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# Separable equations (old)

## Video transcript

we hopefully know at this point what a differential equation is so now let's try to solve some and this first class of differential equations I'll introduce you to they're called separable equations and I think what you'll find is that we're not learning really anything you using just your your first year calculus derivative and integrating skills you can solve a separable equation and the reason why they're called separable is because you can actually separate the x and y terms and integrate them separately to get the solution of the differential equation so let's separable separable equations so let's do a couple and I think you'll get the point these often are really more of exercises in algebra than anything else so the first separable differential equation is dy over DX is equal to x squared over one minus y squared and actually this is a good time to just review our terminology so first of all what is the order of this differential equation well the highest derivative in it is just the first derivative so the order is equal to 1 so it's order first order it's Ordinary because we only have a regular derivative no partial derivatives here and then is this linear or nonlinear well at first you say oh well you know this looks linear I'm not multiplying the derivative times anything else but if you if you look carefully something interesting is going on first of all you have a Y squared and Y is the dependent variable Y is a function of X so to have the Y squared that makes it nonlinear and even if this was a Y if you were to actually multiply both sides of this equation times 1 minus y and get it in the form that I showed you in the previous equation you would have 1 minus y squared actually this is actually the first step of what we have to do anyway so I'll write it down 1 minus so if I'm just multiplying both sides equation times 1 minus y squared you get 1 minus y squared times dy/dx is equal to x squared and then you immediately see that you're actually even if this wasn't a squared here you'd be multiplying the Y times dy DX and that also makes it nonlinear because you're multiplying the dependent variable times it Vivat of of itself so that also makes this a nonlinear equation but anyway let's get back to solving this so this was the first step I'm I just multiplied the both sides by 1 minus y squared and the real end goal is just to separate the Y's and the X's and then integrate both sides so I'm almost there so now what I want to do is I want to multiply both sides of this equation times DX so I have a DX here and get rid of this DX there we go here and I want to waste too much space so you get 1 minus y squared dy is equal to x squared DX I have separated the x and y variables and the differentials right all I did is I multiplied both sides of this equation times DX to get here all right now I can just integrate both sides so let's do that so whatever you do to one side of the equation you have to do the other that's true with regular equations or differential equations so we're going to integrate both sides so what's the integral of this expression with respect to Y let's see the integral of 1 is y the integral of y-squared well that's minus y to the third over 3 and well I'll write the plus C here just just to kind of show you something but you really don't have to write a plus C on both sides I'll call that plus the constant due to Y the Y integration you'll never see this in a calculus class but I just want to make a point here is equal to I just want to show you this our plus C's never disappeared from when we were taking our traditional anti derivatives and what's the derivative of this well that's it's X to the third over 3 X to the third over 3 this is also going to have a plus C a plus C due to the X variable now the reason why I did this magenta one in magenta and I like labeled it like that is because you really just have to write a plus C on one side of the equation and if that doesn't make a lot of sense let's subtract this C from both sides and we get Y minus let me scroll down a little bit why my Y looks like a G y minus y to the 3rd over three is equal to X to the third over 3 plus the constant when we took the antiderivative of the X minus the constant of the antiderivative we took the Y but these two constants they're just common I mean we don't know what they are they're arbitrary constants so we could just write a general C here so you could have just you have to have a constant but it doesn't have to be on both sides of this equation because they're arbitrary C X minus C Y well that's still just another constant and then if we want to simplify this equation more we can multiply both sides of this by 3 just to make it look nicer and you get 3y minus y to the third is equal to X to the third plus well I could write 3c here but once again C is an arbitrary constant right so 3 times an arbitrary constant that's just another arbitrary constant so I'll write the C there and there you have it we have solved this differential equation although it is in it's in implicit form right now and it's fairly hard to get it out of implicit form we could put the C on one side so you could the solution could be 3y minus y to the third minus X to the third is equal to C some people might like that a little bit better but that's the solution and notice the solution just like when you take an antiderivative the solution is a class of implicit functions in this case and why is it a class because we have that constant there depending on what number you pick there it will be another solution but any constant there will satisfy the original differential equation which was up here this was your original differential equation and if you want to solve for that constant someone has to give you an initial condition someone has to say well when X is you know when X is 2 y is 3 and then you could solve for C anyway let's do another one that gives us an initial condition so this one's a little bit I don't want to I'll start over clear image invert so I have optimal space so this one is the first derivative of Y with respect to X is equal to 3x squared plus 4x plus 2 over 2 times y minus 1 this is a parenthesis not an absolute value and they give us initial conditions they say that Y of 0 is equal to negative 1 so once we solve this differential equations and this is a separable differential equation then we could use this initial condition when x is 0 Y is 1 to figure out the constant so let's first separate this equation so let's multiply both sides by 2 times y minus 1 and you get 2 times y minus 1 times dy/dx is equal to 3x squared plus 4x plus 2 multiply both sides times DX this is really just an exercise in algebra you get and I could multiply this one out to you get 2y minus 2 that's just this dy I multiply both sides times DX so that equals 3x squared plus 4x plus 2 DX I have separated the equations I've separated the the independent from the dependent variable and the relative differentials and so now I can integrate and I can integrate in magenta I can integrate what's the antiderivative of this expression with respect to Y well this is C it's y squared minus 2y I want to write the plus C I'll just do it on the right hand side that is equal to 3x squared well the antiderivative enter is 2x squared plus 2x plus C and that C kind of takes care of the constant for both sides of the equation and hopefully you understand why from the last example but we can solve for C using the initial condition y of 0 is equal to negative 1 so let's see when X is zero Y is negative one so let's put Y is negative one so we get negative one squared minus two times negative one right that's the value of y is equal to it is equal to when X is equal to zero so when X is equal to zero that's zero to the third plus two times zero squared plus two times zero plus C so this is fairly straightforward all of these this is this is all zero this is a C negative one squared that's 1 minus 2 times minus 1 that's plus 2 is equal to C and we get C is equal to 3 so the implicit exact solution the solution of our differential equation remember now it's not a class because we got it and we got we they gave us an initial condition is y squared minus 2y is equal to X to the third plus 2x squared plus 2x plus 3 we figured out that's what C was and actually if you want you could write this in an explicit form by completing the square this is just algebra at this point you're done this is an implicit form if you wanted to make it explicit you could add 1 to both sides I'm just completing the square here so Y squared minus 2y plus 1 if I add one of that side have to add one to this side so becomes X to the third plus 2x squared plus 2x plus 4 I just added one to both sides equation why did I do that because I wanted this side to be a perfect square in terms of Y then I could write rewrite this side as Y minus 1 squared is equal to X to the third plus 2x squared plus 2x plus 4 then I could say Y minus 1 is equal to the plus or minus square root of x to the third plus 2x squared plus 2x plus 4 I can add 1 to both sides and I get Y is equal to 1 plus or minus the square root of x to the third plus 2x squared plus 2x plus four and it has plus or minus here and if we have to pick one of the two we would go back to the initial condition the initial condition right well our initial condition our initial condition is told us that Y of zero is equal to negative one so if we put 0 here for X we get Y is equal to 1 plus or minus 0 plus 4 so 1 plus or minus 4 so if Y is going to be equal to negative 1 so we get Y is equal to 1 plus or minus sorry - if this is going to be equal to negative 1 then we have then this has to be 1 minus 2 so the core the explicit form that satisfies our initial condition we're getting a little geeky here you could get rid of the plus it's 1 minus this whole thing that's what satisfies our initial condition and you could figure out where it's satisfied because in order for this what what over what domain is it satisfied well that's satisfied when this term is positive is this becomes negative and you get its undefined in reals and all of that but anyway I've run out of time see you in the next video