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## Exact equations and integrating factors

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# Exact equations example 2

## Video transcript

Let's do some more examples
with exact differential equations. And I'm getting these problems
from page 80 of my old college differential equations books. This is the fifth edition of
Elementary Differential Equations by William Boyce
and Richard DiPrima. I want to make sure they get
credit, that I'm not making up these problems. I'm getting
it from their book. Anyway, so I'm just going to
give a bunch of equations. We have to figure out if they're
exact, and if they are exact, we'll use what we know
about exact differential equations to figure out
their solutions. So the first one they have is,
2x plus 3, plus 2y minus 2, times y prime is equal to 0. So this is our M of x and y--
although, this is only a function of x-- and then
this is our N, right? You could say that's
M, or that's N. You could also say that, if this
is exact-- well, first let's [? test ?] this exact, before we start
talking about psi. So what's the partial of this,
with respect to y? The partial of M, with
respect to y. Well, there's no y
here, so it's 0. The rate of change that
this changes with respect to y is 0. And what's the rate of
change this changes, with respect to x? The partial of N, with respect
to x is equal to-- well, there's no x here, right? So these are just constants from
an x point of view, so this is all going to be 0. But we do see that
they're both 0. So M sub y, or the partial with
respect to y, is equal to the partial with respect to x. So this is exact. And actually, we don't have to
use exact equations here. We'll do it, just so that
we get used to it. But if you look here, you
actually could have figured out that this is actually
a separable equation. But anyway, this is exact. So knowing that it's exact, it
tells us that there's some function psi, where psi is
a function of x and y. Where psi sub x is equal to this
function, is equal to 2x plus 3, and psi-- I shouldn't
say sub x. I say the partial of psi,
with respect to x. And the partial of psi, with
respect to y, is equal to this, 2y minus 2. And if we can find our psi, we
know that this is just the derivative of psi. Because we know that the
derivative, with respect to x of psi, is equal to the partial
of psi, with respect to x, plus the partial
of psi, with respect to y, times y prime. So this is this just the
same form as that. So if we can figure out y,
then we can rewrite this equation as dx, the derivative
of psi, with respect to x, is equal to 0. Let me switch colors, or it's
going to get monotonous. This right here, if we can find
a psi, where the partial with respect to x, is this, the
partial with respect to y, is this, then this can
be rewritten as this. And how do we know that? Because the derivative of psi,
with respect to x, using the partial derivative chain
rules, is this. This partial with respect
to x, that's this. This partial with respect to
y, is this, times y prime. So this is the whole point
of exact equations. But anyway, so let's figure
out what our psi is. Actually, before we figure out,
if the derivative of psi, with respect to x, is 0, then
if you integrate both sides, you just-- the solution
of this equation is psi is equal to c. So using this information, if we
can solve for psi, then we know that the solution of this
differential equation is psi is equal to c. And then if we have some initial
conditions, we could solve for c. So let's solve for psi. So let's integrate both sides
of this equation, with respect to x. And then we get psi is equal
to x squared plus 3x, plus some function of y. Let's call it h of y. And remember, normally when you
take an antiderivative, you have just a plus
c here, right? But you can kind of say we
took an anti-partial derivative. So when you took a partial
derivative, with respect to x, not only do you lose constants--
that's why we have a plus c, normally-- but you
also lose anything that's a function of just y, and not x. So for example, take the partial
derivative of this with respect to x, you're going
to get this, right? Because the partial derivative
of a function, purely of y, with respect to x,
is going to be 0. So it will disappear. So anyway, we take the
antiderivative of this, we get this. Now, we use this information. Well, we use this information. We take the partial of this
expression, and we say, well, the partial of this expression,
with respect to y, has to equal this, and then we
can solve for h of y, then we'll be done. So let's do that. So the partial of psi, with
respect to y, is equal to-- well, that's going
to be 0, 0, 0. This part is a function of x. If you take the partial with
respect to y, it's 0, because these are constants, from
a y point of view. So you're left with
h prime of y. So we know that h prime of y,
which is the partial of psi, with respect to y,
is equal to this. So h prime of y is equal
to 2y minus 2. And then if we wanted to figure
out what h of y is, we get h of y-- just integrate both
sides, with respect to y-- is equal to y squared
plus-- sorry-- y squared minus 2y. Now, you could have a plus c
there, but if you watched the previous example, you'll see
that that c kind of merges with the other c, so you
don't have to worry about it right now. So what is our psi function,
as we know it now, not worrying about the plus c? It is psi of x and y is equal
to x squared plus 3x, plus h of y-- which we figured
out is this-- plus y squared, minus 2y. And we know a solution of our
original differential equation is psi is equal to c. So the solution of our
differential equation is this is equal to c. x squared plus 3x, plus
y squared, minus 2y is equal to c. If you had some additional
conditions, you could test it. And I encourage you to test
this out on this original equation, or I encourage you to
take the derivative of psi, and prove to yourself that if
you took the derivative of psi, with respect to x, here,
implicitly, that you would get this differential equation. Anyway, let's do another one. Let's clear image. So the more examples you
see, the better. So let's see, this one says 2x
plus 4y, plus 2x minus 2y, y prime is equal to 0. So what's the partial of
this with respect to y? So M, the partial of M, with
respect to y-- this is 0-- so it's equal to 4. What's the partial of this, with
respect to x, just this part right here? The partial of N, with
respect to x, is 2. This is 0. So the partial of this, with
respect to y, is different than the partial of N,
with respect to x. So this is not exact. So we can't solve this using
our exact methodology. So that was a fairly
straightforward problem. Let's do another one. Let's see. I'm running out of time, so I
want to do one that's not too complicated. Let's see, 3x squared minus
2xy-- actually, let me do this in the next problem. I don't want to rush
these things. I will continue this
in the next video. See you soon.