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    • Thread Starter
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    Hello
    just wanted to check a rule - when you have something where you take ln of both sides do you also take ln of the constant of integration?

    example:
    I started with

     (x-3) \frac {dy}{dx} = y
     \int y^{-1} dy = \int (x-3)^{-1} dx
     ln y = ln (x-3) + ln C
     y = C (x-3) I know this is the answer from the sheet, just not sure if this is how you get to it?

    thank you in advance!
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    It's correct but to go through it bit by bit you need an extra line of working


    From

     ln y = ln (x - 3) + ln c

     ln  y = ln  C(x-3)

     y = C(x - 3)
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    What is this I don't even
    • PS Helper
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    What you did is fine. It doesn't matter what you do with the constant of integration; if you wrote +C instead of +\ln C then it'd be fine, but when you go to simplify you'd get y=e^C(x-3). However, since C is just an arbitrary constant, so is e^C, so writing A=e^C you'd have y=A(x-3) (which is the same as you have, just with C relabelled as A).
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    (Original post by nuodai)
    What you did is fine. It doesn't matter what you do with the constant of integration; if you wrote +C instead of +\ln C then it'd be fine, but when you go to simplify you'd get y=e^C(x-3). However, since C is just an arbitrary constant, so is e^C, so writing A=e^C you'd have y=A(x-3) (which is the same as you have, just with C relabelled as A).
    Brilliant thanks for the help guys, I need to spread rep around before repping again.
 
 
 
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