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# population differential eq problem?

I'm given the eqn dP/dt= cln(K/P)P where c= .15, K= 1000 and P(0)=100 and I'm asked to solve the diff eq for P(t) with the given conditions and to find at what P does P grow fastest?

Any help would be great thanks!

c is a constant and K is the carrying capacity

### 1 Answer

- Amar SoniLv 71 decade agoFavorite Answer
dP/dt= c ln(K/P)P

dP/{P ln (k/P)} = c.dt.......................(i)

put ln (k/P) = y

(k/P) = e^y

P= k e^-y

differentiate

dP = -k(e^-y) dy......................(ii)

integrate[ -k dy/y]

-k ln y = ct + C

-k ln{ln (k/P)} = ct +C...................(iii)

ln{ ln(k/P)} = -ct/k -C/ k

ln (k/P) = e^(-ct-C)

ln k - ln P = e^(-ct-C)

ln P = ln k + e^(-ct -C).......................(iv)

Put K= 1000, c = 0.15 when t= 0 P(0) =100

ln P(0)= ln 1000 + e^(-0-C)

ln P(0) = 3 + e^-C

ln P(o) -3 = e^-c...................................(v)

ln P = ln k + e^(-ct){ln P(o) - 3}.......................(iv)..................Ans

since dP/dt= c {ln(K/P)}P > 0 because P=100, c=0.15 and ln(k/P) also positive,

therefore P is increasing...................................ans