Learn about Separable Differential Equations Video

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Learn about Separable Differential Equations

We hopefully at this point what a differential equation is, so now let’s try to solve some. And this first class of differential equation, I'll introduce to. There called separable equations and I think were deal, you’ll this that were not learning really anything new using just your first year calculus, derivative and integrating skills. You can solve a separable equation and the reason why they 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 lets separable, separable equations. So lets 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/DX=X²/1-Y² 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 and it is just the first derivative. So the order is equal to one, so its order, first order. It’s ordinary because we only have a regular derivative, no partial derivatives here and then it is linear or non-linear. Well at first you say, oh well you know this looks linear, I'm not multiplying the derivative times anything else but if you look carefully something interesting is going on. First of all, you have an Y² and Y is the dependent variable. Y is a function of X, so they have the Y² that makes it non-linear and even if this was a Y, if you were to actually multiply both sides of this equation times 1-Y and get it in the form that actually in the previous equation you would have 1-Y². Actually this is actually the first step of what we have to do anyway, so I'll write it down. 1-, so if I'm just multiplying both sides of this equation times 1-Y². You get 1-Y²×DYDX=x² and then you immediately see that you’re actually even is this wasn’t the squared here. You’ll be multiplying the Y×DYDX and that also makes it non-linear because you’re multiplying the dependent variable times the derivative of itself. So that also makes this a non-linear equation. But anyway, let’s back to solving this, so this was the first step. I just multiply the both sides by 1-y² and the real end goal is just to separate the Y’s and X’s and then integrates 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. I'm going to go here. I don’t want to waste too much space. So you get 1-y²DY=x²DX I have separated the X and Y variable and the differentials, right. All I did is I multiply both sides of this equation times DX to get here, right. Now I can just integrate both sides, so lets do that. Whatever you do to one side of equation, you have to do the other that’s true it. Regular equations or differentials we’ll integrate both sides. So what’s the integral of this expression with respect to Y? Let’s see, the integral of one is Y. The integral of y², well that’s -y³/3 and well I'll write the plus C here just too kind of show you something but you really don’t have to write plus C on both sides. I'll call the plus the constant to do the Y. The Y integration, you’ll never see this in a calculus class but I just wan to make a point here, is equal to. I just want to show you these are plus C’s never disappear from where were taking our traditional anti-derivatives. And what’s the derivative of this? Well that’s X³/3 and this also going to have plus C, a plus C due the X variable. Now the reason why I drive this magenta one and magenta, and I'll like labeled it like that because you only just to have write a +C on one side of the equation and if that doesn’t make a lot of sense,. Let’s subtract to this C from both sides and we get Y- of me, just scroll down a little bit. Y, my Y looks like a G, Y-y³/3=x³/3 plus the constant when we took the anti-derivative of the X minus the constant of the anti-derivative. We took the Y but these two constants there just can’t do. I mean we don’t know what they are. They are betrayed 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 there are obituary, CX-CY, well that still just another constant. And then if we wanted to simplify this equation more, we can multiply both sides of this by three just to make it look nicer and you get 3y-y³=x³+, well I could write 3C here but once again C is an obituary constant right. So 3x an obituary constant than just another obituary constant, so I'll write the C there and there you have it. We have solved this differential equation although it is in implicit form right now that’s fairly hard to get it out of implicit form. We could put C on one side, so you could, the solution could be 3y-y³-x³=C. Some people might like that a little bit better but that’s the solution. Anyway, the solution just like when you take anti-derivative. 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 would be another solution but any constant there will satisfy the original differential equation which was up here. This was the original differential equation and if you want to solve for that constant. Someone has to give you the initial condition. Someone has to say, well when X is you know, when is 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 is a little bit, I don’t want to, I'll start over. See clear image, invert colors, so I have optimal space. So this one is the first derivative of Y with respect to X is equal to 3x²+4x+2/2×y-1. This is a parenthesis not an absolute value and they give us initial conditions. They say that Y(0)=-1. So once we solve this differential equation and this is a separable differential equation. Then we can use this initial condition. When X is zero, Y is one to figure out the constant. So let’s first separate this equation. So lets multiply both sides by 2xy-1 and you get 2×y-1×DYDX=3x²+4x+2, multiply both sides times DX. This is really does an exercise in algebra. You get and I can multiply this one out too. You get 2y-2 that’s just this, DY. I multiply both sides times DX, so that equals 3x²+4x+2DX. I have separated the equations and I have separated the independent from the dependent variable and their relative differentials. And so now I can integrate and I can integrate in magenta. What’s the anti-derivative of this expression with respect to Y? Well let just see. Its y²-2y, I won’t write the +C. I'll just do it on the right hand side that is equal to 3x². Well the anti-derivative is x³+, enter this is 2x²+2x+C and that C kind of takes care of the constant for both sides of the equation and hopefully you understand Y from the last example. But we can solve for C using the initial condition, Y(0)=-1. So lets see, when X is zero, Y is -1. So let’s put Y’s and -1, so we get -1²-2×-1. Right that’s the value of Y=, when X=0. So when X=0 that’s the 0³+2×0²+2×0+C. So this is fairly straightforward. All of these, these are all zero. This is; let see -1² that’s 1-2×-1 that’s +2=C and we get C=3. So the implicit exact solution. The solution of our differential equation, remember now it’s not a class because where they gave us an initial condition is y²-2y=x³+2x²+2x+3. We figured out that’s what C was. And actually if you want you could write this in implicit 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 one to both sides. I'm just completing the square here, so y²-2y+1, if I add one to that side. I have to add one to these sides, so here comes x³+2x²+2x+4. I just added one to both sides of this 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-1²=x³+2x²+2x+4 then I could say y-1= the plus or minus square root of x³+2x²+2x+4. Now I can add one to both sides and then I get y=1+ or minus the square root of x³+2x²+2x+4. 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, right or our initial condition told us that Y(0)=-1. So if we put zero for X, we get y=1+ or minus 0+4. So 1+ or minus 4, so if Y=-1. So we get y=1+ or minus, sorry 2, if this is going to be equal to -1. Then this have to be 1-2, so the explicit form that satisfies our initial condition and were getting a little geeky here. You can get rid of the plus, it 1- this whole thing that’s what satisfy our initial condition. And you could figure out where it satisfied because in order for this over what domain is it satisfied. Well that’s satisfied when this term is positive. This is becomes negative and you get it’s undefined in real and all of that. But anyway I’ve ran out of time, see you in the next video.

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We hopefully at this point what a differential equation is, so now let’s try to solve some. And this first class of differential equation, I'll introduce to. There called separable equations and I think were deal, you’ll this that were not learning really anything new using just your first year calculus, derivative and integrating skills. You can solve a separable equation and the reason why they 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 lets separable, separable equations. So lets 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/DX=X²/1-Y² 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 and it is just the first derivative. So the order is equal to one, so its order, first order. It’s ordinary because we only have a regular derivative, no partial derivatives here and then it is linear or non-linear. Well at first you say, oh well you know this looks linear, I'm not multiplying the derivative times anything else but if you look carefully something interesting is going on.

First of all, you have an Y² and Y is the dependent variable. Y is a function of X, so they have the Y² that makes it non-linear and even if this was a Y, if you were to actually multiply both sides of this equation times 1-Y and get it in the form that actually in the previous equation you would have 1-Y². Actually this is actually the first step of what we have to do anyway, so I'll write it down. 1-, so if I'm just multiplying both sides of this equation times 1-Y². You get 1-Y²×DYDX=x² and then you immediately see that you’re actually even is this wasn’t the squared here. You’ll be multiplying the Y×DYDX and that also makes it non-linear because you’re multiplying the dependent variable times the derivative of itself. So that also makes this a non-linear equation.

But anyway, let’s back to solving this, so this was the first step. I just multiply the both sides by 1-y² and the real end goal is just to separate the Y’s and X’s and then integrates 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. I'm going to go here. I don’t want to waste too much space. So you get 1-y²DY=x²DX I have separated the X and Y variable and the differentials, right. All I did is I multiply both sides of this equation times DX to get here, right.

Now I can just integrate both sides, so lets do that. Whatever you do to one side of equation, you have to do the other that’s true it. Regular equations or differentials we’ll integrate both sides. So what’s the integral of this expression with respect to Y? Let’s see, the integral of one is Y. The integral of y², well that’s -y³/3 and well I'll write the plus C here just too kind of show you something but you really don’t have to write plus C on both sides. I'll call the plus the constant to do the Y. The Y integration, you’ll never see this in a calculus class but I just wan to make a point here, is equal to. I just want to show you these are plus C’s never disappear from where were taking our traditional anti-derivatives. And what’s the derivative of this?

Well that’s X³/3 and this also going to have plus C, a plus C due the X variable. Now the reason why I drive this magenta one and magenta, and I'll like labeled it like that because you only just to have write a +C on one side of the equation and if that doesn’t make a lot of sense,. Let’s subtract to this C from both sides and we get Y- of me, just scroll down a little bit. Y, my Y looks like a G, Y-y³/3=x³/3 plus the constant when we took the anti-derivative of the X minus the constant of the anti-derivative. We took the Y but these two constants there just can’t do. I mean we don’t know what they are. They are betrayed 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 there are obituary, CX-CY, well that still just another constant. And then if we wanted to simplify this equation more, we can multiply both sides of this by three just to make it look nicer and you get 3y-y³=x³+, well I could write 3C here but once again C is an obituary constant right. So 3x an obituary constant than just another obituary constant, so I'll write the C there and there you have it. We have solved this differential equation although it is in implicit form right now that’s fairly hard to get it out of implicit form.

We could put C on one side, so you could, the solution could be 3y-y³-x³=C. Some people might like that a little bit better but that’s the solution. Anyway, the solution just like when you take anti-derivative. 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 would be another solution but any constant there will satisfy the original differential equation which was up here. This was the original differential equation and if you want to solve for that constant. Someone has to give you the initial condition. Someone has to say, well when X is you know, when is 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 is a little bit, I don’t want to, I'll start over. See clear image, invert colors, so I have optimal space. So this one is the first derivative of Y with respect to X is equal to 3x²+4x+2/2×y-1. This is a parenthesis not an absolute value and they give us initial conditions. They say that Y(0)=-1. So once we solve this differential equation and this is a separable differential equation. Then we can use this initial condition. When X is zero, Y is one to figure out the constant. So let’s first separate this equation. So lets multiply both sides by 2xy-1 and you get 2×y-1×DYDX=3x²+4x+2, multiply both sides times DX.

This is really does an exercise in algebra. You get and I can multiply this one out too. You get 2y-2 that’s just this, DY. I multiply both sides times DX, so that equals 3x²+4x+2DX. I have separated the equations and I have separated the independent from the dependent variable and their relative differentials. And so now I can integrate and I can integrate in magenta. What’s the anti-derivative of this expression with respect to Y? Well let just see. Its y²-2y, I won’t write the +C. I'll just do it on the right hand side that is equal to 3x². Well the anti-derivative is x³+, enter this is 2x²+2x+C and that C kind of takes care of the constant for both sides of the equation and hopefully you understand Y from the last example.

But we can solve for C using the initial condition, Y(0)=-1. So lets see, when X is zero, Y is -1. So let’s put Y’s and -1, so we get -1²-2×-1. Right that’s the value of Y=, when X=0. So when X=0 that’s the 0³+2×0²+2×0+C. So this is fairly straightforward. All of these, these are all zero. This is; let see -1² that’s 1-2×-1 that’s +2=C and we get C=3. So the implicit exact solution. The solution of our differential equation, remember now it’s not a class because where they gave us an initial condition is y²-2y=x³+2x²+2x+3. We figured out that’s what C was.

And actually if you want you could write this in implicit 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 one to both sides. I'm just completing the square here, so y²-2y+1, if I add one to that side. I have to add one to these sides, so here comes x³+2x²+2x+4. I just added one to both sides of this 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-1²=x³+2x²+2x+4 then I could say y-1= the plus or minus square root of x³+2x²+2x+4.

Now I can add one to both sides and then I get y=1+ or minus the square root of x³+2x²+2x+4. 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, right or our initial condition told us that Y(0)=-1. So if we put zero for X, we get y=1+ or minus 0+4. So 1+ or minus 4, so if Y=-1. So we get y=1+ or minus, sorry 2, if this is going to be equal to -1. Then this have to be 1-2, so the explicit form that satisfies our initial condition and were getting a little geeky here. You can get rid of the plus, it 1- this whole thing that’s what satisfy our initial condition. And you could figure out where it satisfied because in order for this over what domain is it satisfied. Well that’s satisfied when this term is positive. This is becomes negative and you get it’s undefined in real and all of that.

But anyway I’ve ran out of time, see you in the next video.

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