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PROFESSOR: Welcome back.

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So in this session,
we're going to look

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at sinusoidal inputs for
ordinary differential

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equations of order one.

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So here in question one, you're
asked to use complex techniques

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to solve x dot plus k*x
equals cosine of omega*t.

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Here k and omega are constants.

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This is of ODE of first order.

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And the sinusoidal
input is referring

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to the right-hand
side of the ODE

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where basically
the we are forcing

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a system with a
function, and it's

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sinusoidal of angular frequency
omega, which means basically

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a period of 2pi over omega.

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So you're asked to
use complex techniques

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to solve this in question a.

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In question b, we're asked to
use what we had in question a

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to solve this modified
function, where again, we

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have a sinusoidal input
on the right-hand side

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with an additional amplitude
F, which is now constant.

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In the third part, we're asked
to use superposition principle

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to solve this combined
equation, where now we're

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introducing a value for
example for F, which is 3.

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And you can see also that
the right-hand side is also

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a sinusoidal input, because
it's a linear combination

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of sinusoidal functions.

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All these equations are linear
with constant coefficients,

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and hence superposition
principle would hold.

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So why don't you stop the
video, take a few minutes,

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and work through the problems.

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And we'll be back.

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Welcome back.

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So what is it that
we're asked to do here?

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We're asked to solve this
using a trick that you learned

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in class to basically
convert a real-valued ODE

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into a complex form.

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So the first thing to do is
to realize that the cosine

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of omega*t is simply the real
part of the imaginary complex

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exponential i*omega*t.

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So when we realize
that, and we see

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that the ODE is
real-valued linear,

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we can convert
this real-value ODE

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into this complex-valued ODE.

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And we're going to label
this equation star.

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From this point, we can
go back to the techniques

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that we learned in class,
namely the integrating factor,

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to realize that we can
rewrite the right-hand side

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in this form, introducing
a new function u,

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so that we need
to seek a function

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u that will then recover
this function, this equation.

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And from previous
recitations and problem,

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we saw that clearly the u that
we will need to pick is just

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exponential of k*t.

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So now from this, basically
we're back to i*omega plus k,

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the whole thing t, on
the right-hand side,

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and here z*u prime that we
just need to integrate on both

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sides.

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So this is simply k*t*z
on the left-hand side.

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And on the right-hand side,
we're integrating-- again,

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it's still an exponential,
even though it's complex.

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And we need to
introduce, of course,

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constant of integration.

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So the solution for the
equation star is then,

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I'm going to write it up
here, so that we keep that

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for the rest of the problem,
minus t minus k*t when I divide

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by this.

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And here, the minus k*t of
this exponential is going to be

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canceled out by the
integrating factor.

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So we need to keep that here.

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So here I gave the
general solution

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with the solution
that would come

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from the homogeneous
equation, which we could refer

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to also as a transient
here, because basically

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after a long time t, this
exponential, if k is positive,

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would decay.

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And the part of
the solution that

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comes from the sinusoidal
input, or the forcing, which

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would be particular solution.

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So to go back to the
original question,

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we were asked to solve
this real-valued equation.

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So from what we
noticed above, we

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saw that the
right-hand side is just

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a real part of the right-hand
side of the complex value

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equation star.

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And similarly then,
we can do the same

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for the actual
solutions themselves.

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x is also just the real
value of this complex number.

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So here, for this
general solution,

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I'm going to just use c bar for
general complex constant here.

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And in this case, we
will take the real value.

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So at this point, it doesn't
really matter what t is.

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It's just we're going to
keep it as a constant.

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And now it's just real valued.

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That's for the one part.

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And then we need to take the
real part of this expression.

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Come back to the line here.

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So to take the real
part of this expression,

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you learned that
basically you just need

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to multiply the denominator
or the numerator

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by the complex conjugate.

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And that will only give us--
so we have k minus i*omega over

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the squares, and then again,
Euler formula that we saw

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in a previous recitation.

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So x, from this
expression, is just

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the real part of all of this.

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So it's going to be this term.

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Let me factorize
for a moment here,

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and then the result
of multiplying the two

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complex parts as
well, i to i, minus 1.

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And so that answers
part a of the problem.

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So if we look at
part b-- I'm just

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going to do it
here-- part b we're

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asked to do to solve a
very, very similar equation.

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And I'm just going to
leave the F out for now.

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It's basically
the same equation,

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except that the input now is
a sine instead of a cosine.

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So we can use the same trick
as we used for question a

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by realizing that
now the sine is just

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an imaginary part
of the exponential,

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the complex exponential.

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So we don't need to
redo all the work.

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We only need to, if we were
considering this equation,

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to just take the imaginary
part of the solution

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that we just found here.

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And we can just read off
the solution from this line.

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Note that in this
expression, I left out

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the homogeneous part, which
I should add here from here.

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And this was just
a complex part.

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So to come back to
what I was saying,

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the sine is just
the imaginary part

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of the complex exponential.

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So we're just going to
write down the solution

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by reading off here.

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So again, we have
the homogeneous part,

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which would be another
real-valued constant,

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and then the real part of
this expression, 1 over k

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squared plus omega squared.

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And its imaginary part then
would give us a k sine omega*t

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and a minus omega cos omega*t.

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So this would be the
solution for part

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b, if we had this equation.

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But we actually have an equation
with an additional amplitude F.

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And what this
leads to do is just

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an additional constant
that would appear then

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for this solution.

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And we could see that
if we rewound the way

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we solved in part a,
and saw that introducing

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constant F to the equation is
equivalent to just multiplying

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the full equation
with F. And then

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in the integrating
factor part, we

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would end up with an F
in front of our solution

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when we are doing
the integration.

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And so basically,
that amplitude could

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be absorbed in the
constant of integration

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for the homogeneous part.

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And the F just remains in
the particular solution here.

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So for the last
part of the problem,

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we were now asked to use
the superposition principle

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to combine solutions of two
previous ODEs that we could

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split from this equation, cosine
omega*t plus 3 sine omega*t.

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So you saw from
previous recitations

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that superposition principle
applies for linear equations.

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This is clearly a
linear equation.

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And here on the
right-hand side, we

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have two sinusoidal functions.

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Because it's linear
we can look at this

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by splitting it
into two equations.

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And I'm just introducing
x_1, x_2 as notation

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to distinguish between the two.

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And here we can
recognize that we already

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did this work that
was in part a,

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and we already did
this work in part

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b, where F now has a value of 3.

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And so what this is telling us,
because it's a linear equation,

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we can use
superposition principle,

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the solution, if you were to
do the addition of these two

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equations, would just be the
sum of the previous solutions

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that we found.

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So this would be our x_3, if
I had labeled this with a 3.

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And so basically, the
solution is simply the sum

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of the two previous forms.

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And so we would end up
still with our c minus k*t,

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which is this basically the
homogeneous part that we

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obtained in the two previous
parts, which is common.

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And then we just need to
add each particular solution

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introduced in part a and part b.

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So if I have room I
will write it all out.

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So we would have k cos omega*t
plus omega sine omega*t from

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the first part of the cosine.

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And then we're introducing a
3k sine omega*t minus 3 omega

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cosine omega*t.

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And that would be then the
solution of the combined two

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equations.

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And so you don't need to
redo the full work, by using

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the superposition principle.

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So here the key was just to
recognize that the cosines

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and the sines are basically
real parts and imaginary parts

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of the complex number, of
the exponential of i*omega*t.

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Again, it's just Euler formula.

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And using that as
a shortcut to be

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able to kill two
birds with one stone

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and solve the two equations by
using only one approach, which

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is just the integrating factor.

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And this ends the
session for today.