- Separable equations introduction
- Addressing treating differentials algebraically
- Worked example: identifying separable equations
- Worked example: finding a specific solution to a separable equation
- Worked example: separable equation with an implicit solution
- Particular solutions to separable differential equations
- Separable equations (old)
- Separable equations example (old)
Separable equations (old)
An old introduction video to separable differential equations. Created by Sal Khan.
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- At8:15, when Sal throws the constant of integration in on the right with the 'x' terms, could he have just as well put it on the left and come up with -3 as an answer so C would be +/- 3 instead of just 3?(7 votes)
- If someone puts the C on the RHS, like Sal did, then the answer will be C=3, and this will be correct for the way that person wrote it, i.e., with C on the RHS. Writing C=-3 would be incorrect in this case.
If you put the C on the LHS, then the answer will indeed be C=-3, and this will be correct for the way you wrote it, i.e., with the C on the LHS. Writing C=3 would be incorrect in this case.
So technically, you can't write C=+/-3, because 3 would only be correct if you introduced C on the RHS, and -3 would only be correct if you had introduced C on the LHS.(14 votes)
- What does it actually mean to 'multiply both sides by dx'? It makes sense algebraically, but in the first place, dx is not a number.(5 votes)
- Does whether an equation is linear impact the method of solution to that equation, or are linear and nonlinear simply ways to classify types of differential equations?(4 votes)
- It matters. You would use a different method to solve a nonlinear than you would to solve a linear one. If you watch more of the videos you will see why! Hope this helps!(2 votes)
- Why can't I play this video?(3 votes)
- Skip to the first second. It worked for me!(4 votes)
- when finding the constant we are given that y(0)=-1. Why does Sal then plug in -1 for all the y's and 0 for the x's? shouldn't it be the other way around?(1 vote)
- y(0) = -1 is written in f(x) notation. x is independent variable and y is dependent variable. In this case 0 is x and y is -1 and the variables are substituted with their respective values.(11 votes)
- PLEASE HELP! how do you solve dy/dx=sqrt(y/x)(2 votes)
- dy/dx = sqrt(y)/sqrt(x) => 1/sqrt(y) dy = 1/sqrt(x) dx => 2 * y^(0.5) = 2 *x^(0.5) + c=> y = x + 2 * x ^0.5*C + C(4 votes)
- I need a little help with a physical DGL describing a capacitor:
R*Q'(t) + Q(t)/C = 0, where Q(t)=C*U(t), C is capacity, U is voltage and R resistance
I can write this as RC*U'(t) = - U(t), which then leads to U'(t)/ U(t) = -1/RC. Then I integrate on both sides, which gives, according to my books, log(absolute value(U(t))) = -t/RC + constant.
My question is, how do you get the logarithm in there? Is integral[ f'(t)/f(t) ] = log [ f(t) ] and therefore the same as integral[ 1/f(t) ]?(2 votes)
- Yeah I figured it out by myself, haven't logged into Khan Academy since, but thank you anyway :)(1 vote)
- what is the convention that groups the 2 to the (y-1)? In other words, why is it incorrect to simply multiply both sides by (y-1) instead of 2(y-1)?(2 votes)
- Both ways are correct, you can leave the 2 in the denominator of the right hand side and integrate it with the rest of the x dependant parts. The result will be the same regardless of where that 2 ends up at the moment of separation.
I guess Sal decided to take the 2 with the (y-1) to avoid having a denominator in the right hand side.(2 votes)
- what is implicit form of a differential equation?(2 votes)
- While solving the differential equation you used the relation (x/y)^1/2 = (x^1/2)/(y^1/2) which is not true as when x is negative and y is negative the first expression is defined but the second is not?(2 votes)
- That's solved by quickly introducing and getting rid of imaginary numbers.
(-9)^(1/2) / (-16)^(1/2) =
i * 9^(1/2) / (i * 16^(1/2)) =
9^(1/2) / 16^(1/2) =
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 new. Using just 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 that'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 1 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 first order. It's ordinary, because we only have a regular derivative, no partial derivatives here. And then, is this linear or non-linear? Well, 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 a y squared. And y is a dependent variable. y is a function of x. So to have the y squared, 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 minus y, and get in the form that I showed you in the previous equation, you would have 1 minus y squared. This is actually the first step of what we have to do anyways, so I'll write it down. So if I'm just multiplying both sides of this 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 even if this wasn't a squared here, you'd be multiplying the y times dy dx, 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 get back to solving this. So this is the first step. I just multiply 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. Let me go here, I don't 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. All I did is I multiplied both sides of this equation times dx to get here. Now, I can just integrate both sides. So let's do that. 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 I'll write the plus c here just to kind of show you something, but you really don't have to write a plus c on both sides. I'll call the 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. I just want to show you that our plus c has never disappeared from when we were taking our traditional antiderivatives. And what's the derivative of this? Well that's x to the third over 3. And this is also going to have a plus c due to the x variable. Now, the reason why I did this magenta one in magenta and I labelled 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, my y looks like a g. y minus y to the third over 3 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 when we took the y. But these two constants, they're just constants. I mean, we don't know what they are. There are 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 the equation, because they're arbitrary. cx minus cy, 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 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. 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 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 the solution could be 3y minus y to the third minus x to the third is equal to c. Some people might like that 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 the 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 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'll start over. Clear image, different colors, 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 parentheses, 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 equation, and this is a separable differential equation, then we can 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. And I can multiply this one out, too, you get 2y minus 2, that's just this, dy. I multiplied both sides times dx, so that equals 3x squared plus 4x plus 2 dx. I have separated the equations. I've 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 antiderivative of this expression with respect to y? Well, let's just see. It's y squared minus 2y. I won't write the plus c, I'll just do it on the right hand side. That is equal to 3x squared. Well, the antiderivative is x to the third, plus 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 0, y is negative 1. So let's put y as negative 1, so we get negative 1 squared minus 2 times negative 1, that's the value of y, is equal to when x is equal to 0. So when x is equal to 0, that's 0 to the third plus 2 times 0 squared plus 2 times 0 plus c. So this is fairly straightforward. All of these, this is all 0. This is, let's see, negative 1 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 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 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 1 to that side, I have to add 1 to this side, so it becomes x to the third plus 2x squared plus 2x plus 4. I just added 1 to both sides of the equation. Why did I do that? Because I wanted this side to be a perfect square in terms of y. Then I can 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 then I can get y is equal to 1 plus or minus the square root of x to the third plus 2x squared plus 2x plus 4. And it has plus or minus here, and if we have to pick one of the two, we'd go back to the initial condition. Well, our initial condition told us that y of 0 is equal to negative 1. 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, 2. If this is going to be equal to negative 1, then this has to be 1 minus 2. So the explicit form that satisfies our initial condition, and 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, over what domain is it satisfied. Well, that's satisfied when this term is positive, this becomes negative, and you get it's undefined in reals, and all of that. But anyway, I've run out of time. See you in the next video.