问一道高数题?
2个回答
展开全部
y'' + y' = 2x^2e^x
特征方程 r^2 + r = 0, r = 0, r = -1
则设特解 y = (ax^2+bx+c)e^x
则 y' = [ax^2+(b+2a)x+c+b]e^x
y'' = [ax^2+(b+4a)x+c+2b+2a]e^x
代入微分方程 ax^2+(b+4a)x+c+2b+2a + ax^2+(b+2a)x+c+b = 2x^2
a = 1
b+4a+b+2a = 0 , b = -3a = -3
c+2b+2a+c+b = 0, c = (3b+2a)/2 = 7/2
特解 y = (x^2-3x+7/2)e^x
通解 y = C1 + C2e^(-x) + (x^2-3x+7/2)e^x
特征方程 r^2 + r = 0, r = 0, r = -1
则设特解 y = (ax^2+bx+c)e^x
则 y' = [ax^2+(b+2a)x+c+b]e^x
y'' = [ax^2+(b+4a)x+c+2b+2a]e^x
代入微分方程 ax^2+(b+4a)x+c+2b+2a + ax^2+(b+2a)x+c+b = 2x^2
a = 1
b+4a+b+2a = 0 , b = -3a = -3
c+2b+2a+c+b = 0, c = (3b+2a)/2 = 7/2
特解 y = (x^2-3x+7/2)e^x
通解 y = C1 + C2e^(-x) + (x^2-3x+7/2)e^x
本回答被提问者采纳
已赞过
已踩过<
评论
收起
你对这个回答的评价是?
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