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# Need help ASAP with differential equation watch

1. 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
2. (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:
.
This however, is patently false, as , which is not what you've said.
3. (Original post by joostan)
I agree with you right up until the point where you exponentiated.
You seem to claim that:
.
This however, is patently false, as , 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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