Question

求解常系数线性微分方程x''-2x'+2x=te^tcost,特解那里的思路就可以了,谢谢

Answer 1

x''-2x'+2x=0
特征方程r^2-2r+2=0
r=1+i或 r=1-i
x=C1e^tcost+C2e^tsint

设方程
特解x=(mt+n)e^tcost+(ut+v)e^tsint
x'=me^tcost+ue^tsint +(mt+n)e^t(cost-sint)+(ut+v)e^t(sint+cost)
=(m+n+v)e^tcost+(u-n+v)e^tsint+(m+v)te^tcost+(-n+u)te^tsint
x''=2me^t(cost-sint)+2ue^t(sint+cost)+(mt+n)e^t[cost-sint-sint-cost)+(ut+v)e^t[sint+cost+cost-sint] =(2m+2u+2v)e^tcost+(-2m+2u-2n)e^tsint+2ute^tcost-2mte^tsint

[(2m+2u+2v)-(2m+2n+2v)+2n ]e^tcost +
[(-2m+2u-2n)-(2u-2n+2v)+2v] e^tsint +
[2u-(2m+2v)+2m] te^tcost +
[-2m-(-2n+u)+u]te^tsint =te^tcost

n=0
-2m+2n=0
2u-2v=1
2n-2m=0
m=n=0,u=1/2+v
特解x=[(1/2+C)t+C]e^tsint
x'=(1/2+C)e^tsint +[(1/2+C)t+C]e^t(sint+cost)
x''=(1/2+C)e^t(sint+cost)+(1/2+C)e^t(sint+cost)+[(1/2+C)t+C]e^t(sint+cost+cost-sint)
=(1+2C)e^t(sint+cost)+(1+2C)te^tcost+Ce^tcost

通解
x=C1e^tcost+C2e^tsint+[(1/2+C)t+C]e^tsint