 Welcome back to our lecture series math 1220 calculus 2 for students at Southern Utah University as usual I'm your professor today. Dr. Andrew Misalign This is the first video For lecture 26, which is continuing on in our unit about differential equations. We had saw on lecture 25 The definition differential equation. What is a differential equation? What's a solution differential equation? A lot of the mechanics that go involved get involved with that In section 9.3 From jane stewart's calculus textbook We're going to introduce the notion of a separable differential equation and talk about this technique of separation variables to solve differential equations And in fact, we saw in the previous video that well, we will see that the examples we saw in the previous video Videos we actually were using this technique of separation variable without knowing it So we say that a differential equation is of the form Uh the differential equation that's of the form dy over dx equals f of x over gy We say this is a separable differential equation and the reason we can do that is because if you cross multiply this thing You end up with the equation of the form g of y dy equals f of x dx And the reason why this is significant is because when you separate the variables The left hand side only depends on y and the right hand side only depends on x You can integrate the left hand side with respect to y And you can integrate the right hand side with only with x and that can help you solve your Uh differential equation and this technique I just illustrated to you is referred to as separation of variables And so let's see some examples on how one can use that So let's find the general solution for the differential equation d or y prime equals x square over y squared and we're going to find The particular solution when y of zero equals two So to begin with we're going to separate the variables and so I in this situation We have a separable differential equation I actually do prefer to write the y prime as dy over dx because if this is kind of helps inform me as I Separate my variables. So as you cross multiply take the y's together and take the x's together You're going to end up with the equation y squared dy equals x squared dx And so the equations have been separated the left hand side only consists of The variable y including the differential dy don't forget about that And the right hand side only consists of the variable x including the differential dx right there So we are able to separate the variables And so once you've separated the variables then integrate both sides The left hand sides are going to integrate with respect to y and the right hand sides You'll integrate with respect to x which shouldn't be so bad Bring it up over here. The left hand sides are going to get y y cubed Over three one third y cubed and on the right hand side, you're going to get x cubed over three plus a constant right here Now you don't need a plus c on both sides even though there's two integrals You only need a plus c on the right hand side because again, this is an arbitrary constant and so as you As you add an arbitrary constant on the left hand side and the right hand side You can combine them on the right. It wasn't make much of a difference. We've seen that before Now to solve for y I would times both sides of the equation by three So if you times the left hand side by three And this times the right hand side by three On the left you'll get y cubed equals on the right x cubed plus c right so what happens is times like three and everything disappears basically because three times a third We'll just become one and what happened over here is we took three c And we just replaced it with c instead because plus c is gelatinous cube here It's just an arbitrary constant multiplied by three is still just as arbitrary as it was before Now to find the general solution. We're going to take the cube root of both sides We get the y equals the cube root of x cubed plus a constant Now this right here is our general solution and be cautious with the right hand side Oftentimes we get tempted because we have an x cubed inside of a cube root. We can simplify that That's a no. No. There's no simplification. We can do right here Uh expander rules do not allow us to simplify the cube root in there and don't forget the plus c right here as well So this kind of illustrates that when we were integrating if we had forgotten the plus c We would end up with just the cube root of x cube in which case we'd simplify this to be x And although that is a solution y equals x is a solution to this differential equation Notice if you take the derivative of y you're going to get one and notice that x squared over y squared of y equals x you're going to get x squared over x squared versus one that is a solution But that's just a particular solution. We're looking for a different one. That is When the initial value is two So this gives us our general solution. It's important not to forget the plus c right there Using the initial value that we had above We're going to see that when y is two when x equals zero And so solving for this The right hand side will just become the cube root of c The cube root of c equals two if you cube both sides, you get c equals two cube That is c equals eight and that's going to give us our particular solution y equals the cube root of x cube plus eight and again, even though that c is itself a perfect cube We cannot just reduce the right hand side. This is as simple as we can get right now And this is the particular solution we were looking for So separation of variables works out really nicely In this example here, we solve this initial value problem I want to take a look at another one here y prime equals x squared over y So this equation has the form dy over dx equals x squared. Oh, sorry x squared times y Now at first you might think this is not a separable differential equation because I don't have a function of x divided by a function of y But if I have a product here, this can be turned into a fraction Which in case you get x squared over one over y And so in that situation you can separate the variables the idea is if you take this first the first equation divide both sides by y The y's will cancel over here and also divide both sides by dx I asked actually dx is in the bottom you want to just times both sides by dx And it'll cancel on the right leaving a dy You can just treat this just treat this like it's like it's an equation of any other kind You take this equation right here Divide both sides by y And then times both sides by dx We separate the variables and we end up with on the right dy over y Sorry, that's the left on the right. We're going to get x squared dx So we've separated the variables and once the variables are separated because the left hand side only consists of y including the dy The right hand side only consists of x including the dx Once you've separated the variables then integrate with respect to the variables on that side of the equation On the right hand side, we have to integrate the respect of y the function one over y That is going to give us the natural log of the absolute value of y. Don't forget the absolute value We need that on the right hand side. We're going to integrate x squared with respect to x We did that in the previous example. We get x cubed over three plus a constant And so now we're going to start to solve for y here. This one didn't give us an initial value So we're not going to try to solve the initial value problem. We just want a general solution here Now to get rid of the natural log on the left hand side notice We'll take its inverse function, which is the power of e Well, that is base e we're exponentiating here and so then Our function then becomes the absolute value of y equals e To the x cubed over three plus a constant. So this right here is the exponent of the e Now, how does one get rid of the absolute value? Well, the absolute value is not a one-to-one function So it doesn't have a genuine inverse function But the idea is since absolute value forgets the sign absolute value Forgets whether we're positive or negative we have to consider both cases We sort of take positive negative both sides to compensate So we have y equals plus or minus e to the x cubed over three plus a constant And although this is a general form. I want to kind of rewrite this into a simpler form Now one thing to notice here is when you have exponents like a to the n plus a to the m Sorry a to the n plus m. This is the same thing as a to the n times a to the n That is addition in the exponent means multiplication And so since we have a plus sign right here, I want to utilize that fact We can write this as plus or minus e to the c times e to the x cubed Over three like so And so what can we do here now notice the expression? Let's first start with the expression e to the c e to the c right here now e to the c is going to be some Positive number any power of e you take will always be positive in fact you can get any positive number you want Um e to the c will be greater than zero, but it's you can get all get arbitrary large So since c is unspecified e to the c represents an arbitrary positive number All right Now if you take an arbitrary positive number and you times it by a positive that'll still be a positive But if you get any positive number and you times it by a negative that actually gives you any negative number So this so this e to the c represents any positive number here Right this plus or minus e to the c this represents any positive or negative number Like so what about zero well if we come back to the original equation It kind of got all Covered up all this stuff right here. Remember y prime is supposed to equal x squared y Now if you take y to equal zero then y prime equals zero the left hand side would equal zero And the right hand side would equal zero as well x squared times zero And so it turns out y equals zero is perfectly a kosher answer to this example to this equation right here Now notice we came over here and we were dividing by y Well, you can't divide by y of y is equal to zero So by dividing by y we're assuming y is not zero and thus ignoring the fact that y could be zero So when we assume y is not zero we get that this coefficient Plus or minus e to the c would represent any real number except for zero um But if you stick a zero in front of this expression right here, you're going to get zero which gives us the correct y value so what i'm trying to say here is that We can actually write our answer as y equals c e to the x cubed over three and this right here represents the general solution to this differential equation Because if c is equal to zero we get this case We were talking about above here Or if c is any positive a real number the plus or minus e to the c or it took care of that And so our gelatinous blob is added again plus or minus e to the c is just an arbitrary Real number right here. And so our general solution turns out to be y equals c e to the x cubed over three There's a little bit of understanding there. I can understand if you scratch your head a little bit here But this situation just show up enough that it's worth mentioning That um, if you get the natural log of y equals something plus c Then you're actually going to solve for y and get y equals c times that something right? And so this gives some examples of solving separable differential equations using this technique of separation of variables