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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

    e^[kat+c] +a = q
    2

    My issue is that when I substitute this and it's derivative back into the original differential to check, I can't get it to solve. Can anyone see where I've gone wrong?
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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

    e^[kat+c] +a = q
    2

    My issue is that when I substitute this and it's derivative back into the original differential to check, I can't get it to solve. Can anyone see where I've gone wrong?
    When you've taken the exponential of both sides, e^(ln(q) - ln(a-q)) is not equal to 2q - a. You should combine the two logs so that you just have e^ln(j) where j is something and that gives you j.

    And if you post this in the maths forum rather than this umbrella forum you will get a faster response.
 
 
 
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