- [Narrator] Welcome to a lesson
on Bernoulli differential equations.
A differential equation in this form here where n
is any real number is a Bernoulli differential equation.
Notice if n is equal to zero or one,
we would have a linear first order differential equation.
So for this video, we'll be looking at values of n
that are greater than one.
So if we recognize that a differential equation fits
this form then to solve the Bernoulli equation,
we're going to perform a substitution
which will be v equals y raised to the power of one minus n.
After performing this substitution, we'll solve
the resulting linear first order differential equation.
Then, after we solve for v, we can determine the solution
to the original differential equation which will be
to solve for y.
So let's take a look at our first example.
The first step is to recognize
that the given differential equation does fit the form
of a Bernoulli differential equation or this form here.
Also notice here we have y to the n.
So in this case n is equal to two.
So once we know n is equal to two,
we can determine v using the formula v equals y
raised to the power of one minus n.
So if n is two, we would have v equals y
raised to the power of one minus two
or y to the negative one which is the same as one over y.
Notice that v and y are reciprocals and since we have
to perform a substitution in the differential equation
for both y here and here as well as y prime,
we actually want to solve this equation for y.
Again, since they're reciprocals, we can say that y
is equal to one over v or v to the negative one.
So again we know y equals v to the negative one
and now we also have to determine y prime
which is actually dy dx, not dy dv.
So we'll have to perform implicit differentiation here.
So y prime is going to be equal to negative one times v
to the negative two or negative v
to the negative two times dv dx.
Again, we're applying the chain rule here
because we want the derivative in terms of x not just v.
Now we'll go ahead and perform the substitution.
So y prime or dy dx is going to be all of this
plus x times y which is v to the negative one
equals x times y squared which would be v
to the negative two.
Now that we've performed this substitution,
this is now a linear first order differential equation
which we'll now solve for v but we do need the equation
in this form here in order to determine
the integrating factor given here in green.
So the first step here, we want
this first term to be dv dx.
So now we'll divide everything
by negative v to the negative two.
So this simplifies to dv dx.
Next notice here we'll have plus a negative,
so we'll just write minus x.
Subtracting the exponents we would have negative one
minus negative two which will give us v to the positive one.
And on the right side, this just simplifies to negative x.
Now that we have your linear first order
differential equation, we want to find
our integrating factor which we'll multiply both sides
of this equation by which is going to be e
raised to the power of the integral of P of x,
integrated with respect to x.
So in this case, P of x is going to be just negative x.
So the integrating factor mu of x is going to be equal
to e raised to the power of the integral of negative xdx.
So mu of x is just equal to e raised to the power of...
This would be negative x squared divided by two.
Now we'll multiply both sides of this differential equation
by the integrating factor.
Let's go ahead and do that on the next slide.
Now let's go ahead and distribute.
Now remember when solving a linear first order
differential equation, the left side of this equation here
is always going to be equal to the derivative of the product
of the integrating factor and y or in this case v.
So this is equal to the derivative of e
raised to the power of negative x squared divided by two.
Again, normally it's times y but in this case it's times v.
We can easily verify this if we apply the product rule here.
But because it's equal to the left side,
it's also equal to the right side of this equation.
So we can say it's also equal to negative x e
raised to the power of negative x squared divided by two.
And now if we integrate both sides of this equation
with respect to x we can then solve for v
and then go back and solve for y.
So now we'll integrate this with respect to x.
On the left side, the integral and derivative
undo each other so we're just left with e
raised to the power of negative x squared
divided by two times v equals.
On the right side this may look like it's going
to require integration by parts
but we can actually just use substitution.
If we let u equal negative x squared divided by two
notice that differential u is equal to just negative xdx.
So in terms of u, this is just the integral of e to the udu
giving us e raised to the power of negative x squared
divided by two plus c.
Now we can solve this for v by dividing everything by e
raised to the power of negative x squared divided by two.
So this simplifies to v.
This simplifies to one plus, if we move this up
into the numerator it's going to change the sign
of the exponent so we would have c times e
raised to the power of x squared divided by two.
Now be careful here, we're not done.
Remember, we're trying to solve for y not v.
So let's finish this on the next slide.
Remember we said earlier that v was equal to one over y
or that these two are reciprocals of one another
so we can also say that y is equal to one over v.
So, if they're reciprocals, then y
is equal to the reciprocal of this
which would just be one divided by
the quantity one plus c times e raised to the power
of x squared divided by two.
So this would be the solution to
the original Bernoulli differential equation.
Now let's finish by taking a look at this graphically.
If we were to graph the original differential equation,
we would produce the following red slope field
and here's our general solution that we just found
so I selected the following values of c
to graph just a few of the solutions
of the family of solutions for this differential equation.
So each c produced one of the graphs below
which represents one of the many solutions
to the given differential equation.
Okay, we'll take a look at another example
in the next video.
I hope you found this helpful.