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※ 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
wil0829ly:s大 解答yp是 3(ax+b)/-8a + b/6a是一樣的嗎@@ 04/19 21:48
ntust661:記得把自己令的東西消去喔= = 04/19 21:48
※ 編輯: shinyhaung 來自: 112.105.113.239 (04/19 22:23)