- [Narrator] Now we'll take a look at another example
of solving a linear first-order differential equation.
First we want the differential equation in this form here,
which is the standard form of a linear first-order DE.
Now what we're going to do is we're going to find
an integrating factor or a function
that we're going to multiply both sides of the equation by
that's going to help us solve this differential equation.
The integrating factor in this case
is going to be mu of x equals e
raised to the power of the integral of P of x
integrated with respects to x.
Notice P of x, this function here
is the function in our differential equation
that's being multiplied by y.
Now we're going to take the differential equation
and multiply everything by mu of x, which is shown here.
And once we have the differential equation in this form,
what we're going to notice is that the left side
of this differential equation, or this side here
is equal to the derivative of the product
of our integrating factor of mu of x, and y.
So this left side here is equal to this derivative.
But notice how this is also equal to mu of x times f of x.
Therefore this derivative is equal
to our integrating factor times f of x.
So once we have the equation in this form here,
we'll integrate both sides and then solve for y.
And that'll give us our solution.
So here's the given differential equation.
We first want to make sure that
it's going to fit this form here.
It looks like it will, and therefore
it's a linear first-order differential equation.
But we do need to change the form of the given equation.
Notice how we have to have the equation in this form.
Remember y prime is equal to dy/dx.
So what we're going to do here is divide everything
by the quantity x squared plus nine.
So if we do this, we would have y prime plus x
divided by the quantity x squared plus nine.
I'll go ahead and write over here times y
so it fits the form that we're looking for.
Equals zero.
Now the next step is to identify the function P of x
so that we can then find the integrating factor.
Again P of x is the function here multiplied by y.
So in this case, P of x
is equal to this quotient here.
X divided by the quantity x squared plus nine.
Which means now we can find our integrating factor mu of x,
which is going to be equal to e
raised to the power of the integral of P of x,
which is this function here,
x divided by x squared plus nine, dx.
Now to integrate this, notice how
the denominator is degree two, the numerator is degree one.
So we can integrate this using substitution.
We'll go ahead and let u equal the denominator,
x squared plus nine.
So differential u is going to be equal to two x dx.
Notice how we don't have a two x dx here, we have x dx.
Let's go ahead and divide both sides by two
so we can say that 1/2 d u is equal to x dx.
So if we want to write this in terms of u,
we would have equals e raised to the power
of the integral of one over u.
And then again, x dx is 1/2 d u.
So here we're going to have e raised
to the power of 1/2 times natural log u,
which would be natural log of x squared plus nine.
Now we can continue simplifying here.
Next step is going to be to apply the power rule of logarithms.
We're going to take this 1/2
and move it to the position of the exponent.
So we have e raised to the power of natural log
of the quantity x squared plus nine to the 1/2.
This simplifies even further now.
We have base e, and natural log is base e,
so all of this simplifies to the quantity
x squared plus nine raised to the 1/2,
which is our integrating factor.
So we have mu of x equals quantity x squared plus nine.
Now we're going to multiply everything
in this differential equation here
by the integrating factor.
So we're going to have the quantity
x squared plus nine to the 1/2
times y prime
plus x squared plus nine to the 1/2
times x divided by the quantity x squared plus nine
times y.
And then multiplying the right side,
we would still have zero.
Let's go ahead and simplify this here.
This is over one.
And this would be the quantity
x squared plus nine to the first.
So you move this quantity down,
it's going to be to the negative 1/2 power.
Then we'd add the exponents.
Negative 1/2 plus one is equal to 1/2.
So we'll still have an x in the numerator.
Our denominator is going to be the quantity
x squared plus nine to the 1/2 power
times y equals zero.
Now that we have the equation in this form,
the idea is the left side of this differential equation
is equal to the derivative
of the integrating factor times y.
So the derivative
of the integrating factor, which was the quantity
x squared plus nine to the 1/2 times y
is equal to this sum here.
Which means this derivative must also equal the right side.
So this is equal to zero.
And now what we're going to do is integrate
both sides of the equation and solve for y.
So now we'll integrate the left side with respects to x
and the right side with respects to x.
On the left side, the integral and the derivative
are going to undo each other.
So we'll be left with just the quantity x squared plus nine
to the 1/2 times y
must equal the integral of the right side.
Well the integral of zero is going to be any constant,
so this is just going to be equal to C.
Now we can solve for y by dividing by this quantity here.
So our general solution to the differential equation
is y equals a constant divided by the quantity
x squared plus nine raised to the 1/2 power.
Let's take a look at several solutions of this
on the slope field for the given differential equation.
So the given differential equation
will produce this given red slope field.
And then to show a few of the solutions,
I've picked the following values for C
and then graphed the functions.
So each of these functions represents
one possible solution to the family of solutions
to the given differential equation.
Okay, I hope you found this helpful.