- [Instructor] Welcome back to a third example of solving
a Bernoulli Differential Equation.
In this example we'll take a look at an initial value
problem.
A differential equation in this form here is a Bernoulli
differential equation where N, the exponent of Y
is any real number.
To solve our Bernoulli differential equation,
we're going to perform a substitution using the formula
V equals Y raised to the power of N minus 1,
after performing the substitution the result will be
a linear first order differential equation,
which we'll solve by using integrating factor,
then we'll solve for V then determine the solution
to the original differential equation.
So let's take a look at our example.
The first step is to recognize that the given
differential equation does fit the form of a Bernoulli
differential equation and that it's in the correct form
meaning the first term is Y prime, or DY DX.
Next, we need to recognize that N is equal to 5,
based upon the power of Y,
so if N is equal to 5,
then we're going to use the formula V equals Y raised to the
power of N minus 1 to perform our substitution,
so V is going to be equal to Y raised to the power of 1 minus 5
or negative 4.
But we're going to perform a substitution into the
differential equation for Y here and here,
as well as Y prime or DY DX, so we actually want to solve
this equation for Y.
So if we raise both sides of this equation to the power
of negative 1 fourth,
notice how in the right we would just have Y to the first
or Y equals V to the negative 1/4th.
Now we want to find Y prime or DY DX, this will require
implicit differentiation because we have Y in terms of V,
not in terms of X, so Y prime is going to be equal to
negative 1/4th times V raised to the power of
negative 1/4th minus 1, negative 5/4ths times DV DX.
Again we're apply the chain roll here because with the
derivative in terms of X, not just V.
So now using these two equations here,
we can perform substitution into the original
differential equation.
So Y prime is going to be all of this,
plus 2 divided by X times Y which is V to the negative 1/4th
equals negative X to the 9th,
times Y to the 5th, well if we raise V to the negative 1/4th
to the 5th, we would have V to the power of negative 5/4ths.
Now this differential equation is a linear first order
differential equation because it fits this form here,
that we do need to manipulate this in order to have
this first term be DY DX or in this case DV DX.
We'll go ahead and do this in two steps to keep it clear,
we'll go ahead and divide everything by V to the
negative 5/4ths to begin with.
This will simplify to negative 1/4th, DV DX plus 2
divided by X, here we'll subtract the exponents,
negative 1/4th minus negative 5/4ths would be plus 5/4ths,
this just becomes V, and on the right side,
this simplifies the 1 so we have negative X to the 9th.
But we still want the first term to be DV DX,
so now we're going to multiply through by negative 4
giving us DV DX, if we multiply by negative 4 here we'll
have minus 8 divided by XV equals positive 4 X to the 9th.
Now this is a linear first order differential equation,
in the form that we need, so now we'll find the integrating
factor which is X equals E raised to the power of the
integral of P of X DX, where P of X is the function
being multiplied by Y in this case P of X is negative 8
divided by X, let's go ahead and find the numerating
factor on the next slide.
So again, P of X equals negative 8 divided by X,
so the integrating factor which we're going to multiply
this differential equation by is going to be E raised to
the power of the integral of negative 8,
1 over X DX, notice how I did factor out the negative 8,
this is going to give us E raised to the power of negative 8
natural log X, applying the power of log rhythms we can
move this coefficient to the exponent position.
Giving us E raised to the power of natural log,
X to the negative 8th,
and this simplifies nicely just to X to the negative 8th.
This is the integrating factor which we'll use
to solve this differential equation.
Remember here if our base is E,
raised to the power of log based E or natural log,
that simplifies just to the number part of the log rithm.
So again now we're going to multiply everything in this
equation by X to the negative 8th.
So we'll have X to the negative 8th, DV DX,
remember this is the same as negative 8 X to the negative 1,
so we'll have minus 8 X to the negative 9 V equals
here adding the exponents, we just have 4 X.
Now remember when solving a linear differential equation,
once we multiply through by the integrating factor,
the left side of this differential equation or this side
here is 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 the integrating
factor which was X to the negative 8th,
times normally Y, but in this case it's going to be V.
And because these are equal, this is also equal to the
right side or 4 X.
Which means now we can solve for V if we integrate both
sides with respects to X.
The integral and derivative undo each other leaving us with
X to the negative 8th times V is equal to,
this would be 4 times X squared divided by 2,
or just 2 X squared plus C.
And now to solve this for V, we can divide both sides
by X to the negative 8th, or multiply by X to the
positive 8th, let's go ahead and do that.
We're going to multiply both sides by X to the 8th,
remember adding the exponents this would be X to the zero
times V which is just V, X to the zero is equal to 1,
here we would have 2 X to the 10th, plus C X to the 8th.
But remember we're not done, we're actually trying to solve
this for Y not V, so let's finish this on the next slide.
Going back to the first slide, remember V is equal to
Y to the negative 4th,
or we can say V is equal to 1 divided by Y to the 4th,
so performing this substitution, we would have 1 over Y
to the 4th must equal 2X to the 10th,
plus C X to the 8th, but let's go ahead and take
the reciprocal of both sides, we have Y to the 4th
equals 1 divided by 2X to the 10th,
plus CX to the 8th.
So this is the general solution to the differential
equation, but since we have this initial condition,
we want to find the particular solution,
which means you want to find the value of C.
So if we know Y of 1 equals 1, this tells us that when
X equals 1, our function value or Y must also equal 1.
So using Y of 1 equals 1,
we'll go ahead and use this form of the solution to find C,
so if Y is equal to 1, the left side is 1.
On the right side, if X is 1, we would have 2 times 1
to the 10th which is 2, plus if X to the 8th would also
be 1 so we have C, so we have C equals, subtracting 2
on both sides, C equals negative 1.
So the particular solution to this initial value problem
is Y to the 4th equals 1 divided by 2X to the 10th
and then because C is negative 1 we'll have
minus X to the 8th.
To get a visual for this let's take a look at a graph.
The given differential equation can be used to create
the red slope field, and then because we're given Y of 1
equals 1 we know our particular solution must contain
this point here, and if we graph our particular solution
it would be this green graph, and notice how it does
fit the slope field nicely and passes through the point
which is our intitial condition.
Okay, I hope you found this helpful, thank you for watching.