>> Welcome to the Cypress College Math Review
on Cauchy-Euler homogenous and nonhomogeneous equations.
Equations of the type x cubed y triple prime minus x squared y
double prime minus 2xy prime minus 4y is equal 0 are called
Cauchy-Euler or equidimensional equations because the power of x
in each term is the same as the order of the derivative.
To solve these differential equations,
you make the substitution x equals e to the t
and your new function cap Y of t is equal to little y of e of t.
With this substitution, t is equal to natural log of x.
So the derivative of t with respect
to x is equal to 1 over x.
Little y of x is equal to cap Y
of t. Let's take the derivative of both sides.
So the derivative of little y of x is equal to the derivative
of cap Y-- of t times the derivative of t with respect
to x by the [inaudible] we're taking the derivative
with respect to x. Since the derivative of t with respect
to x is 1 over x, we get this.
Solving for the derivative of y with respect to t,
we get x times the derivative of y with respect to x,
therefore x times our original y prime is equal to the derivative
of our new function Y.
Let's take a look at what the second derivative will
look like.
So the first derivative was 1 over x times the derivative
of cap y. Let's take the second derivative.
So the second derivative of little y
by the product rule would be negative x
to the negative 2 times dy/dt plus the second derivative
of cap y with respect to t times dt/dx,
since we're taking the derivative with respect to x,
times 1 over x with our product rule.
Now the derivative of t with respect to x is 1 over x. Making
that substitution and reversing the order of our terms,
we have 1 over x squared times the quantity,
the second derivative minus the first derivative.
Multiplying by x squared.
Therefore x squared times little y double prime is equal
to cap D, D minus 1 operated on Y.
The pattern continues.
So if you have Cauchy-Euler equation,
each term in your differential equation matches one of these.
You replace it with the appropriate substitution
from this chart.
You'll be able to transform your equation into an equation
with constant coefficients and then you go through the steps
that you've learned to solve the differential equation.
Let's solve this Cauchy-Euler equation.
It's Cauchy-Euler because the exponent on x matches the order
of the derivative in each term.
It's homogenous differential equation since it's equal to 0.
Make the substitution.
Let x equal e to the t. So our new function cap Y of t is equal
to our original function, y of x which is y of e
to the t. We make our substitutions based
on the chart.
So we have, for this term,
we have the differential operator times D minus 1 times D
minus 2.
For this term, we have the differential operator times D
minus one.
For this term, we have minus 2 times the differential operator.
And for this term, we simply just have minus 4.
That expression is operated on the function cap Y equals 0.
So now we do some algebra.
This expression becomes D squared minus 3D plus 2 minus D
squared plus D, all that is operated
on Y. D cubed minus 3D squared plus 2D minus D squared minus D
minus 4 operated on Y. Finally,
we get D cubed minus 4D squared plus D minus 4 operated
on Y is equal to 0.
From that, we write our auxiliary equation.
P of r is equal to r cubed minus 4r squared plus r minus 4 is
equal to 0.
Factor out an r squared, we're left with r minus 4.
So if we factor that, we have the first factor is r squared
plus one, the second factor is r minus 4.
And our solutions are plus or minus i and 4.
The general form for real solutions looks like this.
So in our case, we're going to have Y of t equals c1 e
to the 0t cosine of 1t plus c2 e to the 0t sine
of 1t plus c3 e to the 4t.
Simplifying, we have c1 cosine of t plus c2 sine
of t plus c3 e to the 4t.
Now we need to substitute back.
Recall that x is equal to e to t. If x is equal to e to the t,
then t is equal to natural log of x. So little y of x,
our original function that we were trying to solve
for would be c sub 1 times cosine of what t is,
which is natural log of x, plus c sub 2 times sine of t,
which is natural log of x, plus c sub 3 times e to the power
of 4t but recall that t was natural log of x. We need
to simplify that final term.
That final term simply becomes x to the fourth.
Recall that x was positive.
And we have our solution to the differential equation.
Here's another Cauchy-Euler differential equation.
If you look at the right-hand side, it's nonhomogeneous.
Let's make our substitution.
So we let x equal e to the t. So cap Y
of t is equal to little y of x.
Looking at our chart, the first term becomes cap D times D
minus 1.
Second term just becomes 6 but the whole thing is operated
on the function cap Y. Looking at the right-hand side
that we have to replace x with e to the t.
So the differential operator becomes D squared minus D minus
6 and that's operated
on the function Y. That's equal to 6e to the 4t.
So the auxiliary equation is P
of r equals r squared minus r minus 6 equals 0.
So our r minus 3 times r plus 2.
So the solutions are 3 and negative 2.
It's the complementary function.
Cap Y sub c of t then is c sub 1 e
to the three t plus c2 e to the negative 2t.
So we got the homogenous solved.
Now, we have to deal with the nonhomogeneous.
Our tentative trial solution cap Y sub p of t is A times e
to the 4t because of this term here.
There are two derivatives in our differential equation.
So we take two derivatives of this function.
There's the first derivative and the second.
Now, we substitute those
into our nonhomogeneous differential equation.
The second derivative minus the first derivative minus 6 times
the function equals, the right-hand side, 6e to the 4t.
So the left-hand side is 6Ae to the 4t
which gives us, that A is 1.
So Y sub p of t is equal to 1e to the 4t.
So our solution, cap Y of t is equal to Y sub c plus Y sub p.
So that's c sub 1 e to the 3t plus c sub 2 e
to the negative 2t, so there's our homogenous solution.
Plus the particular solution to the nonhomogeneous,
so that's plus 1e to the 4t.
Not done yet, we still have to substitute back.
So we substitute back little y of x is equal to c sub 1 e
to the 3 times natural log of x plus c sub 2 e
to the negative 2 natural log of x plus e
to the 4 natural log of x.
Simplifying, we get c sub 1 x cubed plus c sub 2 x
to the negative 2 plus x to the fourth
where x was positive as our answer.