[微積] 微分方程 homogeneous Linear ODEs of …
y''+p(x)y'+q(x)y=0 (1)
we set
y = y2 = uy1
y' = u'y1+uy'1
y'' = u''y1+2u'y'1+uy''1
into (1)
u''y1 + 2u'y'1 + uy''1 + p(u'y1 + uy'1) + quy1 = 0
Collecting terms in u'',u',u,we have
u''(y1) + u'(2y'1+py1) + u(y''1+py'1+qy1) = 0
Since y1 is a solution of (1) , the expression in the last
parentheses is zero.
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我不懂最後一句為什麼這麼說???
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05/28 22:47, , 1F
05/28 22:47, 1F
喔喔，我看懂了，謝謝。
※ 編輯: linbanden 來自: 140.135.42.156 (05/28 22:54)