Learn about Exact Equations Example 2

Khan Academy Presents: Some more exact equation examples

Let’s do some more examples with exact differential equations and I’m getting these problems with page 80 of my old college differential equations book. This is the fifth edition of elementary differential equations book by William Boyce and Richard de Prima. I wanna 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 gonna give a bunch of equations to figure out if they are 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 two x plus three plus two y minus 2 times y prime is equal to zero. So this is our M of x and y, although this is only a function of x. and this is our N. that's M, that’s N. well first, let's test if it’s 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 zero. And what’s the rate of change with respect to y is zero. What's the rate of change of this 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 zero. But we do see that they’re both zeros, 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 we can get used to it but if you look here you actually could have figured it out that this is actually a separable equation. But anyway, this is exact so knowing that it’s exact it tells us there’s some function psi where psi is a function of x and y, where the partial of psi with respect to x is equal to this function, two x plus three. And the partial of psi with respect to y is equal to two y minus 2. And if we can find our psi, we know that this is just a 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 just the same form as that. So if we can figure out y, then we can re-write this equation as the derivative of psi with respect to x is equal to zero. Let me switch colors instead it gets monotonous. This right here, if we can find the psi where the partial with respect to x is this, the partial with respect to y is this, then this can be re-written as this right. 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. 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 zero, then if you integrate both sides, 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 if we have some initial conditions then we can 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 three x plus some function of y, let’s call it H of y. And remember normally when you take an anti derivative you just have a plus C here right? But you can kinda 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 a partial derivative of this with respect to x, you’re going to get this because the partial derivative of a function purely of y with respect to x is going to be zero so will disappear. So anyway, we take the anti derivative of this, we get this. Now we use this information. We take the partial of this expression and we say the partial of this expression with respect to y has to equal this and then we can solve for H of y and we’ll be done. So let’s do that. So the partial of psi with respect to y is equal to 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, two y minus two. And then if we wanted to figure out what H of y is, we get H of y is y squared minus two y. You could have a plus C there but if watched the previous example you can see that that C kinda 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 now, not worrying about the plus C? It is psi of x and y is equal to x squared plus three x plus H of y which we figured out as this, plus y squared minus two y. And we know 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. So x squared plus three x plus y squared minus two y is equal to C. 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 that you would get this differential equation. Let’s do another one. So more examples you see the better. This one says two x plus four y plus two x minus two y prime is equal to zero. So what’s the partial of this with respect to y? So the partial of M with respect to y is equal to four. What’s the partial of this with respect to x? Just this part right here, the partial of N with respect to x is two. 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 straight forward problem. Let’s do another one. Actually, let me do this in the next problem. I don’t wanna rush these things. I will continue this in the next video. See you soon.