WEBVTT

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>> Welcome to the Cypress
College Math Review

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on Variation of Parameters.

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We wish to solve 2nd order
nonhomogeneous equations

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[variable] a sub 0, y double
prime plus [variable] a sub 1 y

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prime plus [variable] a sub
2 y equals F of x where F

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of x is not a linear combination
of the type x to the j e

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to the x cosine of
b x or x to the j e

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to the [variable] a x sine
of b x in which the method

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of undetermined coefficients
would have worked well for us.

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Let u sub 1 and u sub 2 be
linearly independent solutions

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to the homogeneous
equation L of y equals 0.

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Then, y sub c equals c sub 1
times u sub 1 plus c sub 2 times

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u sub 2 is the general solution
to the homogeneous equation.

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Let y sub p equal g sub 1 times
u sub 1 plus g sub 2 times u sub

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2 be a particular solution to
the nonhomogeneous equation L

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of y equals cap F of x. To
find g sub 1 and g sub 2,

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you solve the following system
of equations here in this box.

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I chose to use g sub
1 and g sub 2 instead

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of the more standard cap c sub
1 and c sub 2 simply because,

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in my opinion, c sub
1 and c sub 2 look

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like constants from calc 1.

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You should use what your
instructor prefers you to use.

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This is our first example.

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First, we'll solve the
homogeneous equation.

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Here's our auxiliary equation.

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So r is equal to
1 multiplicity 2.

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So y sub c is equal
to c sub 1 e to the x,

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and then since it's
multiplicity 2, of course,

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we'd have c sub 2 x e to the x.

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If it was multiplicity 3 we
would have x squared e to the x.

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So y sub p is going to be
g sub 1 times e to the x

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because our u sub
1 function is e

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to the x. Our u sub
2 function is x e

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to the x. Here's the system
that we have to solve.

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So now we have to take
the first derivative,

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so g sub 1 prime times the
derivative of e to the x, which,

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of course, is just e to
the x plus g sub 2 prime,

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and then here we have to take
the derivative of x e to the x,

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so we have the product rule.

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So 1 times e to the x plus x
e to the x, and that's equal

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to our function cap F of x,
which is e to the x over 2 x

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over [variable] a sub naught.

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And [variable] a sub naught
in this case is just 1.

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And then we have g sub
1 prime times u sub 1,

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and u sub 1 is just
e to the x plus,

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plus g sub 2 prime
times u sub 2,

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which is just x e
to the x equals 0.

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This is the system of
equations that we have to solve.

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If we subtract, these terms
on the left disappear,

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these two terms disappear,

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and we end up with simply
g sub 2 prime times e

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to the x equals e to the
x over 2 x. Dividing,

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we get g sub 2 prime equals
1 over 2 x. All right,

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let's integrate to find g sub 2.

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So g sub 2 is equal to 1/2 times
the antiderivative of 1 over x,

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which is 1/2 natural log
of x plus c, correct?

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But we don't put the plus c on.

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Why? In the homogeneous solution
you have plus c times u sub 1.

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And our solution is going to
have y sub p plus y sub c,

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so we don't need to write
plus c times u sub 1 again.

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So erase that.

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We'll rewrite that
as natural log

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of the square root
of x. All right.

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Now, the bottom line of
our box is g sub 1 prime e

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to the x plus g sub 2 prime
x e to the x equals 0.

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Solving for g sub 1 prime, we
get negative g sub 2 prime x e

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to the x over e to the x, where
those e to the x's cancel.

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We can replace g sub 2
prime with 1 over 2 x.

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So I have 1 over 2 x there.

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And there's an x there.

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The e to the x's cancel, so
we simply get negative 1/2.

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So g sub 1, after I integrate,
is negative 1/2 x. We're ready

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to write our solution.

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So y sub p, our particular
solution to our nonhomogeneous,

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is g sub 1 e to the x
plus g sub 2 x e to x,

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which in this case would
be negative 1/2 x e

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to the x plus natural log
of the square root of x,

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x e to the x. Y, of
course, is always equal

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to y sub c plus y sub p. So we
get y sub c, which is c sub 1 e

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to the x plus c sub 2 x
e to the x plus y sub p,

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which is negative 1/2 x e
to the x plus natural log

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of the square root of x, x e
to the x, and then we notice

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that the middle two terms are
like terms, so we add them up.

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We have a new constant
times x e to the x,

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and this gives us the solution
to our differential equation.

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For our next example,

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the auxiliary equation is
r squared plus 1 equals 0.

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So the solutions are plus
or minus i. So the solution

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to the homogeneous equation
is c sub 1 times cosine

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of x plus c sub 2 times
sine of x. So now we look

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at the nonhomogeneous.

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So y sub p is going to
be g sub 1 times u sub 1.

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And u sub 1 is cosine of x
plus g sub 2 times u sub 2,

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and u and sub 2 is going to be
sine of x. Here's the system

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of equations that
we have to solve

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to find g sub 1 and g sub 2.

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So we have to take
the derivative

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of g 1 times the derivative
of u 1, and u sub 1 is cosine.

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So the derivative of
cosine, of course,

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is negative sine
plus the derivative

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of g sub 2 times the derivative
of sine, which is cosine,

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equals cap F of x,
which is tangent

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over [variable] a sub
naught, which is 1.

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In the next equation we have g
sub 1 prime times u 1 plus g sub

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2 prime times u sub 2 equals 0.

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So this is the system of
equations that we have to solve.

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Looking at the second equation
and solving for g sub 1 prime,

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if we move the second
term to the other side

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and divide both sides
by cosine of x,

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we get negative g sub
2 prime times tangent

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of x. Let's substitute that
value in for g sub 1 prime

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in the first equation.

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So we get g sub 2 prime tangent
of x. That's going in for,

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that's going in for
g sub 1 prime.

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The negatives are canceling.

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I still have that times the sine
of x plus g sub 2 prime cosine

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of x equals tangent of x. So
we factor out a g sub 2 prime,

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and we have tangent of
x sine of x plus cosine

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of x equals tangent of x. So g
sub 2 prime is equal to tangent

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of x divided by tangent of x
times sine of x plus cosine

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of x. Let's multiply
that by cosine of x

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over cosine of x. Why?

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Because I can see that
tangent of x in the denominator

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over here is sine
of x over cosine

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of x. Let's see what we get.

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When I multiply tangent
times cosine, I get sine.

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So in the numerator I get sine.

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In the denominator
tangent becomes sine,

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so I get sine squared.

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So I have sine squared
plus cosine squared,

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which is simply 1, which means

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that whole function
just becomes sine of x

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because my denominator is 1.

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[ Silence ]

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Let's integrate to find
g sub 2, the function.

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So that would be negative
cosine of x. Recall from above

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that g sub 1 prime is
equal to the opposite

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of g sub 2 prime
times tangent of x.

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So that would be negative
sine of x times tangent of x.

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So now g sub 1 is equal to
the antiderivative of that.

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So we're integrating c. We
have sine times tangent,

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so that's what we're
trying to integrate.

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So sine squared in
the numerator.

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So sine squared is the same
thing as 1 minus cosine squared,

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and the opposite of 1 minus
cosine squared is cosine squared

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minus 1.

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[ Silence ]

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Our goal here, of course,
is to get the cosine

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out of the denominator.

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So we're going to divide now.

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So the numerator divided
by the denominator.

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So I have cosine of x minus
1 divided by cosine of x,

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which is, of course, secant.

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And now I can integrate.

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So I have sine of x minus
the antiderivative of secant,

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which is natural log the
absolute value of secant

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of x plus tangent of
x. And we're ready

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to solve this differential
equation.

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So y sub p is equal to g sub 1
times u sub 1 plus g sub 2 times

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u sub 2, which, in this case,
is going to be the function sine

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of x minus natural log of
the absolute value of secant

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of x plus tangent of x,

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that quantity times the function
u sub 1, which is cosine of x,

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plus the function g sub 2,
which is negative cosine,

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times the function u sub 2,
which is the function sine

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of x. Those cancel out, and that
simply gives me negative cosine

00:16:55.786 --> 00:17:00.996 A:middle
of x times natural log of
the absolute value of secant

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of x plus tangent of
x. So the solution

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to our differential equation
is y sub c plus y sub p,

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which is c sub 1 times u
sub 1, which is cosine,

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plus c sub 2 times u sub 2,
which is sine, plus y sub p,

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which is minus cosine
of x times natural log

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of the absolute value of
secant of x plus tangent of x.

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And that's the solution to
our differential equation

00:17:51.246 --> 00:17:51.976 A:middle
by variation of parameters.

00:17:52.516 --> 00:18:03.176 A:middle
[ Silence ]

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Here's our next example.

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Our auxiliary equation is
r squared minus 3 r plus 2

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equals 0.

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So our solutions are 2 and 1.

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So the solution to our
homogeneous is c sub 1 e

00:18:26.476 --> 00:18:33.856 A:middle
to the 2 x plus c sub 2 e to the
x, so u sub 1 is going to be e

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to the 2 x, u sub 2 is going to
be e to the x, y sub p, then,

00:18:42.616 --> 00:18:50.016 A:middle
is going to be g sub 1 times
e to the 2 x plus g sub 2 e

00:18:50.076 --> 00:18:53.416 A:middle
to the x. Our next task is
to find g sub 1 and g sub 2.

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Here's the system of
equations we have to solve

00:18:56.216 --> 00:18:57.826 A:middle
to find g sub 1 and g sub 2.

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So we take the derivative of
g sub 1 times the derivative

00:19:04.386 --> 00:19:08.346 A:middle
of u sub 1, and the derivative
of e to the 2 x is 2 e

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to the 2 x plus the derivative
of g sub 2 times the derivative

00:19:14.436 --> 00:19:18.536 A:middle
of e to the x equals cap F of x

00:19:22.626 --> 00:19:24.316 A:middle
over [variable] a sub
naught, which is just 1.

00:19:25.986 --> 00:19:33.036 A:middle
The next equation has g sub 1
prime times u sub 1 plus g sub 2

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prime times u sub 2 equals 0.

00:19:37.336 --> 00:19:39.136 A:middle
This is the system
we have to solve.

00:19:40.946 --> 00:19:49.076 A:middle
Looking at the second
equation, moving the first term

00:19:49.076 --> 00:20:02.636 A:middle
over to the other side, divide
both sides by e to the x,

00:20:03.136 --> 00:20:05.636 A:middle
I get that g 2 prime is equal

00:20:05.636 --> 00:20:18.786 A:middle
to negative g 1 prime
e to the x. All right.

00:20:18.786 --> 00:20:20.576 A:middle
I've copied down
the second equation

00:20:20.576 --> 00:20:21.586 A:middle
so we can see what we're doing.

00:20:27.816 --> 00:20:29.616 A:middle
So we've solved for
g sub 2 prime.

00:20:29.946 --> 00:20:30.976 A:middle
Let's substitute that in.

00:20:43.816 --> 00:20:47.326 A:middle
So we have 2 times
g sub 1 prime e

00:20:47.326 --> 00:20:53.236 A:middle
to the 2 x minus g sub
1 prime e to the 2 x

00:20:53.236 --> 00:20:54.456 A:middle
on the left-hand side.

00:20:55.086 --> 00:20:58.766 A:middle
So now we have one of
those, g sub 1 prime e

00:20:58.766 --> 00:21:02.796 A:middle
to the 2 x. Divide both
sides by e to the 2 x,

00:21:02.796 --> 00:21:08.606 A:middle
and we have g sub 1 prime
equals 1 over 1 plus e

00:21:08.606 --> 00:21:14.806 A:middle
to the 2 x. G sub
2 prime is equal

00:21:14.806 --> 00:21:21.936 A:middle
to negative g sub 1 prime times
e to the x, which gives us

00:21:22.446 --> 00:21:27.626 A:middle
that g sub 2 prime is equal
to negative e to the x

00:21:27.876 --> 00:21:33.846 A:middle
over 1 plus e to the 2 x. So now
we have the derivative of each,

00:21:33.976 --> 00:21:36.976 A:middle
and we need to integrate to
find g sub 1 and g sub 2.

00:21:37.516 --> 00:21:44.736 A:middle
[ Silence ]

00:21:45.236 --> 00:21:46.546 A:middle
Here's g sub 1 prime.

00:21:46.906 --> 00:21:48.646 A:middle
Let's integrate to find g sub 1.

00:21:56.846 --> 00:22:00.746 A:middle
A common trick to integrate this
function would be to multiply

00:22:00.746 --> 00:22:02.926 A:middle
by e to the negative
2 x over itself.

00:22:09.106 --> 00:22:13.516 A:middle
So we have e to the negative
2 x over the quantity e

00:22:13.516 --> 00:22:15.516 A:middle
to the negative 2 x plus 1.

00:22:16.966 --> 00:22:19.146 A:middle
And now u substitution
will work.

00:22:19.776 --> 00:22:27.786 A:middle
U equals the denominator,
d u equals negative 2 e

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to the negative 2 x d x. So we
get negative 1/2 antiderivative

00:22:39.666 --> 00:22:48.556 A:middle
of 1 over u d u, which gives
us negative 1/2 natural log

00:22:49.566 --> 00:22:55.476 A:middle
of the quantity 1 plus
e to the negative 2 x.

00:22:56.346 --> 00:22:57.986 A:middle
And that's our function g sub 1.

00:23:04.756 --> 00:23:06.396 A:middle
Here's my function
g sub 2 prime.

00:23:06.806 --> 00:23:08.486 A:middle
Let's integrate to find g sub 2.

00:23:16.786 --> 00:23:19.956 A:middle
We're going to do the same
thing we did a minute ago

00:23:19.956 --> 00:23:22.396 A:middle
because a simple u substitution
is not going to work.

00:23:22.396 --> 00:23:26.386 A:middle
We're going to multiply by e to
the negative 2 x over itself.

00:23:26.966 --> 00:23:35.916 A:middle
In the numerator we have the
opposite of e to the negative x.

00:23:37.596 --> 00:23:40.956 A:middle
In the denominator we have e
to the negative 2 x plus 1.

00:23:46.506 --> 00:23:53.886 A:middle
Let u equal e to the negative
x, so d u is equal to negative e

00:23:54.446 --> 00:24:07.286 A:middle
to the negative x d x. So we
get d u over u squared plus 1,

00:24:08.486 --> 00:24:10.826 A:middle
which hopefully you
recognize as arc tangent.

00:24:16.306 --> 00:24:23.026 A:middle
So we have arc tangent
of e to the negative x.

00:24:24.596 --> 00:24:26.406 A:middle
And there's our function
g sub 2.

00:24:31.746 --> 00:24:32.796 A:middle
All right.

00:24:32.796 --> 00:24:33.906 A:middle
Let's put this all together.

00:24:35.546 --> 00:24:41.446 A:middle
We know that y sub c, the
solution to the homogeneous,

00:24:42.076 --> 00:24:47.716 A:middle
is c sub 1 e to the 2 x
plus c sub 2 e to the x.

00:24:47.816 --> 00:24:53.186 A:middle
And the particular solution to
the nonhomogeneous is g sub 1 e

00:24:53.186 --> 00:24:59.826 A:middle
to the 2 x plus g sub 2 e to
the x. We solved and found

00:25:00.336 --> 00:25:02.866 A:middle
that g sub 1 was equal

00:25:02.866 --> 00:25:07.696 A:middle
to negative 1/2 times
the natural log

00:25:07.696 --> 00:25:11.786 A:middle
of the quantity 1 plus
e to the negative 2 x.

00:25:12.626 --> 00:25:20.176 A:middle
And g sub 2 was equal
to arc tangent of e

00:25:20.176 --> 00:25:26.736 A:middle
to the negative x. Y is always
equal to y sub c plus y sub p,

00:25:27.506 --> 00:25:28.266 A:middle
so the solution

00:25:28.266 --> 00:25:33.266 A:middle
to our differential
equation is c sub 1 e

00:25:33.266 --> 00:25:41.776 A:middle
to the 2 x plus c sub 2 e to
the x minus 1/2 natural log

00:25:41.776 --> 00:25:47.656 A:middle
of the quantity 1 plus e
to the negative 2 x times e

00:25:47.656 --> 00:26:02.336 A:middle
to the 2 x plus arc tangent
of e to the negative x times e

00:26:02.336 --> 00:26:07.126 A:middle
to the x. And there's
the solution

00:26:07.126 --> 00:26:07.936 A:middle
to our differential equation.