xdy-[y+xy^3(1+lnx)]dx=0的通解

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xdy-[y+xy^3(1+lnx)]dx=0的通解
xdy-[y+xy^3(1+lnx)]dx=0的通解

xdy-[y+xy^3(1+lnx)]dx=0的通解
令z=1/y²,则dy=-y³dz/2
代入原方程,化简得xdz+2zdx=-2x(1+lnx)dx
==>x²dz+2xzdx=-2x²(1+lnx)dx
==>d(x²z)=-2x²(1+lnx)dx
于是,x²z=-2∫x²(1+lnx)dx
=-2[x³(1+lnx)/3-(1/3)∫x²dx] (应用分部积分法)
=C/9-2x³(2+3lnx)/9 (C是任意常数)
==>x²/y²=C/9-2x³(2+3lnx)/9
==>9x²=[C-2x³(2+3lnx)]y²
故原方程的通解是9x²=[C-2x³(2+3lnx)]y².