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PROFESSOR: Welcome
to this recitation.
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In this recitation, we're
going to look at linear systems
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with complex roots.
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So the system we're
examining is the one given
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as x dot equals minus 3x minus
2, and y dot equals 5x minus y.
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And you're asked
to use the matrix
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methods to solve this system.
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So why don't you
take a pause here
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and try to solve this problem?
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And I'll be right back.
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Welcome back.
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So the first step is to write
this system in matrix form.
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So we introduced
a vector [x, y],
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the matrix multiplying
column vector [x, y] again.
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The coefficients are going to
be minus 3, minus 2, 5, minus 1.
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So the first step in
solving this system
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is to find the eigenvalues
of the matrix A.
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So the eigenvalues of
matrix A are basically
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the solutions to this following
determinant equal to 0, minus 3
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minus lambda, minus 2,
5, minus 1 minus lambda,
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determinant equals to 0.
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Here, the lambda are
the unknown eigenvalues.
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And to get this
determinant, we're
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basically multiplying these
two terms, minus minus 2 dot 5,
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which gives us a plus sign.
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So here, we're going
to have lambda squared.
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3*lambda plus 1*lambda
gives us 4*lambda.
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And 3 plus 10 gives us 13.
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So this second-order
polynomial in lambda
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will give us the two
eigenvalues for this matrix.
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So let's examine
the discriminant.
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So we have b squared minus 4a*b.
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And this gives us minus 36.
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So the discriminant is negative.
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And that tells us
that we are going
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to have two complex roots, which
is the title of the recitation.
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And these two complex
roots are going
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to be complex conjugate
of each other.
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So the formula gives
us plus or minus i of 6
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for the root of
the discriminant.
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Here, we have minus 4 over 2.
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So the two roots are basically
minus 2 plus or minus i*3
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or 3i.
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So these are our two roots.
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So now, let's focus
on one of the roots
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to get the eigenvector
associated with the eigenvalue.
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So let's focus on the
positive one, for example.
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And we could do all
the following again--
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AUDIENCE: Minus 2.
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PROFESSOR: Minus 2.
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Thank you.
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We can do all the
following calculation
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that I'm going to go do now
for the complex conjugate,
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and I will explain at the
end how that basically
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not change the result.
So for this eigenvalue,
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we need to compute
now the eigenvector.
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So to do that, we basically have
to use minus 3 minus-- I'm just
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going to write the system here,
let you see what I'm doing.
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And we are solving this system.
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So where does this
system come from?
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It comes from the fact
that we're looking
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for an eigenvector, v_plus, that
is defined as a*v_plus equals
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lambda_plus v_plus.
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And you can then
bring everything
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on the left-hand
side, a minus lambda_i
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applied to v_plus gives
us the zero vector.
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So that's what we have here.
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The unknowns are
a_1, a_2, and we're
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going to try to solve for this.
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So if we plug in now for
the value of lambda_plus
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that we have, we have minus 3
plus 2, which gives us minus 1.
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And then, we have a
minus 3i and minus 2.
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And for the second line,
second entry of this matrix
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you have 5, minus
1 minus minus 2.
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So we have 2 minus
1, which is 1.
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And then, we have minus 3i.
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[a 1, a 2] equals to [0, 0].
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So here, you can
check for yourself
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that these two equations
given by the first line
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and the second line
are actually the same.
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And so basically,
to get a_1 and a_2,
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it is sufficient to just solve,
for example, the first one,
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where here, I just wrote
minus 1 minus 3i multiplied
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by a_1 minus 2a_2 equals to 0.
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And I just brought the
minus 2 on this side.
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So here, you can see that
if we pick a_1 equals to 0--
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equals to plus 2, which
would be our first entry,
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we can then cancel out
these two and just have
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a_2 equals to minus 1 minus 3i.
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So this would be one
eigenvector associated
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with this eigenvalue.
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We could have picked other ones.
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They're basically
parallel to this one.
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So now what?
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So what we need to remember
is the meaning of all of this.
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Seeking the eigenvalues and
the eigenvectors is basically
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equivalent to seeking a
solution in the form exponential
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lambda*t with the direction of
the eigenvector associated with
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this eigenvalue.
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So now that we actually
have this eigenvector
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and this eigenvalue, we can
write down the solution.
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And I'm just going to write
the solution in x, which
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has entries basically x and y.
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And one way of writing
it would be just
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to basically first start by
writing what we have there.
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I'm just going to spell it out.
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So we have this
multiplied by 2 minus 3i.
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So what do we do with this?
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Well, we remember
our earlier formula.
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So this is exponential minus
2t plus exponential 3i*t.
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So we can split the exponential
3i*t into a cosine and a sine.
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And this, we're going to
also be able to split it
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into the complex part
and the real part.
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And then, we're going
to combine the real part
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and the complex part.
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So let's do that.
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Exponential minus
2t multiplying,
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basically, cosine 3t plus i
sine 3t for the entry 2 minus 1
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minus 3i.
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So we have an i
here and an i here.
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So things can be combined
into a real part.
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So in the first entry here,
what are we going to have?
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We're going to have
cos 3t multiplying 2.
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That's going to be
in the real part.
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And I'm going to
keep some space.
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And another entry here at the
second entry of this vector
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is going to give us cosine
3t multiplied by minus 1.
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Oops, here, it should be a 3t.
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Sorry.
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So minus cosine 3t.
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Now, where are we going to
have another real part here?
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It's going to come from a
multiplication of i sine 3t
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by 3i.
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So the two i's together
gives a minus 1.
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And we end up with
a plus 3 sine 3t.
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So we're done for the real part.
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Now let's focus on
the imaginary part.
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What do we have?
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We have an i sine
3t multiplying a 2.
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And we have a minus 3i
here multiplying cosine 3t.
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So we want to have
a minus 3 cosine 3t,
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and finally, this minus 1
multiplying this sine 3t.
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So now, we did--
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AUDIENCE: [INAUDIBLE].
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PROFESSOR: Oh yeah.
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Thank you.
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2, from this operation.
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So now, we did
split our solution
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into a real part and
an imaginary part.
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So how can we write the
general solution of the system?
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Well, we knew that for this
linear system of equations,
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if we have a complex
number that is
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a solution to the
linear equation,
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then its real part
and its imaginary part
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are also two
independent solutions.
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So we can write the general
solution of the system
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as a linear combination
of the real part
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and the imaginary part.
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And I can just label this u_1 of
t and u_2 of t here and vector.
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And we can then write
the general solution
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in terms of any constant--
that would be determined
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by the initial condition if
we had one-- exponential minus
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2t along u_1 plus c_2
exponential minus 2t
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along vector u_2, which
are also functions of t,
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just the difference
from what we had before.
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So here, basically we seek
for the eigenvalues values
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of the matrix.
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We looked for the
eigenvector associated
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with the complex eigenvalue.
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We were able to write
the full solution.
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And then, because of
the linearity property,
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we were able to
just then extract
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two linearly independent
solutions, the real part
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of the solution we had and the
imaginary part of the solution
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we had.
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So what I mentioned
earlier was that we
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could do this whole calculation
for the other eigenvalue
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with a minus.
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If you try to do it and
trickle down your minus,
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you would see that
basically you would just
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end up with minus
signs here basically
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in front of the sines.
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And what you could do
then is just simply
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absorb that minus sign
for the general solution
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in c_1 and c_2.
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And basically, it gives
you exactly the same form
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for the general solution.
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So you don't need to redo
it for the second one.
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You would still end up with
only two linearly independent
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solutions, not four.
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OK, so that ends
this recitation.