Exact equations and integrating factors
Integrating factors 2 Now that we've made the equation exact, let's solve it!
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- And, in the last video, we had this differential equation.
- And it at least looked like it could be exact.
- But when we took the partial derivative of this expression,
- which we could call M with respect to y, it was different
- than the partial derivative of this expression, which is N in
- the exact differential equations world.
- It was different than N with respect to x.
- And we said, oh boy, it's not exact.
- But we said, what if we could multiply both sides of this
- equation by some function that would make it exact?
- And we called that mu.
- And in the last video, we actually solved for mu.
- We said, well, if we multiply both sides of this equation by
- mu of x is equal to x, it should make this into an exact
- differential equation.
- It's important to note, there might have been a function of
- y that if I multiplied by both sides it would
- also make it exact.
- There might have been a function of x and y that would
- have done the trick.
- But our whole goal is just to make this exact.
- It doesn't matter which one we pick, which integrating
- factor-- this is called the integrating factor-- which
- integrating factor we pick.
- So anyway, let's do it now.
- Let's solve the problem.
- Let's multiply both sides of this equation by mu, and mu of
- x is just x.
- We multiply both sides by x.
- So see, if you multiply this term by x, you get 3x squared
- y plus xy squared, we're multiplying these terms by x
- now, plus x to the third plus x squared y, y
- prime is equal to 0.
- Well now, first of all, just as a reality check, let's make
- sure that this is now an exact equation.
- So what's the partial of this expression, or this kind of
- sub-function, with respect to y?
- Well, it's 3x squared, that's just kind of a constant
- coefficient of y, plus 2xy, that's the partial with
- respect to y of that expression.
- Now let's take the partial of this with respect to x.
- So we get 3x squared plus 2xy.
- And there we have it.
- The partial of this with respect to y is equal to the
- partial of this with respect to N.
- So we now have an exact equation whose solution should
- be the same as this.
- All we did is we multiplied both sides of
- this equation by x.
- So it really shouldn't change the solution of that equation,
- or that differential equation.
- So it's exact.
- Let's solve it.
- So how do we do that?
- Well, what we say is, since we've shown this exact, we
- know that there's some function psi where the partial
- derivative of psi with respect to x is equal to this
- expression right here.
- So it's equal to 3x squared y plus xy squared.
- Let's take the antiderivative of both sides with respect to
- x, and we'll get psi is equal to what?
- x to the third y plus, we could write,
- 1/2 x squared y squared.
- And of course, this psi is a function of x and y, so when
- you take the partial with respect to x, when you go that
- way, you might have lost some function that's only a
- function of y.
- So instead of a plus c here, it could've been a whole
- function of y that we lost. So we'll add that back when we
- take the antiderivative.
- So this is our psi.
- But we're not completely done yet, because we have to
- somehow figure out what this function of y is.
- And the way we figure that out is we use the information that
- the partial of this with respect to y
- should be equal to this.
- So let's set that up.
- So what's the partial of this expression with respect to y?
- So I could write, the partial of psi with respect to y is
- equal to x to the third plus 2 times 1/2, so it's just x
- squared y plus h prime of y.
- That's the partial of a function purely of y with
- respect to y.
- And then that has to equal our new N, or the new expression
- we got after multiplying by the integrating factor.
- So that's going to be equal to this right here.
- This is, hopefully, making sense to you at this point.
- And that should be equal to x to the third plus x squared y.
- And interesting enough, both of these
- terms are on this side.
- So let's subtract both of those terms from both sides.
- So x to the third, x to the third, x squared
- y, x squared y.
- And we're left with h prime of y is equal to 0.
- Or you could say that h of y is equal to some constant.
- So there's really no y, the extra function of y.
- There's just some constant left over.
- So for our purposes, we can just say that
- psi is equal to this.
- Because this is just a constant, we're going to take
- the antiderivative anyway, and get a constant on
- the right hand side.
- And in the previous videos, the
- constants all merged together.
- So we'll just assume that that is our psi.
- And we know that this differential equation, up
- here, can be rewritten as, the derivative of psi with respect
- to x, and that just falls out of the partial derivative
- chain rule.
- The derivative of psi with respect to x is equal to 0.
- If you took the derivative of psi with respect to x, it
- should be equal to this whole thing, just using the partial
- derivative chain rule.
- And we know what psi is.
- So we can write-- or actually we don't even have to.
- We could use this fact to say, well, if we integrate both
- sides, that a solution of this differential equation is that
- psi is equal to c.
- I just took the antiderivative of both sides.
- So, a solution to the differential equation is psi
- is equal to c.
- So psi is equal to x to the third y plus
- 1/2 x squared y squared.
- And we could have said plus c here, but we know the solution
- is that psi is equal to c, so we'll just write that there.
- I could have written a plus c here, but then you
- have a plus c here.
- You have another constant there.
- And you can just subtract them from both sides.
- And they just merge into another arbitrary constant.
- But anyway, there we have it.
- We had a differential equation that, at least superficially,
- looked exact.
- It looked exact, but then, when we tested the exactness
- of it, it was not exact.
- But we multiplied it by an integrating factor.
- And in the previous video, we figured out that a possible
- integrating factor is that we could just multiply
- both sides by x.
- And when we did that, we tested it.
- And true enough, it was exact.
- And so, given that it was exact, we knew that a psi
- would exist where the derivative of psi with respect
- to x would be equal to this entire expression.
- So we could rewrite our differential
- equation like this.
- And we'd know that a solution is psi is equal to c.
- And to solve for psi, we just say, OK, the partial
- derivative of psi with respect to x is
- going to be this thing.
- Antiderivative of both sides, and there's some constant h of
- y-- not constant, there's some function of y-- h of y that we
- might have lost when we took the partial with respect to x.
- So to figure that out, we take this expression.
- Take the partial with respect to y, and set that equal to
- our N expression.
- And by doing that, we figured out that that function of y is
- really just some constant.
- And we could have written that here.
- We could have written that plus c.
- We could call that c1 or something.
- But we know that the solution of our original differential
- equation is psi is equal to c.
- So the solution of our differential equation is psi x
- to the third y plus 1/2 x squared y
- squared is equal to c.
- We could have had this plus c1 here, then
- subtracted both sides.
- But I think I've said it so many times that you
- understand, why if h of y is just a c, you can
- kind just ignore it.
- Anyway, that's all for now, and I will see
- you in the next video.
- You now know a little bit about integrating factors.
- See you soon.
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At 5:31, how is the moon large enough to block the sun? Isn't the sun way larger?
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