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2Y(1+x)y'=4+Y^2 [2y/(4+y^2)]dy=[1/(1+x)]dx 兩邊求積分,得: ln(y^2+4)=ln(x+1)+C,C為常數 將x=3,y=2代入,得: C=ln2 所以:ln(y^2+4)=ln(x+1)+ln2 y^2+4=2(x+1) y^2=2x-2 y=+ -根號(2x-2)
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