Re: [理工] [工數]-一階ODE
1. 應該是令ln(y/x)=u這樣算 可是答案不一樣 orz
看了幾次 感覺沒算錯 請高手指教
y
xy' = -------- y(2)=2
lny-lnx
let ln(y/x)=u
y=xe^u
y'=e^u+u'xe^u
e^u
(e^u+u'xe^u)= -------
u
1+u'x=1/u
u 1
-----du = ----- dx
1-u x
(-1+1/1-u)du= dx/x
-u-ln(1-u)=lnx +c'
ce^-u=x(1-u)
c x
-------- ---=x
1-ln(y/x) y
y(2)=2 c=2
y=2+yln(y/x)
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