Question

大一高等数学第九题

Answer 1

k1t - k2v = m (dv/dt)
即 dv/dt + (k2/m)v = (k1/m)t
为一阶线性微分方程
v = e^(-∫(k2/m)dt) [ C + ∫(k1/m)t e^(∫(k2/m)dt) dt ]
= e^(-k2t/m) [ C + (k1/m)∫t e^(k2t/m) dt ]
= e^(-k2t/m) [ C + (k1/k2)∫t de^(k2t/m) ]
= e^(-k2t/m) { C + (k1/k2)[te^(k2t/m) -∫e^(k2t/m)dt ] }
= e^(-k2t/m) { C + (k1/k2)[te^(k2t/m) -(m/k2)e^(k2t/m) ] }
= Ce^(-k2t/m) + (k1/k2)t - mk1/(k2)^2
t = 0 时， v = 0， 则 C = mk1/(k2)^2
得 v = [ mk1/(k2)^2][e^(-k2t/m)-1] + (k1/k2)t

Answer 2

k1t-k2v=ma,其中v=at,代入得整理可得v=k1t^2/(k2t+m)