解 R Ic(t) 1、布列微分方程: U=R(1)+u2(t) du(t RO +(t 2、解微分方程 RCS+1=0 u(t)=Ke r+U S RO l(0)=K+U=0 K=-U uch(t)=Ke Q=Uu(=Ue +Us=U(-e"+ + _ _ U S u C(t) R i C(t) 解：1、布列微分方程：( ) ( ) ( ) ( )u t dt du t RC U Ri t u t c c s c c = + = + 2、解微分方程： 1 1 1 0 = − = − + = RC s RCs t uch t Ke − ( ) = cp Q U s u ( t ) = = ( ) t c s u t Ke U  − = + (0) 0 c s s u K U K U = + = = − ( ) (1 ) t t c s s s u t U e U U e   − − = − + = −