y'ï¼xï¼â(x^2ï¼y) 设â(x^2ï¼y)-x=u, x^2ï¼y=x^2ï¼2xu+u^2 y'=2u+2xu'+2uu' 代å
¥å¾ï¼ u=2u+2xu'+2uu' u'=-u/(2u+2x) æï¼dx/du+2x/u=-2 è¿æ¯xä½ä¸ºå½æ°ãuä½ä¸ºåéçä¸é¶çº¿æ§å¾®åæ¹ç¨ï¼ç±é解å
¬å¼ï¼ x=(1/u^2)(C-(2/3)u^3) xu^2+(2/3)u^3=C 代å
¥â(x^2ï¼y)-x=uï¼ C=(2/3)u^2(3x/2+u) =(2/3)(â(x^2ï¼y)-x)^2(x/2+â(x^2ï¼y)) C=(2/3)[(x^2ï¼y)-2xâ(x^2ï¼y)+x^2](x/2+â(x^2ï¼y)) =(2/3)(x(x^2ï¼y)/2+(x^2+y)^(3/2)-x^2â(x^2ï¼y)-2x(x^2ï¼y)+x^3/2+x^2â(x^2ï¼y)) =(2/3)((x^2+y)^(3/2)-x^3-(3/2)xy)
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