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dy/dx=y/(x+y的平方) 即 dx/dy=(x+y²;)/y=1/y*x+y 即 dx/dy-1/y*x=y 所以 x=e^(∫(1/y)dy)(∫ye^(∫-(1/y)dy)dy+c) =y(∫y*1/ydy+c) =y(∫dy+c) =y(y+c) 即 x=y²;+cy
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