- [Voiceover] Welcome to a lesson in solving
a linear first order differential equation.
If we do have a linear first order differential equation
then it must fit this form here
which is the standard form of a
linear first order differential equation
so once we have the equation in this form
we're going to multiply through by a function
called an integrating factor that's going to make
solving this differential equation a lot easier.
So next, we'll find the integrating factor
which is the function mu of x equals e
raised to the power of the integral of
p of x, integrative respects to x.
Notice that p of x, this function here
is the function being multiplied by y
so once we find the integrating factor, we'll multiply
both sides of the equation by mu of x, given here
and once we have the differential equation in this form
what we should notice is that the left side
of the differential equation, or this side here
is now equal to the derivative
of the product of the integrating factor, mu of x, and y
so this left side here is equal to this derivative here
which means this derivative must also equal
the right side of the equation, given here
which is the integrating factor, times the function f of x
so once we have the equation in this form here
we'll integrate both sides of the equation
and then solve for y
so let's go ahead and take a look at our first example.
The first thing we need to do is make sure
that we do have a linear first order differential equation
that fits this form here
and we do.
Next, we want to identify the function p of x
which again is being multiplied by y here
so in this case, p of x is equal to two divided by x
and once we have p of x
we can find the integrating factor, given here.
So mu of x, our integrating factor, is going to be equal to
e raised to the power of the integral of p of x
which is two divided by x.
I'm going to go ahead and factor the two out
so we have two times the integral of one over x dx.
Well, the integral of one over x is just going to be
natural log x, so this is equal to
e raised to the power of two natural log x.
We don't want to leave it in this form, though.
We'll now apply the power property of logarithms
and move this two here to the exponent on this x
so this is equal to
e raised to the power of natural log x squared
and now, this simplifies even further.
Remember, if we have e raised to the power of log e
or natural log, this simplifies nicely just to x squared
so our integrating factor, mu of x, is equal to x squared.
Now, I'll multiply both sides of the differential equation
by x squared, so that would give us
x squared times dy dx
plus x squared times two over x times y
equals x squared times the quantity x minus one.
Now, let's go ahead and simplify this a little bit.
This should be over one, so here we'll have
one factor of x simplify out
so we have x squared dy dx
plus this would now just be two xy
equals, let's go ahead and distribute here.
We have x cubed minus x squared.
Now that we've multiplied by the integrating factor
what's special about this is
that the left side of the equation, this side here
is now equal to the derivative
of the integrating factor, which is x squared
multiplied by y.
Let's go ahead and check this by applying the product rule.
We'd have the first function, x squared
times the derivative of y with respects to x
which is dy dx, plus the second function, which is y
times the derivative of x squared, which would be two x
so that's true, and since this is equal to
the derivative here, the derivative must also be equal to
the right side, so we can set this equal to
x cubed minus x squared.
Now that we've done this
we'll now integrate both sides of the equation
with respects to x and solve for y.
Looking at the left side
the integral and the derivative are going to undo each other
so we're left with just x squared y equals
integrating with respects to x, we would have
x to the fourth divided by four
or one fourth x to the fourth
minus x to the third divided by three
or one third x to the third
plus our constant of integration.
Now, to solve this for y
we just need to divide both sides
of the equation by x squared.
Let's go ahead and finish this on the next slide.
So on the left side, we just have y.
On the right side, we're dividing each term by x squared
so we'll be subtracting the exponents
so we'll have one fourth x to the second
and we'll have minus one third x
plus C divided by x squared
and this should be the general solution
to our differential equation.
Let's take a look at a second example.
So the first step, again, is to make sure
that this linear first order differential equation
fits this form here.
Notice how we just want dy dx here
and we have x times dy dx
so our first step is going to be to divide everything by x
so we'd have dy dx
minus four divided by x times y
equals, we'd have x to the fifth, e to the x.
Again, we divided everything by x.
Next step, we want to identify our function, p of x
which is being multiplied by y, so p of x
is equal to negative four divided by x
so now we can find our integrating factor, mu of x
is going to be equal to e
raised to the power of the integral of p of x dx.
We'll go ahead and factor out the negative four
so we'll have negative four times the integral
of one over x dx.
The integral of one over x, just like last time
is natural log x, so we'll have e
raised to the power of negative four natural log x.
We'll go ahead and apply the power property here
so we have e raised to the power of
natural log x to the negative four
and again, the base here is e.
The base of the log is also e
so this simplifies nicely to x to the power negative four.
So our integrating factor, mu of x
is equal to x to the negative fourth.
So now we're going to multiply both sides
of our differential equation here by x to the negative four
so we'll have x to the negative four dy dx
minus x to the negative four
times four over x times y
equals x to the negative four
times x to the fifth, e to the x.
Let's go ahead and simplify this.
X to the negative four, dy dx
minus, now here, if we move this down to the denominator
it'd be x to the fourth
so we'll have four all over x to the positive fifth
times y must equal
here, we'll add the exponents because we're multiplying.
That would just be x e to the x
so again, the big connection here is that
now, on the left side of this equation, this side here
is equal to the derivative
of the integrating factor times y.
Well, the integrating factor was x to the negative four
then times y
so these two are equal.
Let's go ahead and check it by applying the product rule.
The first times the derivative of the second
that looks good
plus the second, which is the y
times the derivative of the first
and that's the reason why it's minus
because if we multiply by negative four
and then subtract one, that'd be x to the negative fifth
which, in the denominator, is x to the positive five
so that looks good, and since these are equal
this derivative must also equal the right side here
which is x times e to the x.
Now, the last step is to integrate both sides
and solve for y.
Let's go ahead and do that on the next slide.
So again, now we're going to integrate both sides
with respects to x
so the integral and derivative will undo each other
so we'll have x to the negative four, y
and to integrate here, we have to apply
integration by parts, so just to review
here's the formula for integration by parts
so to integrate here, we would let u equal x
so du is equal to one dx, and dv
is equal to e to the x dx
so if we integrate to find v
the integral of e to the x with respects to x
is just e to the x
so now we can apply integration by parts.
We have u times v, which is x times e to the x
minus the integral of v du.
Well, v is e to the x, and du is just dx
so we have e to the x dx
and now this is much easier to integrate
so we have x to the negative four, y
equals x, e to the x minus e to the x plus C.
Now, we could divide both sides by x to the negative four
but instead, let's go ahead and multiply by x to the fourth.
Remember, we add these exponents.
X to the zero would be equal to one.
So we have y equals x to the fifth, e to the x
minus x to the fourth, e to the x
plus C times x to the fourth.
Okay, I hope you found this video helpful.
We'll take a look at some more examples
in the next few videos.