>> Welcome to the Cypress College math review
on Bernoulli differential equations.
A Bernoulli differential equation is one
that can be written in the form Y prime plus function
of X times Y equals function of X times Y to the Nth power
where N is any number other than zero or one.
After you have the equation
in that form make the substitution U equals Y
to the one minus N power.
That will transform the equation
into a first order linear differential equation.
You'll also need to find U prime.
Before you actually make your substitution divide the equation
by Y to the Nth, then substitute out for all of your Y's.
Now you have a first order linear differential equation.
Obtain and simplify the integrating factor mu of X,
which is equal to E to the power of an anti-derivative
of the function P. Leave off the constant
of integration and simplify.
Solve for U by integrating
in the equation U times mu equals the antiderivative
of Q times mu.
And then substitute back and solve for Y.
Let's put this in standard form, see what we got.
So we'll divide by three X Y squared.
So we have Y prime minus one over X Y equals X cubed cosine
of X over three times.
Well in this nominee we have Y squared so we'll move that up
and make that Y to the negative two power.
So we do have a Bernoulli differential equation.
And N negative two.
So we're going to make a substitution.
We're going to let U equal Y to the power of one minus N
and N is negative two.
U prime then would be three Y squared Y prime.
Now we divide by Y to the negative two,
same thing as multiplying by Y squared.
So we end up with Y squared Y prime minus one
over X Y cubed equals X cubed cosine X over three.
Now we make our substitution.
So Y squared Y prime is equal to U prime over three.
Here we have Y cubed which is U.
And now you can see we have our first order linear
differential equation.
If we multiply it by three we have it in standard form.
Let's find the integrating factor.
So mu of X is equal to E to the power of the antiderivative
of P. P is negative three over X. So we get E to the power
of negative three natural log of X, which gives us X
to the negative three.
So we have X to the negative three times U equals the
derivative of X to the negative three times our function Q. Our
function Q is X cubed times cosine of X,
and we're integrating with respect to X.
So we're actually just integrating the function cosine.
So we get sine of X plus C. So U then is equal
to C cubed times the quantity, sine of X plus C. What was U?
U was Y cubed.
And if we take the cube root of both sides we have our answer.
Once again standard form is our first goal.
So we have Y prime plus three
over two X times Y. We'll move the other term
to the other side.
Equals negative X cubed over two times Y cubed.
So that's our N, N is three.
We're going to make the substitution, but U equal Y
to the power of one minus.
Once again N was three.
So that would be negative two.
U prime would be negative two.
Y to the negative three times Y prime.
And then divide.
Divide by Y cubed.
So we have Y to the negative three Y prime plus three
over two X Y to the negative two equals negative X cubed
over two.
Now we make our substitutions.
We look back over here.
Y to the negative three Y prime is equal
to U prime over negative two.
So this is U prime over negative two.
Y to the negative two is equal to U. Multiply by negative two
and we'll have it in standard form.
So we have U prime minus three over X times U equals X cubed.
Now we find our integrating factor mu,
which is E to the power of the antiderivative
of the function negative three over X. And we get E
to the power of negative three natural log absolute value of X
which just gives us X to the negative three power.
Therefore X to the negative three times U is equal
to the antiderivative of X to the negative three times
or function Q, which is also X to the positive three.
So we're just integrating the function one.
So we get X plus C when we integrate.
So U is equal to X cubed times the quantity X plus C.
And what was U?
U was equal to Y to the negative two power.
So Y to the negative two is equal
to X cubed times the quantity X plus C. So Y is equal to plus
or minus one over the square root
of X cubed times the quantity X plus C.
So divide by X. So we have Y prime minus the quantity three X
plus six over X times Y equals negative nine E
to the negative X times Y to the four third's.
There's N, four thirds.
So this is Bernoulli.
So we're going to make the substitution.
U equals Y to the power of one minus four third's.
So that's Y to the negative one third.
So U prime is equal to negative one third Y
to the negative four third's times Y prime.
Divide by Y to the four thirds power.
And we have Y to the negative four third's times Y prime minus
three X plus six quantity over X times Y
to the negative one third equals negative nine E
to the negative X. Now we make our substitution.
We notice over here that Y
to the negative four third's times Y prime is equal
to negative three U prime.
This is U. Divide by negative three
and we'll have our first order linear differential equation
in standard form.
Find mu. So mu of X is equal to E to the power
of the antiderivative of the function quantity X plus two
over X with respect to X. We split
that up that's one plus two over X that we're integrating.
So we have E to the power
of X plus two natural log the absolute value of X.
That gives us E to the X times X squared.
Now apply it.
So we have mu times U equals the antiderivative
of mu times Q. Now we're integrating with respect to X.
So we're simply integrating the function X squared.
So we have X cubed plus C. So U ends up being X cubed plus C all
over E to the X times X squared.
Now we substitute back.
What was U?
U was Y to the negative one third.
So we invert and cube that expression.
There's our answer.
There is another problem to try.
Pause the video, solve this differential equation.
When you're finish restart the video and check the answer.
Here's the answer to the practice problem.