s+1)(s+3)s=-2=-1(s+1)(s+3)A2=(s+3)s+1s=-3=2F(s)=-2s+31s+2f(t)=2e-3t-e-2tF(s)=s(s+1)2(s+2)2-3-2函数的拉氏变换。解:f(t)=est+lim[est]s(s+1)2s=-2ddsss+2s-1=-2e-2t+lim(est+est)s-1sts+22(s+2)2=-2e-2t-te-t+2e-t=(2-t)e-t-2e-2tF(s)=2s2-5s+1s(s2+1)2-3-3函数的拉氏变换。解:F(s)(s2+1)s=+j=A1s+A2s=+jA1=1,A2=-5A3=F(s)s=1s=0∴f(t)=1+cost-5sintF(s)=++1ss2+1s-5s2+12-3-4函数的拉氏变换。(4)F(s)=s+2s(s+1)2(s+3)解:f(t)=est+ests+2(s+1)2(s+3)s=0s+2s(s+1)2s=-3+lims-1d[est]s+2s(s+3)ds=+e-3t+lim[+]23112s-1(-s2-4s-6)est(s2+3)2(s+2)tests2+3s=+e-3t-e-t-e-t2311234t2(2-4-1)求下列微分方程。d2y(t)dt2+5+6y(t)=6,初始条件:dy(t)dty(0)=y(0)=2。·解:s2Y(s)-sY(0)-Y(0)+5sY(s)-5Y(0)+6Y(s)=1s′A1=sY(s)s=0∴y(t)=1+5e-2t-4e-3tA2=(s+2)Y(s)s=-2A3=(s+3)Y(s)s=-3A1=1,A2=5,A3=-4∴Y(s)=6+2s2+12ss(s2+5s+6)(2-4-2)求下列微分方程。d3y(t)dt3+4+29=29,d2y(t)dt2dy(t)dt初始条件:y(0)=0,y(0)=17,y(0)=-122···解:
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