CP1,n不等的负实根(0! R+ u=Aepittaepat C LDL c(0+)=U0→A1+A2=U0 i=-C-C du i(0+) (0+)=—=0→>f1A1+P2A2=0ci dt C dt C P A,= U 0 A 2 U 0 分(2en-p% 2 U C一. 2 p 1 , p 2 不等的负实根 CL R  p t p t C u e e 1 2 = A 1 + A 2 0 A 1 A 2 0 u C ( 0 + ) = U → + = U 0 A A 0 (0 ) (0 ) dd = → 1 1 + 2 2 = − = + + P P Ci tuC 0 2 1 1 0 2 2 1 2 A1 A U P PP U P P P −− = − = ( ) 1 2 2 1 2 1 0 p t p t C P e P e P P U u − − = Ci dt du d t du i C C C = − = − R L C +- i u c u L+- ( t=0)