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4由y^2dx+xydy=0,得d(xy)=0,xy=c.
设y=c(x)/x,则dy=[c'(x)/x-c(x)/x^2]dx,
方程变为x^2+(c(x))^2/x^2+x+c(x)[c'(x)/x-c(x)/x^2]=0,
c(x)c'(x)=-x^3-x^2,
[c(x)]^2=(-1/2)x^4-(2/3)x^3+c,
c(x)=土√[(-1/2)x^4-(2/3)x^3+c],
所以y=土√[(-1/2)x^2-(2/3)x+c/x^2].
设y=c(x)/x,则dy=[c'(x)/x-c(x)/x^2]dx,
方程变为x^2+(c(x))^2/x^2+x+c(x)[c'(x)/x-c(x)/x^2]=0,
c(x)c'(x)=-x^3-x^2,
[c(x)]^2=(-1/2)x^4-(2/3)x^3+c,
c(x)=土√[(-1/2)x^4-(2/3)x^3+c],
所以y=土√[(-1/2)x^2-(2/3)x+c/x^2].
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