第3个回答 2018-01-11
(7)y'=2xy/(x²+y²)=2(y/x)(1+(y/x)²)
设y/x=k,y=kx,y'=k'x+k,k=y/x,k'=(y'x-y)/x²=(yy'/k-y)/(y²/k²)
=(y'-k)/(y/k)=k(y'-k)/y
y'=2k/(1+k²)
k'=k(2k/(1+k²)-k)/kx
k'=k(2/(1+k²)-1)/x
k'x=2k/(1+k²)-k
k'/[2k/(1+k²)-k]=1/x
(1+k²)k‘/(2k-k-k³)=1/x
(1+k²)k'/(k-k³)=1/x
设(1+k²)/k(1+k)(1-k)=A/k+B/(1+k)+C/(1-k)
=[A(1-k²)+B(k-k²)+C(k+k²)]/(k-k³)
=(A+(B+C)k+(C-B)k²)/(k-k³)
A=1,B+C=0,C-B=1,C=1/2,B=-1/2
[1/k-1/2(1+k)-1/2(k-1)]k'=1/x
两边积分:
lnk-(1/2)ln(1+k)-(1/2)ln(k-1)=lnCx
ln[k/√[(1+k)(k-1)]=lnCx
k/√(k²-1)=Cx
k²/(k²-1)=C²x²
k²/(-1)=C²x²/(1-C²x²)
k²=C²x²/(C²x²-1)
y/x=Cx/√(C²x²-1)
y=Cx²/√(C²x²-1)