- [Instructor] Welcome back to part three
of solving a first-order homogeneous differential equation
written in differential form.
Now we'll look at an initial value problem.
In the previous two videos, I went over this slide
on how to determine if the first-order DE is homogeneous
and then also how to solve the
first-order homogeneous differential equation.
For a quick overview, before we do our example,
the first step is to make sure that we do have
a homogeneous differential equation
by making sure function M and N
are both homogeneous functions of the same degree
and then if it is homogeneous,
we can solve it by performing a substitution
and then using separation of variables.
The substitution is going to be either
Y equals X times V,
and therefore DY is equal to XDV plus VDX
or we can use X equals Y times V,
and therefore DX is equal to YDV plus VDY.
Let's take a look at our example.
The first step is to put the given differential equation
in the form given here below,
so let's go ahead and subtract XY squared DY on both sides,
so we can write this as the quantity
Y cubed minus X cubed
DX minus XY squared DY equals zero.
Notice how we do have this initial condition,
and therefore, after we find the general solution,
we can actually find the particular solution
given that Y(1) is equal to two
but first things first, we need to make sure
this is homogeneous, so we need to recognize that
M of X comma Y is equal to Y cubed minus X cubed
and N of X comma Y is going to be equal to negative XY squared.
Now we need to find M of TX comma TY
as well as N of TX comma TY.
So here we'd have T cubed Y cubed
minus T cubed X cubed.
If we factor out the T cubed,
we'd have T cubed times the quantity Y cubed minus X cubed
which is function M, so we can write this as
T to the third times M of X comma Y.
So M is a degree three homogeneous function
and if N is also a degree three homogeneous function,
then we know our DE is also homogeneous.
So for function N, we would have
negative TX times T squared Y squared.
Notice how we do have a factor of T cubed
so we can write this as T cubed times negative XY squared.
Notice how again, this is function N,
so we can write this as T to the third times N of X comma Y.
Since N is also a degree three homogeneous function,
we know the original differential equation
is a homogeneous differential equation.
So now we can solve this using substitution.
Let's go ahead and do that on the next slide.
Remember, we do have a choice of substitutions,
but notice how this function here,
function N is simpler than function M,
so it's going to be easier if we perform our substitution
for DY, which means we're going to let Y equal X times V
and therefore differential Y is equal to X times DV
plus V times DX.
So now we'll go ahead and perform the substitution.
So for Y cubed we'd have X cubed V cubed
minus X cubed DX
minus X times Y squared which would be X squared V squared
times DY, which is XDV
plus VDX.
Now we'll clear the parentheses and simplify.
So we'll have X cubed V cubed DX
minus X cubed DX.
Here we're going to have
negative X to the fourth V squared DV
or minus X to the fourth V squared DV
and then minus X to the third V to the third DX.
Notice how the first term and the last term are opposites.
This is going to simplify to zero,
and since this term contains DV,
let's move it to the other side by adding
X to the fourth V squared DV to both sides.
So we'd have negative X cubed DX
equals positive X to the fourth V squared DV.
We want all the Xs on the left side with DX,
so now we'll divide both sides by X to the fourth.
This leaves us with negative one over X DX
equals V squared DV
and now we'll integrate both sides of the equation.
So here we'll have negative natural log absolute value of X
plus C equals, here we'd have V to the third
divided by three or one third V to the third.
Let's go ahead and finish this on the next slide.
We need to find our general solution in terms of X and Y,
so we need to perform our substitution here for V,
and where we started with the substitution
Y equals X times V, so V is equal to Y divided by X.
So now we can write this as
negative natural log absolute value of X plus C equals,
we'd have Y cubed over 3X cubed,
let's go ahead and write this as
C equals Y cubed over 3X cubed
plus natural log absolute value of X.
Now let's clear this fraction
by multiplying both sides by 3X cubed.
So we have 3X cubed C equals,
we'd have Y cubed plus 3X cubed
natural log absolute value of X.
So this would be our general solution,
but remember, since we have this initial condition,
we can now find the particular solution
since Y(1) equals two, that means when X is one,
Y must be equal to two.
So using...
Y(1) equals two, we would have three times one cubed C
must equal two cubed plus three times one cubed
times natural log absolute value of one,
when natural log absolute value of one is equal to zero,
so we have three C equals eight,
divide both sides by three,
we have C equals eight thirds.
Now that we know that C is equal to eight thirds,
we can go back to our general solution here
and replace C with eight thirds.
So we'd have 3X cubed times eight thirds
equals Y cubed plus 3X cubed
natural log absolute value of X
and notice how that these three simplify out,
so our particular solution is going to be 8X cubed
equals Y cubed plus 3X cubed
natural log absolute value of X.
Let's finish this by taking a look at this graphically.
We can use the original differential equation
to generate the red slope field shown below.
Y(1) equals two represents the point
with coordinates one comma two, or this point here.
Notice if you graph our solution,
it does fit nicely in the slope field
and passes through the point,
which is our initial condition.
So this does verify our solution is correct.
Okay, I hope you found this helpful.