dP/dt= -aP*ln(P/b),
所以
1/b *dP/dt= -aP/b *ln(P/b),
即d(P/b)/dt= -aP/b *ln(P/b),
故 1/(P/b) * d(P/b)= -a *ln(P/b) *dt
而显然 d[ln(P/b)] = 1/(P/b) * d(P/b),
所以d[ln(P/b)] = -a *ln(P/b) *dt
故 1/ln(P/b) * d[ln(P/b)] = -a dt
对等式两边同时积分,
得到
ln | ln(P/b) | = -at +C (C为常数)
我就不继续化简了啊