Re: [理工] [工數]-ODE
看板Grad-ProbAsk作者shinyhaung (我是Shiny)時間15年前 (2010/04/19 21:38)推噓1(1推 0噓 1→)留言2則, 2人參與討論串237/280 (看更多)
※ 引述《wil0829ly (汪汪)》之銘言：
: Solve the differential equation
: 2
: (ax+b) y"(x)-2(ax+b)y'(x)-6ay(x)=3ax+2b in terms of (ax+b),
: where a and b are constants.
: 請問這題要怎麼做呢@@
令 ax+b = e^t , t = ln|ax+b|
則 (ax+b)y' = a(Dt)y , (ax+b)^2y" = a^2(Dt)(Dt-1)y , 其中Dt = d/dt
(ax+b)^2y"(x)-2(ax+b)y'(x)-6ay(x) = 3ax+2b
=> [a^2(Dt)(Dt-1) - 2a(Dt) - 6a]y = 3e^t - b
=> [(Dt)^2 - (1 + 2/a)(Dt) - 6/a]y = (3/a^2)e^t - (b/a^2)
<1>yh
令 y = e^mt 帶入
=> m^2 - (1 + 2/a)m - 6/a = 0 , 解出 m = m1 , m2
帶回可得
m1t m2t m1 m2
yh = c1e + c2e = c1(ax+b) + c2(ax+b)
<2>yp
1
yp = ------------------------------(3e^t - b)
(Dt)^2 - (1 + 2/a)(Dt) - 6/a
3e^t b
= --------------------- + -----
1 - (1 + 2/a) - 6/a 6/a
-3a ab
= -----[ln|ax+b|] + ----
8 6
<3>
y = yh + yp
= ............
應該吧
--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 112.105.113.239
推
04/19 21:48, , 1F
04/19 21:48, 1F
→
04/19 21:48, , 2F
04/19 21:48, 2F
※ 編輯: shinyhaung 來自: 112.105.113.239 (04/19 22:23)
討論串 (同標題文章)