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PROFESSOR: Hi everyone.

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Welcome back.

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So today, we're
going to take a look

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at first-order linear
differential equations

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with constant coefficients.

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And specifically, we're going
to use integrating factors

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to solve them.

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So the equation that we're going
to solve today is x dot plus

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k*x equals 1.

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And then in part B we're going
to change the right-hand side

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to e to the minus 5t.

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And then in part C, we're
asked to use the superposition

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principle to solve x dot
plus k*x equals 4 plus 7 e

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to the minus 5t.

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So I'll let you think
about this for a moment,

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and I'll come back in a second.

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Hi, everyone.

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Welcome back.

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So I should mention that every
first-order linear differential

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equation, whether it has
constant coefficients or not,

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can always be solved using
an integrating factor.

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However, in this case, we
have a constant coefficient,

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which is particularly nice.

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And later on in
the course, we're

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going to learn some even
better ways, or quicker ways,

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to solve linear
differential equations

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with constant coefficients.

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But for today, we're asked
to use an integrating factor.

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So for part A, we have the
equation x dot plus k*x equals

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1.

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And the first step is to
compute the integrating factor.

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So the integrating factor,
which I'll call g of t,

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it's always going
to be an exponential

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of the integral of the function
that appears in front of x.

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So in this case, the
function is just a constant.

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It's k.

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So we have k*dt, which
gives us e to the k*t.

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So once we have the
integrating factor,

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we just multiply our
equation through by g of t.

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And by construction, what
the integrating factor does

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is it lets us combine these
two terms on the left-hand side

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into an exact derivative.

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So these two terms are
actually the time derivative

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of the product x times
the integrating factor e

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to the k*t.

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And then on the right-hand
side we just have e to the k*t.

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So we can just go
ahead and integrate

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both sides of the equation.

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And when we do that, the
right-hand side becomes

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the integral of k*t, which
is 1 over k e to the k*t plus

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a constant of integration.

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And now, just to isolate x,
I could divide through by e

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to the k*t.

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And I obtain 1 over k plus a
constant e to the minus k*t.

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So here's the
solution to the ODE.

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OK, so this concludes part A.

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For part B, we have the
equation x dot plus k*x equals e

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to the negative 5t.

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So if we take a look
at this equation,

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the only thing that we've
changed is the right-hand side.

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We haven't changed
the left-hand side.

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And again, if we compute
the integrating factor,

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well, we know that it's the same
integrating factor as in part

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A. And the reason is that
the integrating factor only

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depends on the left-hand side.

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It only depends on
the linear terms.

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So I can multiply the
equation through by

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the integrating factor again.

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And when I do this, I'll
just combine the terms

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on the right-hand side.

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So this is e to the k*t
times e to the minus 5t.

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And again, by construction, the
left-hand side is going to be

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the same as in part A, the time
derivative of x times e to k*t.

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And now we can go ahead
and integrate both sides.

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OK, so if we integrate
this, we end up

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getting 1 over k
minus 5, e to the k

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minus 5t, plus a constant c.

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And if we step back and take
a look at this for a second,

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we see that when k equals
5, we have a problem.

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Particularly, the
denominator vanishes,

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which would give us 1 over 0.

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So this equation,
this right-hand side,

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actually only holds when
k is not equal to 5.

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So this is only valid
for k not equal to five.

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So the question is, what
happens when k equals to five?

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And in this case, we
would have x e to the k*t,

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times the integral of 1 dt,
which would just give us t plus

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a constant c.

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So in this case, we would
have t e to the minus k*t,

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plus c e to the minus k*t.

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And this is when
k is equal to 5.

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Meanwhile, for k not
equal to 5, well, we

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have the solution
worked out already.

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So we can just isolate x, and
divide through by e to the k*t.

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And we have 1 over k minus
5, e to the minus 5t, plus c,

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e to the minus k*t.

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And this concludes part B.

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So the solution for k equal
to 5 is t, e to the minus k*t,

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which would be e
to the minus 5t,

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plus a constant c times
e to the minus 5t.

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And when k is not equal to
5, we have 1 over k minus 5,

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e to the minus 5t, plus
c e to the minus k*t.

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So I'd just like to point
out a few things between

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the solutions for part A
and for part B. First off,

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we note that both part A and
part B share a common solution

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of the form constant c
times e to the minus k*t.

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So this is a term that appears
in the solution for both part A

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and for part B.
The reason is this

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can be thought of as
the homogeneous solution

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to the differential equation.

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This is the term that solves
the differential equation when

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the right-hand side is set to 0.

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Secondly, in part B, if we
take a look, when k is not

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equal to 5, we have the term
which is a constant times

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e to the minus 5t.

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However, when we have k
equal to 5, what happens

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is we have a term which
essentially occurs

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from forcing the differential
equation on resonance,

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which gives us an extra factor
of t times e to the minus 5t.

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And we'll see more about
resonance in the future.

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OK, so for part C, we're
asked to use superposition.

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To solve the differential
equation x dot plus k*x equals

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4 plus 7 e to the minus 5t.

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Now if we take a look at
this differential equation,

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we already know the solution
when the right-hand side is 1

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and when the right-hand
side is e to the minus 5t.

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So we've changed the
right-hand side now

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so it's 4 times 1 plus 7
times e to the minus 5t.

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So what's the total
solution going to be?

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Well, it's going to be four
times our solution when

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the right-hand side was 1,
plus seven times the solution

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when the right-hand side
was e to the minus 5t.

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This is one of the beautiful
things about linear equations.

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When we add two forcings
to the right-hand side,

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our total solution
is just going to be

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the sum of the solutions
to the individual terms.

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OK.

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So what this means
is our solution, x,

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is going to be 4
times the solution

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when the right-hand side was 1.

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And in that case, it was 1 over
k plus c e to the minus k*t,

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plus 7 times the solution
when the right-hand side was e

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to the minus 5t.

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And when k was not equal to 5,
this becomes 1 over k minus 5,

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e to the minus 5t, plus
c e to the minus kt.

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So if we take a look at
the sum of these two terms,

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I'll denote the two
constants as c_1 and c_2.

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We actually have 4 times
c_1 plus 7 times c_2.

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That's just going to give
us a new constant, c_3.

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So in general, this becomes 4
over k plus 1 over k minus 5,

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e to the minus 5t, plus a
constant c e to the minus k*t.

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So I can just recombine
the 4c e to the minus k*t,

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and the 7c e to the minus k*t.

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That together just gives
me a new constant, c_3,

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times e to the minus k*t.

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And this is the solution
when k is not equal to 5.

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When k equals to 5, what we do
is we just replace this term

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with t e to the minus k*t.

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So this is when k
is not equal to 5.

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It should be a 7 here.

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So I'll just conclude there.

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And for summary,
we've taken a look

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at a first-order
linear differential

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equation with a couple
different right-hand sides.

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We've solved them using
an integrating factor.

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And then what we've done is
we've used the superposition

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principle to solve the same
ODE for a right-hand side which

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is the superposition of
multiples of the functions

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that we've had in
part A and for part B.

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So I'd like to conclude here,
and I'll see you next time.