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Where should I put the C in differential equations

I asked My teacher once, months ago, and I remember he said it doesn’t matter. But just to make sure, sometimes my C is of a different sign than the one in the ms. Is that fine ?
Reply 1
The sign of C is important, can you give an example?
Original post by CurrentlyAtLevel
I asked My teacher once, months ago, and I remember he said it doesn’t matter. But just to make sure, sometimes my C is of a different sign than the one in the ms. Is that fine ?
Original post by mqb2766
The sign of C is important, can you give an example?


Ignore this
(edited 4 years ago)
Yep the sign of C doesn't matter because it's a constant, so it could be positive or negative. So even if you put "+C" in your final answer that C could still be negative but that minus sign has been absorbed into the constant as a result of that. Hope that makes sense haha
Original post by Y12_FurtherMaths
Ignore this
Original post by Chlozza27
Yep the sign of C doesn't matter because it's a constant, so it could be positive or negative. So even if you put "+C" in your final answer that C could still be negative but that minus sign has been absorbed into the constant as a result of that. Hope that makes sense haha


Lol I said “ignore this” because I sent a message by mistake so wanted him to ignore the message. I wasn’t disagreeing with him 😂
Reply 5
Perhaps I should have said it better. The sign of the constant is important. But whether it is +C or -C, it will come out in the wash.
Original post by Y12_FurtherMaths
Ignore this
Original post by mqb2766
Perhaps I should have said it better. The sign of the constant is important. But whether it is +C or -C, it will come out in the wash.


Yeah, I sent a message by mistake so I meant ignore my message lol
Ok I’m confused
Here’s an example
6FBE4312-F1A5-4C4C-B286-E168BB5004E2.jpeg
I put the +c on the right side which in this case was on the t side so I got +c on the right side and then I subbed in the x and t values they had given me and got -1/2ln(1/3)
The answer in the solution bank is +ve 1/2ln(1/3)
One little tip about which side to initially put C is where you end up with ln of something after integration on one side.
In this case, it makes sense to write ln(A) instead of C on the side with ln on it, and then combine this with the other ln. Especially useful if you then decide to take anti-logs.

For example, if after integrating you end up with something like:

ln(y) = x^2,

then writing ln(A) on the left gives

ln(A) + ln(y) = x^2,

ln(Ay) = x^2

Ay = e^(x^2)

Which is a bit more elegant than having
y = e^(x^2 + c)
Reply 9
Could you upload your solution pls? Note that a sign difference with a log is equivalent to flipping its argument, so
-ln(1/3) = ln(3)
Your answers are still different, but just worth mentioning.

Edit: I got -1/2ln(3) which is +1/2ln(1/3). Stupid question, but did you get the partial fraction correct?


Original post by CurrentlyAtLevel
Ok I’m confused
Here’s an example
6FBE4312-F1A5-4C4C-B286-E168BB5004E2.jpeg
I put the +c on the right side which in this case was on the t side so I got +c on the right side and then I subbed in the x and t values they had given me and got -1/2ln(1/3)
The answer in the solution bank is +ve 1/2ln(1/3)
(edited 4 years ago)

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