2y"+5y'=(cosx)平方，y"+y'-2y=3(e)x-1/2sin2x这两个解微分方程怎么解？

Question

如题所述

Answer 1

2y''+5y'=cos^2(x):
2y''+5y'=(1+cos(2x))/2
2(y_h)''+5(y_h)'=0 => AQE:2m^2+5m=0 => m=0,-5/2 => y_h=Ae^(5x/2)+B
设：y_p=Cx+Dcos(2x)+Esin(2x)
则: 2(y_p)''+5(y_p)'=[-4Dcos(2x)-4Esin(2x)]+5[C-2Dsin(2x)+2Ecos(2x)]
故：5C=1/2, -4D+10E=1/2, -10D-4E=0
故：C=1/10, D=-1/58, E=5/116
即：y=y_h+y_p=Ae^(5x/2)+B+(1/10)x-(1/58)cos(2x)+(5/116)sin(2x)
-----
y"+y'-2y=3e^x-(1/2)sin(2x):
(y_h)''+(y_h)'-2y_h=0 => AQE:m^2+m-2=0 => m=1,-2 => y_h=Ae^x+Be^(-2x)
设：y_p=Cxe^x+Dcos(2x)+Esin(2x)
则: (y_p)'=Ce^x+Cxe^x-2Dsin(2x)+2Ecos(2x)
(y_p)''=2Ce^x+Cxe^x-4Dcos(2x)-4Esin(2x)
(y_p)''+(y_p)'-2(y_p)
=[2Ce^x+Cxe^x-4Dcos(2x)-4Esin(2x)]+[Ce^x+Cxe^x-2Dsin(2x)+2Ecos(2x)]-2[Cxe^x+Dcos(2x)+Esin(2x)]
=3Ce^x+(2E-6D)cos(2x)+(-2D-6E)sin(2x)
故：3C=1,2E-6D=0,-2D-6E=-(1/2)
故：C=1, D=1/40, E=3/40
即：y=y_h+y_p=Ae^x+Be^(-2x)+xe^x+(1/40)cos(2x)+(3/40)sin(2x)