AP®︎/College Calculus AB
- Particular solutions to differential equations: rational function
- Particular solutions to differential equations: exponential function
- Particular solutions to differential 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
Solving a separable differential equation given initial conditions. In this video, the equation is dy/dx=2y² with y(1)=1.
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- At5:00, Sal got arbitrary C = 1 but when I calculated for C I got C = -1 and C = -1/2, and also why at3:40why did he do the reciprocal for both sides instead of just plugin in the value? When I plugged in the value at3:40instead of doing the reciprocal I ended up getting C = -1(48 votes)
- @darkfall20 Not sure how you got C= (-1) but I also got C= (-1/2) when I solved the following equation before Sal simplified it:
-(1/2) y^-1 = x + C
-(1/(2*(-1)) = 1 + C
1/2 = 1 + C
C = -1/2
When you plug C = -1/2 into the equation you started with and solve for y when x=3 you will get the same answer as Sal that y = -1/5 when x =3.(7 votes)
- At3:30Sal keeps the C unchanged because it will still be an arbitrary constant regardless of whether or not you multiply it, I get why one can still simply treat it as C, though I still don't see why one couldn't treat it as 1/y = (-2x-2c) and then solve it. Given that we have 1/y = (-2x-2c) and we solve for y, then;
y = 1/(-2x-2C), where we have (1,-1)
then substituting the values;
-1 = 1/(-2-2C) followed by some algebra
-1(-2-2C) = 1
2+2C = 1
2C = -1
c = -1/2
I'm probably wrong somewhere in my reasoning but I don't see why or at what point.(21 votes)
- You aren't wrong, your answer and Sal's answer are just totally compatible. You just multiplied C by -2, so your answer for C is -1/2 of Sal's answer. Plugging your answer into your version of the solution:
Which gives Sal's answer. Essentially you can do what you did, its just more work.(20 votes)
- It's not completely clear when you are supposed to apply arithmetic on constant "C" and when not to.(10 votes)
- From what I've learned, you'd apply functions like division and multiplication to C, so if 2y = x + C then you'd divide the entire right side by 2. But functions like addition and subtraction are the whole reason for C's existence. C just represents some constant value that we currently don't know, so x + C - 2 will just turn out to x + C, since either way you don't know the actual value of your constant.(2 votes)
- Why is the integral of dx = x? Is dx "short" for 1 dx?(6 votes)
- Yes, dx is the same as 1 dx, because 1 times any quantity is that quantity.
Have a blessed, wonderful day!(7 votes)
- Why is there an arbitrary constant only on one side? In his previous video, he had an arbitrary constant on both sides of the equation. Just wanna make sure I really get this stuff down.
Update: Never mind I figured it out with a little more thinking, but thanks!(7 votes)
- How come we can treat dy/dx as a fraction? I thought that we couldn't separate them without integrating because of chain rule.(5 votes)
- At1:38, why can't you divide by y^2 instead of 2^y^2, this would leave me with only dy/y^2=2^dx?(4 votes)
- You can solve this problem either way. You can divide by y^2 or 2^y^2, it doesn't matter.... You'll still arrive at C = 1 and get the same answer. Hope this helps.(2 votes)
- please explain more why we put (+c) in one side ?
if we have (+c) in each side they will remove each other .. i think
- The two C's don't have to be the same. We can put the C on both sides, but then we just subtract one of the C's, and since one arbitrary constant minus another arbitrary constant is just a third arbitrary constant, that results in a third C on one side.(6 votes)
- I understand that a differential equation represents the relationship between a function to its derivatives. Sometimes you can see that the differential equation actually equals the derivative of the function. For example: dy/dx=1/x.
My question is what is the difference between the derivative to the diff eq in that case? can I say that they are the same?
I hope my question was clear. Please let me know if more details are needed.
- When you say that the differential equation equals the derivative of the function, I assume you mean that it equals the derivative when C=0. If that's what you mean, then you are correct, the derivative at C=0 is equal to that particular solution of the differential equation. Let me know if that helped, or if I misunderstood you. :)(2 votes)
- Let's now get some practice with separable differential equations, so let's say I have the differential equation, the derivative of Y with respect to X is equal to two Y-squared, and let's say that the graph of a particular solution to this, the graph of a particular solution, passes through the point one comma negative one, so my question to you is, what is Y, what is Y when X is equal to three for this particular solution, so the particular solution to the differential equation that passes through the point one comma negative one, what is Y when X is equal to three, and I encourage you to pause the video, and try to work through it on your own. So I'm assuming you had a go at it, and the key with a separable differential equation, and that's a big clue that I'm even calling it a separable differential equation, is that you separate the Xs from the Ys. Or all the Xs and the DXs from the Ys and DYs. So how do you do that here? Well, what I could do, let me just rewrite it. So it's gonna be DY DX is equal to two Y-squared, is equal to two Y, equal to two Y-squared. So let's see, we can multiply both sides by DX, and let's see, so then we're gonna have, that cancels with that if we treat it as just a value, or as a variable. We're gonna have DY is equal to two Y squared DX. Well, we're not quite done yet. We gotta get this two Y squared on the left hand side. So we can divide both sides by two Y-squared. So if we divide both sides by two Y-squared, two Y-squared, the left hand side, we could rewrite this as 1/2 Y to the negative two power, is going to be equal to DY, DY is equal to DX, and now, we can integrate both sides. So we can integrate both sides. Let me give myself a little bit more space. And so, what is, what is this left hand side going to be? Well, we increment the exponent, and then divide by that value, so Y to the negative two, if your increment is Y to the negative one, and then divide by negative one, so this is going to be -1/2 Y to the negative one power, and we could do a plus C like we did in the previous video, but we're gonna have a plus C on both sides, and you could subtract, or you know, you have different arbitrary constants on both sides and you could subtract them from each other, so I'm just gonna write the constant only on one side. So you have that is equal to, well if I integrate just DX, that's just going to give me X, that's just gonna give me X. So this right over here is X, and of course I can have a plus C over there, and If I want I can, I can solve for Y if I multiply, let's see, I can multiply both sides by negative two, and then I'm gonna have, the left hand side you're just gonna have Y to the negative one, or 1/Y is equal to, if I multiply the right hand side times negative two, I'm gonna have negative two times X plus, well it's some arbitrary constant, it's still going to, it's gonna be negative two times this arbitrary constant but I could still just call it some arbitrary constant, and then if we want we can take the reciprocal of both sides, and so we will get Y is equal to, is equal to 1/-2X+C. And now we can use, we can use the information they gave us right over here, the fact that our particular solution needs to go through this point to solve for C. So, when X is negative one, so when X is negative one. Oh sorry, when X is one, when X is one, Y is negative one, so we get negative one is equal to 1/-2+C, or we could say C minus two, we could multiply both sides times C minus two, if then we will get, actually let me just scroll down a little bit, so if you multiply both sides times C minus two, negative one times C minus two is going to be negative C plus two or two minus C is equal to one. All I did is I multiplied C minus two times both sides, and then, let's see, I can subtract two from both sides, so negative C is equal to negative one, and then if I multiply both sides by negative one, we get C is equal to one. So our particular solution is Y is equal to 1/-2X+1. And we are almost done, they didn't just ask for, we didn't just ask for the particular solution, we asked, what is Y when X is equal to three. So Y is going to be equal to one over, three times negative two is negative six plus one, which is equal to negative, is going to be equal to 1/-5, or -1/5. And we are done.