已知微分方程为u(t)=Ri(t)+Ldi(t)\dt+e(t),求电流i(t)的拉氏式

Question

已知微分方程为u(t)=Ri(t)+Ldi(t)\dt+e(t),求电流i(t)的拉氏式

Answer 1

∵对应的齐次方程是LdI(t)/dt+RI(t)=0,
==>此齐次方程的特征方程是Lx+R=0
==>特征根是x=-R/L
∴此齐次方程的通解是I(t)=Ce^(-Rt/L) (C是积分常数)
于是,设原方程的通解是I(t)=C(t)e^(-Rt/L) (C(t)表示关于t的函数)
∵dI(t)/dt=C'(t)e^(-Rt/L)-(RC(t)/L)e^(-Rt/L)
代入原方程,得RI(t)+L[C'(t)e^(-Rt/L)-(RC(t)/L)e^(-Rt/L)]+e(t)=u(t)
==>RC(t)e^(-Rt/L)+LC'(t)e^(-Rt/L)-RC(t)e^(-Rt/L)]=u(t)-e(t)
==>LC'(t)e^(-Rt/L)=u(t)-e(t)
==>C'(t)=[(u(t)-e(t))/L]e^(Rt/L)
==>C(t)=C+∫[(u(t)-e(t))/L]e^(Rt/L)dt (C是积分常数)
∴I(t)={C+∫[(u(t)-e(t))/L]e^(Rt/L)dt}e^(-Rt/L)
故 原方程的通解是I(t)={C+∫[(u(t)-e(t))/L]e^(Rt/L)dt}e^(-Rt/L) (C是积分常数).