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Need help ASAP with differential equation watch

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    dq = kq(a-q)
    dt

    I have to solve this for the general equation, using partial fractions. I have broken it down to (1/ka)((1/q)+(1/a-q)dq = 1dt and have integrated that to get ln(q)-ln(a-q) = ka(t)+c
    I take exponentials and get 2q-a = e^[ka(t)+c]
    Finally adding a and dividing by 2 to isolate q gives

    q=e^[kat+c] +a
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    (Original post by Ryanicholls)
    dq = kq(a-q)
    dt

    I have to solve this for the general equation, using partial fractions. I have broken it down to (1/ka)((1/q)+(1/a-q)dq = 1dt and have integrated that to get ln(q)-ln(a-q) = ka(t)+c
    I take exponentials and get 2q-a = e^[ka(t)+c]
    Finally adding a and dividing by 2 to isolate q gives

    q=e^[kat+c] +a
    I agree with you right up until the point where you exponentiated.
    You seem to claim that:
    e^{\ln(q)-\ln(a-q)}=2q-a.
    This however, is patently false, as e^{x-y}=e^xe^{-y}, which is not what you've said.
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    (Original post by joostan)
    I agree with you right up until the point where you exponentiated.
    You seem to claim that:
    e^{\ln(q)-\ln(a-q)}=2q-a.
    This however, is patently false, as e^{x-y}=e^xe^{-y}, which is not what you've said.
    Thanks! I've used the log laws to simplify ln(q)-ln(a-q) to ln(q/(a-q)) and then taken exponentials, hopefully this has solved my problem
 
 
 
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Updated: November 14, 2015
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