把t和k省略
i' = -λis
s' = λis - αs(s+r)
r' = αs(s+r)
则i'+s'+r' = 0
积分得i + s + r = C
得到s = i'/(-λi)
s' = [i'/(-λi)]'
s + r = C - i
代入第二个方程,得到
[i'/(-λi)]' = -i' - αi'/(-λi) * (C-i)
整理得
i''i - (i')² = (λ+α)i'i² - αCi'i
令i' = p,则i'' = dp/dt = dp/di * di/dt = pdp/di
代入上式得
pidp/di - p² = (λ+α)pi² - αCpi
idp/di - p = (λ+α)i² - αCi
dp/di = p/i + (λ+α)i - αC
设p/i = u,则dp/di - p/i = (idp - pdi)/idi = d(p/i) * i/di = idu/di
代入上式得idu/di = (λ+α)i - αC
即du = [(λ+α) - αC/i]di
积分得u = (λ+α)i - αClni + C'
则p = iu = (λ+α)i² - αCilni + C'i
即di/dt = (λ+α)i² - αCilni + C'i
解这个微分方程得
i(t) =
