.

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So this is a kind of
an interesting point.

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That sometimes we can get
lucky and do a nonlinear

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transformation and
convert a nonlinear

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equation into a linear one.

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So for part 3 we have y prime
plus cos(x)*y equals x cubed.

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So again, we see that
this is a linear equation

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because we can identify
p of x with cosine x, q

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of x with x cubed.

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So this equation has the
general form of a linear ODE.

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For part 4, well if we take
a look at it right away,

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y dot divided by y is
equal to t squared.

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It doesn't appear to be linear.

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But of course, we can
multiply through by y

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and rewrite things.

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And of course, this
equation is equivalent to y

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dot minus t squared y equals 0.

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Which again, is
linear because p of t

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can be identified with
negative t squared and q of t

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can be identified with 0.

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And then lastly,
for part 5, we have

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x squared y y prime
plus 4x equals x cubed.

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And again, if we were to try
to write it in the general form

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of a linear ODE, which we have
here, it would look like...

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4x divided by x squared y minus
x cubed divided by x squared

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y equals 0.

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And because there's a y
on the denominator here,

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this equation doesn't
have the general form

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of a first-order
linear equation.

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Now again, I note that if you
were to make a substitution,

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u equal to y squared,
that substitution

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would make this equation linear.

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So this concludes part a.

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So for part b, we're given
a differential equation

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y dot plus y squared
is equal to q of t.

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And we want to show that
this equation doesn't satisfy

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the superposition principle.

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Now, you might note that
this equation is not linear.

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And second of all, we know
that linear equations satisfy

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the superposition principle.

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So we need to exploit the fact
that it's a nonlinear equation

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to make it fail the
superposition principle.

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And just as an example,
what we can do is I

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can just pick a couple
right-hand sides

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q as examples of this ODE.

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So we can imagine we
have a solution y_1 which

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solves the ODE y_1
dot plus y_1 squared

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equals-- I'm just going to
say 1 on the right-hand side.

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And I can also take
another function

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y_2, which say satisfies this
differential equation, say t.

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So all I've done
is I've just picked

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one q_1 on the right-hand
side and another function q_2.

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And I imagined that I
have a solution y_1 which

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satisfies this differential
equation and y_2 which

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satisfies this
differential equation.

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So what does the
superposition principle say?

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Well, superposition
says that if y_1

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solves this equation and
y_2 solves this equation,

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then the function y equals
y_1 plus y_2 must solve

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the equation y dot
plus y squared equals

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the sum of the right-hand
side, 1 plus t.

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So if we try and
substitute in y_1 plus y_2

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into this equation, we're going
to come to a contradiction.

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So let's do that now.

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So we take d by dt
of y_1 plus y_2,

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and I'm going to add it to
y_1 plus y_2 quantity squared.

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And now what I can do is
I can just simplify this.

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So I get y_1 dot plus y_2 dot.

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And I can expand
out this square.

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So I get y_1 squared plus
y_2 squared plus 2y_1*y_2.

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And now what I'm
going to do is I'm

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going to combine the term
y_1 with y_1 squared.

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And by construction, y_1 dot
plus y_1 squared is equal to 1,

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because I assumed that it
satisfied this differential

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equation.

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And then in addition by
construction, y_2 dot plus y_2

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squared satisfies
the right-hand side

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of the second equation with t.

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So I get 1 plus t, but I'm
left over with this other piece

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2y_1*y_2.

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So in general, if I
take a function which

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solves this differential
equation and another function

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which solves this
differential equation,

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and I add them
together, and plug it

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into the left-hand side of
this differential equation,

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I get something which is 1
plus t plus some other stuff.

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And if it did satisfy the
principle of superposition,

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it must equal 1 plus t.

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So we arrive at a
contradiction because it

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doesn't equal 1 plus t.

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In fact, it fails to equal 1
plus t by this term 2y_1*y_2.

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And this term comes directly
from the fact that we had

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a nonlinearity in the equation.

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So this is just one
illustration of the fact

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that a nonlinear equation
doesn't necessarily satisfy

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the superposition principle.

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Whereas, every linear equation
satisfies the superposition

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principle.

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This is one reason we
love linear equations

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and we study them extensively.

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OK, so I'd just like
to conclude here.

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And I'll see you next time.