y'' + 6y' + 8y = 2t + 1
where y' = dy/dt and y'' = d^2y/dt^2
THX
how to solve this ODE??
2007-02-12 7:41 pm
回答 (2)
2007-02-13 1:15 am
✔ 最佳答案
Please see the deductions below which uses D-operator method:圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Maths/Diffeqn2.jpg
參考: My Maths knowledge
2007-02-12 8:08 pm
From http://hk.knowledge.yahoo.com/question/?qid=7007020701530 (your own question earlier), the homogeneus solution is
y_0=Aexp(-2t) + Bexp(-4t)
The particular sol'n must satisfy y'' + 6y' + 8y = 2t + 1
Easily, you can see that y_p must take polynomial form of order 1. By inspection, you may check that if
y_p = c*t + d
So 6c + 8*(ct+d) = 2t+1
So c = 1/4, and
6c + 8d = 1 implies 8d=1-6/4 = -1/2
So d = -1/16
Therefore, y_p = t/4 - 1/16
That is, the complete solution is
y(t) = y_0 + y_p = Aexp(-2t) + Bexp(-4t) + t/4 - 1/16
y_0=Aexp(-2t) + Bexp(-4t)
The particular sol'n must satisfy y'' + 6y' + 8y = 2t + 1
Easily, you can see that y_p must take polynomial form of order 1. By inspection, you may check that if
y_p = c*t + d
So 6c + 8*(ct+d) = 2t+1
So c = 1/4, and
6c + 8d = 1 implies 8d=1-6/4 = -1/2
So d = -1/16
Therefore, y_p = t/4 - 1/16
That is, the complete solution is
y(t) = y_0 + y_p = Aexp(-2t) + Bexp(-4t) + t/4 - 1/16
參考: me
收錄日期: 2021-04-12 23:48:05
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20070212000051KK01076