Question

e^xy+ln(y/x+1)=0

就想看一下ln(y/x+1)怎么求，手写了一直凑不到答案

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Answer 1

e^(xy) + ln[y/(x+1)] = 0, x = 0 时，1+lny = 0， y = 1/e.
两边对 x 求导， 注意 y 是 x 的函数， 得
e^(xy)(y+xy') + [(x+1)/y][(x+1)y'-y]/(x+1)^2 = 0
(y+xy')e^(xy) + [(x+1)y'-y]/[y(x+1)] = 0
x = 0, y = 1/e 代入， 得
1/e + [y'(0)-1/e]/(1/e) = 0
1/e + ey'(0)-1 = 0, y'(0) = (e-1)/e^2

Answer 2

e^(xy) + ln[y/(x+1)]=0

带入 x=0

e^0 + ln[y(0)]=0

lny(0) = -1

y(0) = e^(-1)

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e^(xy) + ln[y/(x+1)]=0

d/dx { e^(xy) + ln[y/(x+1)] }=0 

(xy'+ y).e^(xy) + [(x+1)/y]. { [ (x+1)y' - y ]/(x+1)^2 } =0

带入 x=0 

(0+ y(0)).e^0 + [1/y(0)].  [ y'(0) - y(0) ]  =0

y(0)+ [1/y(0)].[ y'(0) - y(0) ]  =0

带入 y(0) =e^(-1)

e^(-1) + e.[ y'(0) - e^(-1) ]  =0

e^(-1) + e.y'(0) - 1  =0

e.y'(0) = 1-e^(-1)

y'(0) = (e-1)/e^2