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# Separable equations introduction

AP.CALC:
FUN‑7 (EU)
,
FUN‑7.D (LO)
,
FUN‑7.D.1 (EK)
,
FUN‑7.D.2 (EK)

## Video transcript

so now that we've spent some time thinking about what a differential equation is and even visualizing solutions to a differential equation using things like slope field let's start seeing if we can actually solve differential equations and as we'll see different types of differential equations might require different techniques and solve them we might not be able to solve at all using analytic techniques we'd have to resort to numeric techniques to to estimate the solutions but let's go to what I would argue is the the simplest form of differential equation to solve and that's what's called a separable separable differential equation and we will see in a second why it is called a separable differential equation so let's say that we have the derivative of Y with respect to X is equal to negative x over Y e to the x squared so we have this differential equation and we want to find the particular solution that goes through the point that goes to the point 0 comma 1 and I encourage you to pause this video now I'll give you a hint if you can on one side of this equation through algebra separate out the Y's and the D Y so on the other side have all the X's and DX's and then integrate perhaps you can find the particular solution to this differential equation that contains this point now if you can't do it don't worry because we're about to work through it so as I said let's use a little bit of algebra to get all the Y's and dys on one side and all the XS and DX's on the other side so one way to if I want let's say I want to get all the Y's and dys on the left-hand side and all the XS and DX's on the right-hand side well I can multiply both sides times y so I can multiply both sides times y that has the effect of putting the Y's on the left-hand side and then I can multiply both sides times DX I can multiply both sides times DX and we kind of treat you can treat these differentials as you would treat a variable when you're manipulating it to essentially separate out the variables and so this will cancel with that and so we are left with we are left with y dy Y dy is equal to negative x and actually let me write it this way let me write it as negative x e actually I might a little more space so negative X e to the negative x squared DX D X now why is this interesting because we could integrate both sides and now this also highlights why we call this separable you won't be able to do this with every differential equation you won't be able to algebraically separate the Y's and dys on one side and the x's and DX is on the other side but this one we were able to and so that's why this is called a separable differential equation differential differential equation and it's usually the first technique that you should try hey can I separate the Y's and the X's and I said this is not going to be true of many if not most differential equations but now that we did this we can integrate both sides so let's do that so I'll find a nice color to integrate with so I'm going to integrate integrate both sides now if you integrate the left-hand side what do you get you get and remember we're integrating with respect to Y here so this is going to be Y squared over 2 and we could put some constant there I could call that plus c1 and if you're integrating now that's going to be equal to now the right hand side we're integrating with respect to X and let's see you could do you substitution or you could recognize that look the derivative of negative x squared is going to be negative 2x so if that was a 2 there and if you don't want to change the value of the integral you put the 1/2 right over there and so now you could you could either do u substitution explicitly or you could do it in your head or you said u is equal to U is equal to negative x squared and then D U will be negative 2 X DX or you can kind of do this in your head at this point so I have I have some and it's derivative so I really could just integrate with respect to that something to with respect to that you so this is going to be this is going to be 1/2 this 1/2 right over here the antiderivative this is e to the negative x squared and then of course I might have some other constant I'll just call that c2 and once again if this part over here what I just did seemed strange the u-substitution you might want to review that piece now what can I do here we'll have a constant on the left hand side it's an arbitrary constant we don't know what it is I haven't used this initial condition yet we could call it so let me just subtract C 1 from both sides so if I just subtract C 1 from both sides I have an arbitrary so this is going to cancel and I have C 2 sorry let me so this is C 1 so these are going to cancel and C 2 minus C 1 these are both constants arbitrary constants we don't know what they are yet and so we could just rewrite this as on the left hand side we have Y squared over 2 is equal to on the right hand side all right 1/2 e let me write that in blue just because I wrote it in blue before 1/2 1/2 e to the negative x squared and I'll just say C C 2 minus C 1 let's just call that C so if you take the sum of those two things let's just call that C and so now this is kind of a general solution we don't know what this constant is and we haven't explicitly solved for Y yet but even in this form we can now find a particular solution using this initial condition let me separate it out this wasn't part this wasn't part of this original expression right over here but using this initial condition so it tells us when x is 0 Y needs to be equal to 1 so we would have 1 squared which is just 1 over 2 is equal to 1/2 e to the negative 0 squared well that's just going to be e to the 0 is just 1 so it's going to be 1/2 plus C and just like that we're able to figure out if you subtract one half from both sides see is equal to zero so the relationship between y and X that goes through this point we could just set C is equal to zero so that's equal to zero let's see oh right over there and so we are left with y squared over two is equal to e to the negative x squared over two now we can multiply both sides by two and we're going to get Y squared Y squared let me do that so we're going to get Y squared is equal to is equal to e to the negative x squared now we can take the square root of both sides and you could say well look you know Y squared is equal to this so why Y could be equal to the plus or minus square root of e to the negative x squared of e to the negative x squared but they gave us an initial condition where Y is actually positive so we're finding the particular solution that goes through this point that means Y is going to be the positive square root if this was the point 0 negative 1 then we would say Y is the negative square root but we know that Y is the positive square root it's the principal root right over there so let me do that a little bit neater so we can get rid of whoops that was writing in black so we can get rid of this right over here we're only going to be dealing with the positive square root or we could so we could write Y is equal to e to the negative x squared to the one-half power and that of course is equal to is equal to e to the negative x squared over 2 so this right over here is or y equals y equals e to the negative x squared over 2 is a particular solution that satisfies the initial conditions to this original differential equation so just like that because we were able to just as a review because this differential equation was set up in a way or because we could algebraically separate the Y's dys from the X DX's we're able to just separate them out algebraically integrate both sides and use the information given in the initial condition to find the particular solution