WEBVTT

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>> Welcome to the Cypress
College Math Review

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on Cauchy-Euler homogenous
and nonhomogeneous equations.

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Equations of the type x cubed y
triple prime minus x squared y

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double prime minus 2xy prime
minus 4y is equal 0 are called

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Cauchy-Euler or equidimensional
equations because the power of x

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in each term is the same as
the order of the derivative.

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To solve these differential
equations,

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you make the substitution
x equals e to the t

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and your new function cap Y of t
is equal to little y of e of t.

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With this substitution, t is
equal to natural log of x.

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So the derivative
of t with respect

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to x is equal to 1 over x.

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Little y of x is equal to cap Y

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of t. Let's take the
derivative of both sides.

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So the derivative of little y
of x is equal to the derivative

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of cap Y-- of t times the
derivative of t with respect

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to x by the [inaudible]
we're taking the derivative

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with respect to x. Since the
derivative of t with respect

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to x is 1 over x, we get this.

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Solving for the derivative
of y with respect to t,

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we get x times the derivative
of y with respect to x,

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therefore x times our original y
prime is equal to the derivative

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of our new function Y.

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Let's take a look at what
the second derivative will

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look like.

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So the first derivative was
1 over x times the derivative

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of cap y. Let's take
the second derivative.

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So the second derivative
of little y

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by the product rule
would be negative x

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to the negative 2 times dy/dt
plus the second derivative

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of cap y with respect
to t times dt/dx,

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since we're taking the
derivative with respect to x,

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times 1 over x with
our product rule.

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Now the derivative of t with
respect to x is 1 over x. Making

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that substitution and reversing
the order of our terms,

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we have 1 over x squared
times the quantity,

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the second derivative
minus the first derivative.

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Multiplying by x squared.

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Therefore x squared times
little y double prime is equal

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to cap D, D minus
1 operated on Y.

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The pattern continues.

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So if you have Cauchy-Euler
equation,

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each term in your differential
equation matches one of these.

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You replace it with the
appropriate substitution

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from this chart.

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You'll be able to transform
your equation into an equation

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with constant coefficients and
then you go through the steps

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that you've learned to solve
the differential equation.

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Let's solve this
Cauchy-Euler equation.

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It's Cauchy-Euler because the
exponent on x matches the order

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of the derivative in each term.

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It's homogenous differential
equation since it's equal to 0.

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Make the substitution.

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Let x equal e to the t. So our
new function cap Y of t is equal

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to our original function,
y of x which is y of e

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to the t. We make our
substitutions based

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on the chart.

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So we have, for this term,

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we have the differential
operator times D minus 1 times D

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minus 2.

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For this term, we have the
differential operator times D

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minus one.

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For this term, we have minus 2
times the differential operator.

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And for this term, we
simply just have minus 4.

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That expression is operated on
the function cap Y equals 0.

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So now we do some algebra.

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This expression becomes D
squared minus 3D plus 2 minus D

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squared plus D, all
that is operated

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on Y. D cubed minus 3D squared
plus 2D minus D squared minus D

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minus 4 operated on Y. Finally,

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we get D cubed minus 4D
squared plus D minus 4 operated

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on Y is equal to 0.

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From that, we write
our auxiliary equation.

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P of r is equal to r cubed minus
4r squared plus r minus 4 is

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equal to 0.

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Factor out an r squared,
we're left with r minus 4.

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So if we factor that, we have
the first factor is r squared

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plus one, the second
factor is r minus 4.

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And our solutions are
plus or minus i and 4.

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The general form for real
solutions looks like this.

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So in our case, we're going
to have Y of t equals c1 e

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to the 0t cosine of 1t
plus c2 e to the 0t sine

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of 1t plus c3 e to the 4t.

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Simplifying, we have c1
cosine of t plus c2 sine

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of t plus c3 e to the 4t.

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Now we need to substitute back.

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Recall that x is equal to e to
t. If x is equal to e to the t,

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then t is equal to natural
log of x. So little y of x,

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our original function that
we were trying to solve

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for would be c sub 1
times cosine of what t is,

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which is natural log of x,
plus c sub 2 times sine of t,

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which is natural log of x, plus
c sub 3 times e to the power

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of 4t but recall that t was
natural log of x. We need

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to simplify that final term.

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That final term simply
becomes x to the fourth.

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Recall that x was positive.

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And we have our solution to
the differential equation.

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Here's another Cauchy-Euler
differential equation.

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If you look at the right-hand
side, it's nonhomogeneous.

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Let's make our substitution.

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So we let x equal e
to the t. So cap Y

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of t is equal to little y of x.

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Looking at our chart, the first
term becomes cap D times D

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minus 1.

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Second term just becomes 6 but
the whole thing is operated

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on the function cap Y.
Looking at the right-hand side

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that we have to replace
x with e to the t.

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So the differential operator
becomes D squared minus D minus

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6 and that's operated

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on the function Y. That's
equal to 6e to the 4t.

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So the auxiliary equation is P

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of r equals r squared
minus r minus 6 equals 0.

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So our r minus 3 times r plus 2.

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So the solutions are
3 and negative 2.

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It's the complementary function.

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Cap Y sub c of t
then is c sub 1 e

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to the three t plus c2
e to the negative 2t.

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So we got the homogenous solved.

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Now, we have to deal
with the nonhomogeneous.

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Our tentative trial solution
cap Y sub p of t is A times e

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to the 4t because
of this term here.

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There are two derivatives in
our differential equation.

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So we take two derivatives
of this function.

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There's the first
derivative and the second.

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Now, we substitute those

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into our nonhomogeneous
differential equation.

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The second derivative minus the
first derivative minus 6 times

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the function equals, the
right-hand side, 6e to the 4t.

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So the left-hand
side is 6Ae to the 4t

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which gives us, that A is 1.

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So Y sub p of t is
equal to 1e to the 4t.

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So our solution, cap Y of t is
equal to Y sub c plus Y sub p.

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So that's c sub 1 e to
the 3t plus c sub 2 e

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to the negative 2t, so there's
our homogenous solution.

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Plus the particular solution
to the nonhomogeneous,

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so that's plus 1e to the 4t.

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Not done yet, we still
have to substitute back.

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So we substitute back little
y of x is equal to c sub 1 e

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to the 3 times natural
log of x plus c sub 2 e

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to the negative 2
natural log of x plus e

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to the 4 natural log of x.

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Simplifying, we get c sub
1 x cubed plus c sub 2 x

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to the negative 2
plus x to the fourth

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where x was positive
as our answer.