Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. Separable equations. The function H in terms of XY is a separable equation when H of X, Y equal the derivative DYDX and the function H equals the product of 2 functions G of X * H of Y. Or we could think of that as the product of G of X * 1 / F of Y, or G of X / F of Y where H of Y equal 1 / F of Y. So then we could have the derivative equaling G of X / F of Y. If we cross multiply and integrate, we'd have the integral F of YDY equaling the integral G of X DX. So capital F of Y equal capital G of X + C. Two functions that have the same derivative on an interval if and only if they differ by a constant on that interval. Recall 2 functions have the same derivative, IE were saying that that integral of F of YDY and the integral of G of XDX are equivalent if they differ by some constant on that interval. A graph of a differentiable solution cannot have a vertical tangent anywhere. That one's important implicit solution is a solution that's not easily solved for Y in terms of X. More formally, if we have some function K in terms of the variables XY that equals 0, that's going to be an implicit solution of a differential equation. If it is satisfied on some interval by some solution Y equal Y of X. A particular solution may or may not satisfy a given initial condition. Now it's important that we understand it only has to be true on some interval, doesn't have to be true over all the real numbers we're going to talk about. If it's true around a point, not every possible algebraic solution of an implicit solution satisfies the same differential equation. That's sort of like back in our algebra days when we had extraneous roots. We had to do the problem, and then we needed to look at the solution to decide if it really satisfied the conditions or not. Also, we need to be careful if we divide each side by a term that that term may still be a solution. General general solution has a constant C where a singular solution has a specific value for C. Some of the applications for this topic are growth and decay, specifically population growth or population decay, interest, compound interest, interest on investments, etcetera. Radioactive decay. Drug elimination of a out of a person's system. Cooling and heating. Newton's law of cooling is the method frequently used here. Draining of a Tang Torricelli's law. Let's look at some examples. If I have DYD X + 2 XY squared equaling zero, we're going to start by taking the two XY squared to the other side. And now what we need to do is we need to get all of the YS on one side, so we're going to have D y / y ^2, and we need to get all the XS on the other side. So we get -2 XDX. Then we're going to actually evaluate this. The integral of D y / y ^2 is going to be -1 / y. The integral of -2 XDX is going to be negative X ^2 and we're going to have some plus C That's going to be an equation that is of a general solution. If we take this further and we want to actually solve for Y, we could see that we'd have -1 equaling negative the quantity X ^2 -X ^2 + C all times Y. We could multiply each side by a -1 if we wanted to, so we'd get one equaling X ^2 -, C times Y. So Y would equal 1 / X ^2 -, C This equation we could actually solve for Y fairly easily. This next 1DY DX equal 2X secant Y. Same concept. We're going to get all of the YS on one side, so we'd have DY divided by secant Y. And that's really just cosine YDY equaling the integral of two XDX. And we know that that's really X ^2 plus some constant. The integral of cosine YDY is going to be sine Y equaling X ^2 + C. If we wanted to solve for Y, we'd get Y equals sine inverse of X ^2 + C. Now doing that sine inverse means that we have some restrictions and that's OK. So the X ^2 is going to be positive the plus C. To have the sine Y equaling this, we know that the sine inverse is always an angle, so this would have to be in between negative π halves and Pi halves. X ^2 + 1 tangent YY prime equaling X. Well, in this one, first thing to remember is that Y prime we can really rewrite as DYDX. So we'd have X ^2 + 1 tangent YDYDX equaling X. Same skills. We want to take all the YS to one side. Tangent YDY equal the integral X / X ^2 + 1 DX. The integral of tangent YDY is going to be the natural log of secant Y. We're going to put an absolute value in there equaling. Now if we don't remember, just off to the aside. If we don't remember what the integral of tangent Y is, we could think of tangent as sine Y over cosine YDY and do AU sub. If we let U equal cosine Y, then DU is going to equal negative sine YDY. So we get the integral of negative DU over U which is negative natural log of the absolute value of U, negative natural log of cosine Y. And then if we take that negative up into the exponent, we'd get the natural log of cosine to the -1. But cosine to the -1 is just secant. So over here on the right hand side we might also do AU sub. If I have the integral of X / X ^2 + 1 DX. If you let U equal X ^2 + 1, DU would be two XDX or 1/2 DU equal XDX. So here's my XDX gets replaced by 1/2 DU. The X ^2 + 1 is U, so we get 1/2 natural log of the absolute value of U, or 1/2 natural log of X ^2 + 1. As you do more and more of these, you'll find that you don't need to do EU sub as much. You'll start looking at the denominator and thinking is the denominator's derivative somewhere up in the numerator. So over here, this right hand side is going to turn into 1/2 the natural log of X ^2 + 1 plus some C Let's call that C1 for a minute. Now in order to combine these, we can see a natural log here and a natural log here. We're going to actually let C1 turn into something that might look like natural log of some other constant, let's call it C2. So when we look at this, we could have the natural log of secant Y equaling 1/2, the natural log of X ^2 + 1 plus the natural log of C2. Remember, variables are just variables, and so this constant can be replaced by any constant. Now we can see we have the natural log of secant Y and that's going to equal the natural log. We can bring that 1/2 up X ^2 + 1 to the one half power. Now a natural log plus a natural log, we can rewrite it as multiplication so times C2 of that whole thing. Now we could actually drop the natural logs, so we would get secant Y equaling C two X ^2 + 1 to the one half. I can drop the square or I can drop the absolute values because the square root forces it to be positive on the inside. And we're going to actually be able to drop the absolute values because we're going to make an understanding that the C could be positive or negative. So that constant, if it was a positive, we'd be OK, and if it was a negative we'd have it be the opposite in order to be able to have the absolute value. So we'd get secant Y equaling C2 square root of X ^2 + 1. And if we wanted to solve for Y, we'd get Y equals secant inverse of some constant to square root X ^2 + 1. Sometimes they the book will use C all the way through and understand that you realize we're just changing constants, but they're just constants all the way throughout here. 2Y DYDX equaling X / sqrt X ^2 -, 16 Y of five equaling 2. Now recall that all of the solutions we've done so far have been general solutions because we weren't given some initial condition. So this one's going to be the first time we actually find a singular solution. So when we look at this, we're going to get all the YS to one side and everything that doesn't have AY or everything with the XS to the other side. The left hand side's pretty easy. It's just going to turn into y ^2. The right hand side you may or may not see. If we thought about just looking at this inside piece, the inside pieces derivative would be two X. I'm going to go ahead and do AU sub off to the side just in case you're not seeing it. If we have U equaling X ^2 -, 16, then DU would equal 2X DX, or 1/2 DU would equal XDX. So when we look back here, we would have the XDX right here being replaced by 1/2 DU and we'd end up with 1 / sqrt U. So when we evaluate that, we'd get 1/2 U to the 1/2, and then we would divide by 1/2, which is the same thing as multiplying by the reciprocal plus C. So we'd get y ^2 equaling U to the 1/2 or the square root of U. But U is really X ^2 - 16 + C Now when we have this, we're going to go back and put in the initial condition so we can actually find the value of C So we're going to get 2 ^2 equaling sqrt 5 ^2 - 16 + C So 4 equal 25 - 16, which would be 9 + C Sqrt 9 is 3. So we can see that C is going to be 1. So our final solution here is going to be y ^2 equaling sqrt X ^2 - 16 + 1. Now this can be thought of as an implicit solution, or we could go ahead and solve for Y equalling positive or negative sqrt sqrt X ^2 - 16 + 1. Thank you and have a wonderful day.