I have the equation
x^(3) - x^(2) - x^(1) + x = 0
How do I solve this equation?
The (number) is an indicater of how many time it's differentiated.
Third order differential equation problem?
2016-03-21 12:03 pm
回答 (2)
2016-03-21 12:21 pm
✔ 最佳答案
Covert it to a polynomial by assuming a solution x = a*e^(mt) Then x"' -x" - x' +x = 0 --> m^3 - m^2 - m + 1 = 0
factor m^2(m-1) - (m+1) = 0 --> divide by (m-1)
m^2 - 1 = 0 --> m = +/-1 So you have one degenerate root thus
x = a*e^t + b*e^-t + c*t*e^t
2016-03-21 12:10 pm
m^3-m^2-m+1 = 0
m^2(m-1)-1(m-1)=0
(m-1)(m^2-1) = 0
(m-1) (m+1)(m-1) = 0
(m+1) (m-1)^2 = 0
m=-1 , m=1,1
x = C1 e^(-t) + C2 e^(t) + C3 t e^(t)
m^2(m-1)-1(m-1)=0
(m-1)(m^2-1) = 0
(m-1) (m+1)(m-1) = 0
(m+1) (m-1)^2 = 0
m=-1 , m=1,1
x = C1 e^(-t) + C2 e^(t) + C3 t e^(t)
收錄日期: 2021-04-21 17:23:23
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160321040302AAOMykO