※ 引述《plumeshouse (羽居)》之铭言： : 2 (2) (1) : x y + 2 y + x y =0 : 此题可用换y 级数 拉式转换来解 : 前两项我会算 : 可是换成拉式却卡住 : 请大大帮我解一下 : (-1/2) (-1/2) : y=C1x sinx+C2x cosx xy''+(1-2v)y'+xy=0 y(0)=0 y'(0)=0 -d dY ---(s^2Y)+(1-2v)(sY) - --- =0 ds ds -(s^2+1)Y'-(2S)Y+(1-2v)sY=0 (s^2+1)Y'+(1+2v)sY=0 1 lnY+---(1+2v)ln(s^2+1)=c' 2 c Y=----------------- (s^2+1)^(1+2v)/2 c 1 Y=-------- ------------------ s^(1+2v) (1+1/s^2)^(1+2v)/2 c ∞ r 1 =--------Sum C (---)^k ,r=-(1+2v)/2 s^(1+2v)k=0 k s^2 其中 r (-(1+2v)/2)((-(1+2v)/2)-1)...((-(1+2v)/2)-k+1) C =---------------------------------------------- k k! -1-2v -3-2v -2k+1-2v (------)(------)...(--------) 2 2 2 =---------------------------------------------- k! 1+2v 3+2v 2k-1+2v (------)(------)...(--------)(-1)^k 2 2 2 =---------------------------------------------- k! (-1)^k(1+2v)(3+2v)...(2k-1+2v) =------------------------------ (2^k)k! (-1)^k(1+2v)(2+2v)(3+2v)...(2k-1+2v) =---------------------------------- (2^k)k!(2+2v)(4+2v)...(2k-2+2v) (-1)^k(1+2v)(2+2v)(3+2v)...(2k-1+2v)(2k+2v) =----------------------------------------- (2^k)k!(2+2v)(4+2v)...(2k-2+2v)(2k+2v) (-1)^k(2k+2v)! =-------------------- 带回 (2^k)k!2^(k+v)(k+v)! => c ∞ (-1)^k(2k+2v)! 1 --------Sum -------------------- (---)^k s^(1+2v)k=0 (2^k)k!2^(k+v)(k+v)! s^2 ∞ (-1)^k (2k+2v)! =cSum -------------------- ------------ k=0 k!2^(2k+v)(k+v)! s^(2k+2v+1) y=L^-1(Y) ∞ (-1)^k =cSum -------------------- x^(2k+2v) k=0 k!2^(2k+v)(k+v)! ∞ (-1)^k x^(2k+v) =cx^vSum ------------------- k=0 k!(k+v)! 2^(2k+v) =cx^vJ (x) v imply y=dx^vJ (x) is also solution -v ∴ yh=cx^vJ (x)+dx^vJ (x) , v -v v=-1/2 use bessel to verify answer xy''+2y'+xy=0 Let y=vx^-1/2=vu y'=v'u+u'v y''=v''u+2v'u'+vu'' x(v''u+2v'u'+vu'')+(2)(v'u+u'v)+xvu=0 x^2v''+ xv'+(x^2-0.25)v v=c1J0.5+C2J-0.5 y=x^-1/2(c1J0.5+C2J-0.5) -- 路遥知马力,脸书见人心 http://www.facebook.com/home.php#/iyen.li ---> FB Online XD --

※ 发信站: 批踢踢实业坊(ptt.cc) ◆ From: 123.193.214.165 ※ 编辑: iyenn 来自: 123.193.214.165 (10/15 07:17) ※ 编辑: iyenn 来自: 123.193.214.165 (10/15 09:35)