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### Course: AP®︎/College Calculus AB>Unit 7

Lesson 6: Finding particular solutions using initial conditions and separation of variables

# Worked example: finding a specific solution to a separable equation

Solving a separable differential equation given initial conditions. In this video, the equation is dy/dx=2y² with y(1)=1.

## Want to join the conversation?

• At , Sal got arbitrary C = 1 but when I calculated for C I got C = -1 and C = -1/2, and also why at why did he do the reciprocal for both sides instead of just plugin in the value? When I plugged in the value at instead of doing the reciprocal I ended up getting C = -1
• People are getting different values of C all over the place, and I'm here to say that you're ALL doing it RIGHT. That's right, you aren't wrong. This is because while you're still doing algebra to what you got after integrating, what you do to C doesn't really matter. If you multiply both sides by something, you can pick whether or not you want to say so about C, because it's just a constant you can solve for later. If you DO say "2C," you'll end up with 1/2, which is fine, because you're going to multiply that by 2 and get 1. If you decide to solve for C earlier than Sal did, you might get -1 instead of 1, which is still ok, because if you plug in that value before doing any more algebra, you'll have to do algebra to that value too, and you'll still get the right answer. When I tried this, I got C= -1, and my final equation was y = -1/(2x-1). Notice how this still gives the right answer.
y = -1/(2*3 - 1)
y = -1/(6-1)
y = -1/5
As long as you plug in your C in the same step you solve for it, you'll get the right answer.
• At Sal 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.
y=1/(-2x-2C)
y=1/(-2x-2(-1/2))
y=1/(-2x+1)
Which gives Sal's answer. Essentially you can do what you did, its just more work.
• It's not completely clear when you are supposed to apply arithmetic on constant "C" and when not to.
• 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.
• Why is the integral of dx = x? Is dx "short" for 1 dx?
• Yes, dx is the same as 1 dx, because 1 times any quantity is that quantity.

Have a blessed, wonderful day!
• Are there any separable differential equation exercises?
• yes on the side
(1 vote)
• 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!
• notice in previous videos he write c 1 , c2 then subtracted them and write C , here he directly write C.
(1 vote)
• How come we can treat dy/dx as a fraction? I thought that we couldn't separate them without integrating because of chain rule.
• At , why can't you divide by y^2 instead of 2^y^2, this would leave me with only dy/y^2=2^dx?
• 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.
• Does anyone know why solving for C when the equation in is the form (-1/2y^2) = x + C is incorrect? I get C = -1/2 when I do it that way but I don't understand why that is not also a valid solution to this problem.