高数微分方程,求大佬解答
展开全部
y1= (1/2)e^(2x)+(x-1/3)e^x
y1'= e^(2x)+(x+2/3)e^x
y1''=2e^(2x)+(x+5/3)e^x
y1''+ay1'+by = ce^x
[2+a+(1/2)b]e^(2x) +(1+a+b)xe^x +[5/3 + (2/3)a -(1/3)b]e^x = ce^x
=>
2+a+(1/2)b =0 (1)
1+a+b=0 (2)
5/3 + (2/3)a -(1/3)b=c (3)
(2)-(1)
(1/2)b = 1
b=2
from (2)
1+a+2=0
a=-3
from (3)
5/3 + (2/3)a -(1/3)b=c
5/3 - 2 -2/3=c
c=-1
(a,b,c)=(-3,2,-1)
y1'= e^(2x)+(x+2/3)e^x
y1''=2e^(2x)+(x+5/3)e^x
y1''+ay1'+by = ce^x
[2+a+(1/2)b]e^(2x) +(1+a+b)xe^x +[5/3 + (2/3)a -(1/3)b]e^x = ce^x
=>
2+a+(1/2)b =0 (1)
1+a+b=0 (2)
5/3 + (2/3)a -(1/3)b=c (3)
(2)-(1)
(1/2)b = 1
b=2
from (2)
1+a+2=0
a=-3
from (3)
5/3 + (2/3)a -(1/3)b=c
5/3 - 2 -2/3=c
c=-1
(a,b,c)=(-3,2,-1)
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询