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Studying for a test? Prepare with these 8 lessons on Første ordens differentialligninger.

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# Integrerende faktorer 1

Video transskription

A lot of what you'll learn in
differential equations is really just different
bags of tricks. And in this video I'll show
you one of those tricks. And it's useful beyond this. Because it's always good when,
if maybe one day, you become a mathematician or a physicist,
and you have an unsolved problem. Some of these tricks that solved
simpler problems back in your education might be a
useful trick that solves some unsolved problems. So
it's good to see it. And if you're taking
differential equations, it might be on an exam. So it's good to learn. So we'll learn about integrating
factors. So let's say, we have an
equation that has this form. Let's say this is my
differential equation. 3xy-- I'm trying to write it
neatly as possible-- plus y squared plus x squared plus xy
times y prime is equal to 0. So, especially since we've
covered this in recent videos, whenever you see an equation
of this form where you have some function of xy, then you
have another function of x and y times y prime equals 0, you
said, oh, this looks like this could be an exact differential
equation. And how do we test that? Well, we can take the partial
derivative of this with respect to y, and we
could call this function of x and y, M. So the partial of that, with
respect to y, so M partial with respect to y, would
be 3x plus 2y. And if this function right here,
that expression right there, that's our function
N, which is a function of x and y. We take the partial with respect
to x, and we get that is equal to 2x plus y. And in order for this to have
been an exact differential equation, the partial of this
with respect to y would have to equal the partial of this
with respect to x. But we see here, just by looking
at these two, they don't equal each other. They're not equal. So, at least superficially, the
way we looked at it just now, this is not an exact
differential equation. But what if there were some
factor, or I guess some function that we could multiply
both sides of this equation by, that would
make it an exact differential equation? So let's call that mu. So what I want to do is I want
to multiply both sides of this equation by some function mu,
and then see if I can solve for that function mu that
would make it exact. So let's try to do that. So let's multiply both
sides by mu. And just as a simplification,
mu could be a function of x and y. It could be a function of x. It could be a function
of just x. It could be function
of just y. I'll assume it's just
a function of x. You could assume it's just
a function of y and try to solve it. Or you could just assume it's
a function of x and y. If you assume it's a function
of x and y, it becomes a lot harder to solve for. But that doesn't mean that
there isn't one. So let's say that mu
is a function of x. And I want to multiply it by
both of these equations. So I get mu of x times 3xy plus
y squared plus mu of x times x squared plus
xy times y prime. And then, what's 0 times
any function? Well, it's just going
to be 0, right? 0 times mu of x is just
going to be 0. But I did multiply the right
hand side times mu of x. And remember what we're doing. This mu of x is-- when we
multiply it, the goal is, after multiplying both sides
of the equation by it, we should have an exact equation. So now, if we consider this
whole thing our new M, the partial derivative of this with
respect to y should be equal to the partial derivative
of this with respect to x. So what's the partial derivative
of this with respect to y? Well, if we're taking the
partial with respect to y here, mu of x, which is only
a function of x, it's not a function of y, it's just
a constant term, right? We take a partial with respect
to y. x is just a constant, or a function of x can be viewed
just as a constant. So the partial of this with
respect to y is going to be equal to mu of x, you could just
say, times 3x plus 2y. That's the partial of this
with respect to y. And what's the partial of
this with respect to x? Well, here, we'll use
the product rule. So we'll take the derivative of
the first expression with respect to x. mu of x is no
longer a constant anymore, since we're taking the partial
with respect to x. So the derivative of mu of
x with respect to x. Well, that's just mu prime
of x, mu prime, not U. mu prime of x. mu is the Greek letter. It's for the muh sound, but
it looks a lot like a U. So mu prime of x times a second
expression, x squared plus xy, plus just the
first expression. This is just the product
rule, mu of x. Times the derivative of
the second expression with respect to x. So times-- ran out of space
on that line-- 2x plus y. And now for this new equation,
where I multiplied both sides by mu. In order for this to be exact,
these two things have to be equal to each other. So let's just remember
the big picture. We're kind of saying, this
is going to be exact. And now, we're going to
try to solve for mu. So let's see if we
can do that. So let's see, on this side, we
have mu of x times 3x plus 2y. And let's subtract this
expression from both sides. So it's minus mu of
x times 2x plus y. You'll see a lot of these
differential equation problems that get kind of hairy. They're really just
a lot of algebra. And that equals-- what
do we have left? I'll write it in yellow. That equals-- I'm going
to run out of space. I'm going to do it a
little bit lower. That equals, just this
term right here. That equals mu prime of x
times x squared plus xy. And let's see, if we factor out
a mu of x here, we get mu of x times 3x plus 2y minus 2x
minus y is equal to mu prime of x, the derivative of mu with
respect to x, times x squared plus xy. Now, we can simplify this. So we get mu of x times-- what
is this-- 3x minus 2x is x. 2y minus y, so x plus y, is
equal to-- and I'm just going to simplify this side a
little bit-- is equal to mu prime of x. Let's factor out an x here. And the reason why I'm doing
that is because it seems like if I factor out an x here,
I'll get an x plus y. So this is mu prime of x
times x times x plus y. x times x plus y is
x squared plus xy. So that's why I did it, and I
have this x plus y on both sides equation, which I will
now divide both sides by. So if you divide both sides by
x plus y, we could maybe assume that it's not 0. That simplifies things
pretty dramatically. We get mu of x is equal to
mu prime of x times x. And now, just the way my brain
works, I like to rewrite this expression just in our operator
form, where instead of writing it mu prime
of x, we could write that as d mu dx. So let's do that. So we could write mu of x is
equal to d, the derivative of mu with respect to x, times x. And this is actually a separable
differential equation in and of itself. It's kind of a sub-differential
equation to solve our broader one. We're just trying to figure
out the integrating factor right here. So let's divide both
sides by x. So we get mu over x, this is
just a separable equation now, is equal to d mu dx. And then, let's divide both
sides by mu of x, and we get 1 over x is equal to 1 over mu. That's mu of x, I'll just write
1 over mu right now, for simplicity, times d mu dx. I'm actually going to go
horizontal right here. Multiply both sides by dx, you
get 1 over x dx is equal to 1 over mu of x d mu. Now, you could integrate both
sides of this, and you'll get the natural log of the absolute
value of x is equal to the natural log of the
absolute value of mu, et cetera, et cetera. But it should be pretty clear
from this that x is equal to mu, or mu is equal
to x, right? They're identical. If you look at both sides of
this equation there, you can just change x for mu, and it
becomes the other side. So, this is obviously telling us
that mu of x is equal to x. Or mu is equal to x. So we have our integrating
factor. And if you want, you can take
the antiderivative of both sides with the natural logs,
and all of that. And you'll get the
same answer. But this is just, by looking at
it, by inspection, you know that mu is equal to x. Because both sides of
this equation are completely the same. Anyway, we now have our
integrating factor. But I am running out of time. So in the next video, we're
now going to use this integrating factor. Multiply it times our original
differential equation. Make it exact. And then solve it as
an exact equation. I'll see you in the
next video.