送交者: lala 于 May 31, 2003 13:01:54:
回答: lala同学~~~, 看在党国的份上，拉兄弟一把 由 淇淇 于 May 31, 2003 00:22:42:
From d(y^3)/dt=k1*(y/z-1), we get
3ydy = k1/(z-1)dt.
Note that z^3-k2=k3*t. So dt = (3/k3)(z^2)dz.
It follows that
ydy = (k1/k3)(z^2)/(z-1)dt
= (k1/k3)((z-1) + 2 + (z-1)^{-1})dz
and hence that
y^2 = (k1/k3)( (1/2)z^2 + z + ln(z-1) ) + C
where C is a constant determined by the initial condition.