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    It is given that dP/dt = 2P. At t = 0, P = 3.
    Find the value of P when t = 2.

    My working:
    (Integral) (1/2P) dP = (Integral) (1) dt
    So,
    Ln|2P| = t + C
    Let C = LnA
    2P = e^(t + LnA)
    2P = Ae^t

    This is wrong, because apparently it is essential to leave the 2 on the side of dt and integrate it with respect to t.
    Could someone please explain why you must leave the 2 on the side of dt, and how it breaks the laws of mathematics to put it on the other side?

    How do you know what sides to place constants?

    The differing answers are:
    P = 3e^2 and P = 3e^4.

    I'm sorry if this is a stupid question; this is just something I've never understood.
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    (Original post by JLegion)
    It is given that dP/dt = 2P. At t = 0, P = 3.
    Find the value of P when t = 2.

    My working:
    (Integral) (1/2P) dP = (Integral) (1) dt
    So,
    Ln|2P| = t + C
    Let C = LnA
    2P = e^(t + LnA)
    2P = Ae^t

    This is wrong, because apparently it is essential to leave the 2 on the side of dt and integrate it with respect to t.
    Could someone please explain why you must leave the 2 on the side of dt, and how it breaks the laws of mathematics to put it on the other side?

    How do you know what sides to place constants?

    The differing answers are:
    P = 3e^2 and P = 3e^4.

    I'm sorry if this is a stupid question; this is just something I've never understood.
    Your integral is incorrect.

    Integral 1/(2P) = 1/2 x integral 1/P = 1/2 x ln|P|

    Moving the 2 over to the other side is irrelevant.
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    (Original post by ghostwalker)
    Your integral is incorrect.

    Integral 1/(2P) = 1/2 x integral 1/P = 1/2 x ln|P|

    Moving the 2 over to the other side is irrelevant.
    What a ridiculous mistake.
    Thank you yet again!
    • Study Helper
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    (Original post by JLegion)
    What a ridiculous mistake.
    Thank you yet again!
    You're welcome
 
 
 
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