推 mzhrqoc01 :讓x=a*t =>y'=dy/dt *dt/dx=(1/a)*dy/dt 01/01 00:07
→ mzhrqoc01 :所以原式＝>(1-t^2)y"-2ty'+n(n+1)y=0 01/01 00:08
→ momentaneous:痾...我還是不太懂耶... 01/01 01:20
Find a solution of
(a^2-x^2)y''-2xy'+n(n+1)y=0, a不等於0
by reduction to the Legendre equation.
我算不出來耶
有人可以幫忙一下嗎
謝謝...
--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 140.135.43.92