
求这个题的做法 5
1个回答
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y|x=0 =1
y' -(1/2)y = (1/2)e^x
let
y= Ae^(x/2) + Be^x
y|x=0 =1
A+B=1 (1)
y'=(A/2)e^(x/2) + Be^x
y' -(1/2)y = (1/2)e^x
(A/2)e^(x/2) + Be^x - (1/2)(Ae^(x/2) + Be^x) = (1/2)e^x
(B/2)e^x = (1/2)e^x
=>B=1
from (1)
A=0
ie
y= e^x
y' -(1/2)y = (1/2)e^x
let
y= Ae^(x/2) + Be^x
y|x=0 =1
A+B=1 (1)
y'=(A/2)e^(x/2) + Be^x
y' -(1/2)y = (1/2)e^x
(A/2)e^(x/2) + Be^x - (1/2)(Ae^(x/2) + Be^x) = (1/2)e^x
(B/2)e^x = (1/2)e^x
=>B=1
from (1)
A=0
ie
y= e^x
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