(1/L)∫[e(t)-r(t)]dt+C0=r(t)/R+Cdr(t)/dt
[e(t)-r(t)]/2=r'(t)/2+r''(t)/4
r''(t)+2r'(t)+2r(t)=4e(t)能否給個過程啊流過電感的電流是電阻和電容上的電流之和
電感的電流是il(t),電壓=e(t)-r(t)=Ldil(t)/dt,所以il(t)=(1/L)∫[e(t)-r(t)]dt+C0
電容的電流ic(t)=Cdr(t)/dt
所以(1/L)∫[e(t)-r(t)]dt+C0=r(t)/R+Cdr(t)/dt
兩邊求導則[e(t)-r(t)]/L=r'(t)/R+Cr''(t)
整理得e(t)=r(t)+Lr'(t)/R+LCr''(t)
代入得e(t)=r(t)+r'(t)+r''(t)/2，r''(t)+2r'(t)+2r(t)=2e(t)
我那個還搞錯了耶！