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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
• @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.
• 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?
• 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.
• please explain more why we put (+c) in one side ?
if we have (+c) in each side they will remove each other .. i think
thanks
• 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.