3.2草地水量模型 Mathematical Modeling 2012 模型求解 0<t≤1800时 =10-5-103Q(t) dt Q(0)=0 Q(t)=0.01(1 0.00lt )0<t≤1800 →Q(1800)≈0.00835 Department of Mathematics HUSTMathematical Modeling 2012 Department of Mathematics HUST 0 1800  t 时      = = − − − (0) 0 10 10 ( ) d d 5 3 Q Q t t Q 0.001 ( ) 0.01(1 ) 0 1800 t Q t e t − = −   Q(1800)  0.00835 模型求解 3.2 草地水量模型