- Welcome to a lesson on solving a first-order homogeneous
differential equation written in differential form.
A first-order differential equation
written in differential form as we see here
is homogeneous if function M and N
are homogeneous functions of the same degree,
which means we're going to find M of TX, TY
and N of TX, TY and make sure we can write them both
as T raised to some power times the original function.
And in both cases, the exponent on T must be the same
in order for the functions to be homogeneous
and therefore the differential equation to be homogeneous.
So if we determine if the first-order DE is homogeneous,
we can solve it by performing a substitution.
We have two options for the substitution.
We can let Y equal X times V, which means differential
Y would be equal to this using implicit differentiation
and the product rule, or we can let X equal Y times V
and therefore DX would be equal to this,
again, using implicit differentiation and the product rule.
To decide which substitution to do,
we're going to look at the original differential equation
and determine whether M or N is simpler.
If M is simpler, that means we're going to perform
a substitution for DX, and therefore,
we'll let X equal Y times V.
If N is simpler, then we'll perform a substitution
for DY, and therefore, will let Y equal X times V.
Once we perform the substitution,
we can solve the resulting DE using separation of variables,
and then we'll solve the original DE in terms of X and Y.
So let's take a look at our first example.
The first step is a test to see if this is homogeneous,
so we need to recognize that function M of X, Y
equals X minus Y,
and N of X, Y would be equal to just X.
So now we'll find M of TX, TY,
which would be TX minus TY.
Notice if we factor out the T, we have T times X minus Y,
which means this is equal to T times
the original function M of X, Y.
So because this is T to the first,
M is a first degree homogeneous function,
and that will determine N of TX, TY,
which would just be T times X,
which we can write as T times
the original function N is equal to X.
So this is T times N of X, Y.
Notice how this is also T to the first times N.
Notice both M and N are degree one homogeneous functions.
Therefore, the original differential equation
is also homogeneous, which means we can solve
it by performing a substitution.
Let's perform that substitution on the next slide.
Notice that here, function N is simpler than function M.
So we want to perform a substitution for DY,
which means we're going to let Y equal X times V.
Therefore, differential Y is going to be equal
to X times DV plus V times DX.
Here we applied implicit differentiation
and the product rule.
Now we'll perform the substitution.
So we'll have X minus Y,
which is equal to XV
times DX
plus X times DY,
which is XDV plus VDX.
And this is all going to equal zero.
Notice now, our differential equation
is in terms of X's and V's.
Let's go ahead and clear the parentheses and simplify.
So we'll distribute here.
We're going to have XDX
minus XVDX
plus X squared DV
plus XVDX.
Notice how here we have minus XVDX
and here we have plus XVDX,
so this simplifies to zero.
Now we want to solve this issuing separation of variables.
Since this term has DV, we'll subtract it to the other side.
So we'll have XDX equals negative X squared DV.
We want all the X's on the left side,
so let's go ahead and divide both sides by X squared.
So now we're going to have one over XDX
equals, this would be negative DV.
Now we're going to integrate both sides of the equation.
We'll go ahead and factor out the negative
so the integral symbol would be here.
So we'll have the natural log (mumbles) of X
plus our constant of integration is going to be equal to,
this would just be negative V.
We would have another constant of integration here,
but we already put a constant
on the left side of the equation.
Let's go ahead and finish this on the next slide.
Remember, our goal is to solve this in terms of X and Y,
so we need to perform a substitution for V.
Remember, we started with the substitution Y equals
X times V,
which means V is equal to Y divided by X.
So we have natural log (mumbles) of X plus C
is equal to negative Y divided by X.
Now we'll go ahead and multiply both sides
of the equation by negative X.
So we'll have negative X natural log (mumbles)
of X minus C times X equals Y.
Or if we want, we can switch this around.
Y equals negative X natural log (mumbles)
of X minus C times X.
This would be the general solution to the original
first-order homogeneous differential equation.
Let's finish by taking a look at this graphically.
Here's the original differential equation.
We can use this to create the given red slope field below.
And then given our general solution,
if we select values of C,
for example C equals negative two,
negative one, zero, one, and two,
we can graph some of the solutions
from the family of solutions.
So every graph below represents a possible solution
to the original differential equation.
OK, I hope you found this helpful.
We'll take a look at another example in the next video.