Aplicaciones matemáticas ecuación cosecha de peces
Enviado por watsumh • 30 de Octubre de 2015 • Práctica o problema • 318 Palabras (2 Páginas) • 182 Visitas
dp/dt= p(a-bp)+H
dp/ p(a-bp)+H =dt
∫(dp/(-bp^2+pa+h)) = ∫dt
∫(dp/(bp^2+pa+h)) = -∫dt
P=(a±√ (a^2+4bh))/2b
P=(a/2b) ± H
∫{dp/[(p-(a/2b)-H)( p-(a/2b)+H)]}dp= [A/( p-(a/2b)-H) ]+[ B/(p-(a/2b)+H)]
∫A[p-(a/2b)+H]+B[p-(a/2b)-H]dp
Si P=(a/2b) ± H
A[(a/2b) + H-(a/2b)+H] +B[(a/2b) - H-(a/2b)-H]
A=2H B=-2H
1= A2H 1=-B2H
1/2H=A -1/2H=B
∫[(1/2H)/(p-(a/2b)-H)]dp + ∫[(-1/2H)/(p-(a/2b)+H)]dp
(1/2H) ∫[dp/(p-(a/2b)-H)] -(1/2H) ∫[dp/(p-(a/2b)+H)]
(1/2H) ln[(p-(a/2b)-H)] - (1/2H) ln[(p-(a/2b)+H)] + T + C
Ln[(p-(a/2b)-H) / ((p-(a/2b)+H)]= 2HT+C
(p-(a/2b)-H) / (p-(a/2b)+H)= Ce^2HT C= (po-(a/2b)-H) / ((po-(a/2b)+H)
(1/2H) ln[(p-(a/2b)-H)] - (1/2H) ln[(p-(a/2b)+H)]= T+ = (po-(a/2b)-H) / ((po-(a/2b)+H)
(p-(a/2b)-H) / (p-(a/2b)+H)= = [ (po-(a/2b)-H) / ((po-(a/2b)+H) ] e^2HT
p-(a/2b)-H = (p-(a/2b)+H) (Ce^2HT)
p-(a/2b)-H = pCe^2HT - (a/2b)e^2HT + CeH^2HT
p-(a/2b)-H – pCe^2HT = (-a/2b)e^2HT + CeH^2HT
p - pCe^2HT = (-a/2b)e^2HT + CeH^2HT + (a/2b)+H
p(1-Ce^2HT) = Ce^2HT[ (-a/2b)+H] + (a/2b) + H
p={ Ce^2HT[ (-a/2b)+H] + (a/2b) + H} / (1-Ce^2HT)
p={{Ce^2[√ (a^2+4bh) / 2b] [-(a/2b) + (√ (a^2+4bh) / 2b)]} + (a/2b) + (√ (a^2+4bh) / 2b) } / (1-Ce^2[√ (a^2+4bh) / 2b]
si h=0
p(T)= (a/b) / 1-Ce^(a/b)T
p(T)= 1-(po/po-(a/b) (e^(a/b)T
...