- [Voiceover] Welcome to a lesson on solving a
First-Order Homogeneous Differential Equation
written in Standard Form.
A first order differential equation in the form
dy dx equals f of x comma y is homogeneous if it
does not depend on x and y separately but on the
ratio of x to y or y to x.
So I can say a differential equation is homogeneous
if it can be written in the form of dy dx equals
f of y divided by x or x divided by y.
So we'll first check to see if the given
differential equation is homogeneous.
And then if it is,
we can solve it by performing substitution.
We'll perform the substitution
v equals y divided by x but since we'll also
have to perform a substitution for dy dx,
if we solve this equation for y by multiplying
both sides by x,
we can say that y is equal to x times v.
And then define dy dx, we'll have to perform
implicit differentiation and the product rule.
So dy dx is going to be equal to x
times dv dx
plus v times the derivative of x,
which would just be one.
So for the substitution, we'll use
v equals y divided by x
and dy dx equals x times dv dx plus v.
Once we perform this substitution, we can solve
the resulting differential equation using
separation of variables.
And then we'll solve the original
differential equation in terms of x and y.
So let's take a look at our first example.
So our step is to see if this is homogeneous by
seeing if we can write the right side of this
differential equation as a function
of y divided by x, shown by our notes here below.
Now here, because we're dividing by a monomial
or one term, we could break this up into
a two individual fractions.
But another way to approach this would be to
recognize that if we have a y squared here,
we want this to be a ratio of y to x.
Our denominator of this term would have to be
x squared.
So we can multiply the numerator
by one over x squared as long as you do the same
to the denominator.
Now if you look at the denominator, notice how
we'd x squared over x squared.
So this product simplifies the one.
And now looking at the numerator,
if we distribute here,
we'd have three times y squared over x squared,
which we can write as three times y divided
by x squared.
Plus here we'd have xy over x squared.
Notice that one factor of x would simplify out,
leaving us with just y divided by x.
Notice how the right side of this
differential equation is a function
of y divided by x.
Therefore we know this is a homogeneous
differential equation so now we'll perform
a substitution and solve this using separation
of variables.
And again the two substitutions we are going to use
are v equals y divided by x.
And if we solve this for y, we'd have x times v.
And then applying implicit differentiation,
we would have dy dx equals x times dv dx plus v.
So we'll use these two equation to perform a
substitution into our original
differential equation.
So instead of dy dx, we're going to have
x times dv dx
plus v.
So here's dy dx equals.
Now three times y divided by x squared is now
just going to be three v squared.
And then y divided by x is just going to be v.
And now we can sole this using separation
of variables.
Notice that we have a plus v on both sides
of the equation.
So if we subtract v on both sides,
these two terms would simplify out.
So we'd have x times dv dx
equals 3 v squared.
Now we want the x's and dx on one side
and the v's and dv on the other.
Now because the dx is the denominator,
we want the x terms on the right side.
So we're going to go ahead and multiply both sides
by one over x.
We want the v terms on the left side so let's
also multiply by one over 3 v squared.
When we do this, notice how these x's will
simplify out and on the right side,
the 3 v square's will simplify out.
So let's go ahead and write the left side
as 1/3 v to the negative two
dv dx equals.
On the right side, we just have one over x.
And now we'll go ahead and write this in
differential form which we can think of as
multiplying both sides by dx.
So we'd have 1/3
v to the negative two dv
equals one over x dx.
And now we'll integrate both sides of the
equation.
So playing the power role here, we'd have
1/3 times v to the negative one divided
by negative one
or negative 1/3 v to the negative one.
Now there would be a
constant of integration here,
but we'll go ahead and put it on the right side.
So this is going to be equal to.
The integral of one over x would be natural log
absolute value of x plus c.
Let's go ahead and continue this
on the next slide.
Let's go ahead and write this back
in terms of x and y.
So v equals y divided by x.
V to the negative one power
would be x divided by y.
So we'll have negative 1/3 times x divided
by y equals natural log
absolute value of x plus c.
Let's go ahead and eliminate this fraction
by multiplying both sides by negative three.
So we'd have x divided by y equals negative three
times natural log absolute value of x.
And we'd have minus three 3.
Let's let c sub one equal negative three times c.
That way we can just write plus c sub one.
This would be the solution to the homogeneous
differential equation.
But let's go ahead and solve this for y.
We could think of this as all over one then take
the reciprocal of both sides so we'd have
y over x equals one over,
let's write, c sub one
minus 3 natural log absolute value x.
Now the last step here we can multiply
both sides by x.
So we'd have y equals x divided by the quantity
c sub one minus three natural log
absolute value x.
And this would be our general solution
to the original homogeneous differential equation.
Let's finish by taking a look
at this graphically.
The red slope field that we see here
can be generated by using the original
differential equation.
And then here's our general solution.
And what I did here was graph several
of the possible solutions with c equals
negative two negative one, zero, one, and two.
So each graph below, there's a different color
represents one of the solutions from the family
of solutions.
In part two, we'll take a look at
a second example of solving a first-order homogeneous
differential equation.
I hope you found this helpful.