Re: [微方]降階法
※ 引述《wichanm (wichan)》之銘言：
: Use the method of reduction to find a second solution of the given
: differential equation.
: (x-1)y''-xy'+y = 0 ,x>1 ; y1(x) = exp(x)
: Ans: y2(x) = x
(x-1)y"-xy'+y = 0
y"-[x/(x-1)]y'+[1/(x-1)]y=0
λ^2 - [x/(x-1)]λ+[1/(x-1)]=0
假設x=2
λ^2 - 2λ + 1 = 0
λ = 1
y1(x) = e^(1)x
設 y2(x) = y = u*e^(1)x
y'= u'*e^(x)+u*e^(x)
y"= u"*e^(x)+u'*e^(x)+u'*e^(x)+u*e^(x)
= u"*e^(x)+2*u'*e^(x)+u*e^(x)
{u"*e^(x)+2*u'*e^(x)+u*e^(x)}
-[x/(x-1)]*{u'*e^(x)+u*e^(x)}+[1/(x-1)]*{u*e^(x)}=0
u"*e^(x) = 0
u" = 0
u' = ∫0 = x
u = ∫x = (1/2)*x^(2)
Ans : y2(x) = u*e^(x) = (1/2)*[x^(2)]*e^(x)
通解：y(x) = C1*e^x + C2*(1/2)*[x^(2)]*e^(x)
y'(x) = C1*e^x + C2*x
------------------------------------------------------------------------------
y(0)=1 → 1 = C1 + 0
C1 = 1
C2 = 0
y(x)另解 = y2(x) = y1(x)= e^(x)
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◆ From: 61.224.69.98
※ 編輯: gemini822 來自: 61.224.69.98 (12/29 20:12)
推
12/29 21:51, , 1F
12/29 21:51, 1F
抱歉！看錯了！
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12/29 21:56, , 2F
12/29 21:56, 2F
答案還是跟你不同？
不知道哪裡錯了？麻煩版上前輩們不吝嗇指導，謝謝！
※ 編輯: gemini822 來自: 1.165.168.148 (01/01 20:37)
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