 This lesson is on separable differential equations of the first order. There are other orders, second order, etc. Our first order is using the first derivative for a separation. And when we do this, what we are trying to do is look at a dy part and a dx part and separating everything that is in x's with the dx and everything that is in y's with a dy. Example one. Solve the differential equation dy dx is equal to xy. Well now we have to separate the variables. We'll put the dy with the y and the dx with the x. So this now becomes dy over y is equal to x dx. And when we do this, we integrate it. The dy over y is equal to the integral of x dx. And we get ln the absolute value of y plus some constant. And I'll call that c1. If we take the antiderivative of x, we know that it's x squared over 2 plus another constant. But what we do when we're working with differential equations is to take both constants and put them into one and we normally just put it on the right hand side. So we have ln of the absolute value of y is equal to x squared over 2 plus some constant. This needs to be finished so that we can solve for y. And if you remember, e to the ln of something is equal to that something. So in this case, raise this to the e to the ln of y and raise this e to the x squared over 2 plus c. So now we will have y is equal to e to the c times e to the x squared over 2. This is how we usually end up our natural log. So let's go on. Suppose we are asked, in our example, for y of 0 is equal to 2. And we have to find a particular solution, which means we will be evaluating our c. So in our final answer, we said that y was equal to e to the c times e to the x squared over 2. So if y of 0 is equal to 2, then 2 is equal to e to the c times, well, e to the 0 is a 1. So we see that e to the c is equal to 2. And in these examples, you will see that the e to the c will be substituted by that 2 at 0. So our final answer will become 2 e to the x squared over 2. Let's try another example. Solve the differential equation where y of 2 is equal to 3. And our differential equation is dy dx is equal to negative x squared over y. First of all, separate these. So we get y dy is equal to negative x squared dx. This is slightly different from the last one. Remember, the last one had dy over y, which led to an ln. This one doesn't. So as we take the integrals of each side, we have y squared over 2 is equal to negative x cubed over 3 plus c. Now we need to solve for c. So y of 2 is equal to 3. So we have 9 halves is equal to negative 8 thirds plus c. Putting this all together, we find that c is equal to 7 and 1 sixth. So we have y squared over 2 is equal to negative x cubed over 3 plus 7 and 1 sixth. In this particular problem, I did not solve for y. We can easily do that because of the fact we can just take the square root plus or minus. But this seems like a better form to leave it in. Our next example is solve the differential equation where b of 2 is equal to 100. And we have db dt plus a b is equal to 15. Our db and b are on the same side of the equation, but they can't be put together with plus sign or a minus sign. It really has to be multiplication or division. So what we need to do is put the b on the other side. So make db dt equal to 15 minus b. Then we can put it back on the other side with the db. But before I do that, I do not like a negative in the denominator. Remember this is going to be a udu form when we get it down into that denominator. So what I'm going to do is factor out a negative and have b minus 15. So now db dt is equal to negative b minus 15. This will make everybody's life very easy in comparison to trying to take the antiderivative of the 15 minus b. So now we will have db over b minus 15 equals negative dt. When we take the antiderivative of these, we get ln of the absolute value of b minus 15 is equal to negative t plus c. So we do want to solve for b in this one. Again, we raise everything to the e. And we get b minus 15 is equal to e to the c times e to the negative t. Or b is equal to 15 plus e to the c times e to the negative t. This has an initial condition where b of 2 is equal to 100. So now we will make a substitution in there. And we see if b is 100, then we have 100 equals 15 plus e to the c times e to the negative 2. And we see 85 equals e to the c times e to the negative 2. Now we have to just determine e to the c. So e to the c is equal to 85e squared. So we substitute that back into that equation. And we get b is equal to 15 plus 85e squared times e to the negative t. And if we want to make this a little bit better, we say b is equal to 15 plus 85e to the 2 minus t. A little complicated example, but yet not very difficult. I think you're wondering what happens when I take my absolute value off of the b minus 15. Well, what happens is if b is greater than 15, we're OK. If b is less than 15, then we're going to come out with a negative answer. And it usually works out as we work through these kinds of problems. Let's do another example. This one is a little bit more complicated than the ones we have been doing. It reads solve y squared times the square root of 1 minus t squared dy dt plus t is equal to 0. And we have the initial condition that y of 0 is equal to 1. On this particular one, we need to put the t on the other side. So we can have y squared times the square root of 1 minus t squared dy dt is equal to negative t. Now we can separate our variables and put the square root of 1 minus t squared and the dt on the other side. So now we will have y squared dy, that's simple enough, is equal to negative t over the square root of 1 minus t squared dt. This is more complicated. We see that we have a u du form. Our u is 1 minus t squared. Our du is negative 2t. So when we take those antiderivatives, the right-hand side, we will need to do u du and compensate with a 2 there and a 1 half on the outside. And our du, remember, is negative 2t dt. Always put that in. So the antiderivative of y squared is y cubed over 3. That equals 1 half. Now remember, we have du over the square root of u. So this becomes u to the 1 half times 2 plus our constant. And if we go on, we have y cubed over 3 is equal to the 2's will cross out. The u becomes 1 minus t squared to the 1 half plus c. So now we have found our antiderivative. Now we just have to use the initial condition. Given that y of 0 is equal to 1. If we substitute a 1 in, we have 1 third is equal to 1 plus c. Or c is equal to negative 2 thirds. So that makes our final answer, y cubed over 3 is equal to the square root of 1 minus t squared minus 2 thirds. Separable differential equations can take on different features. We can first have the type dy dx is equal to a function times another function. And that separates to dy over h of y equals g of x dx. So one function times another, 1 in x, 1 in y, simple separation. It can also take on a g of x plus a constant times the h of y. Again, we put the y's with the dy's. So this one would be dy over h of y. And the dx goes with the gx. So g of x plus c dx. Or it can take on a backwards one where dy over dx is equal to some function g of x times h of y plus c. Again, the h of y plus c has to go with a dy. So we have dy over h of y plus c is equal to g of x dx. Another way to look at these is as a fraction. dy dx is equal to g of x over h of y. And we can see that we can separate these to h of y dy is equal to g of x dx. And the other type we have dy dx is equal to h of y over g of x. And of course, these separate to dy over h of y is equal to dx over g of x. As long as you put the y's with the dy's and the x's with the dx's or the same variables on one side of the equation, then you should be able to separate them and take their antiderivatives. This concludes our lesson on separable differential equations.