求(2xy^2+y)dx+(x+2x^2y-x^4y^3)dy=0的通解

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  • ∵(2xy^2+y)dx+(x+2x^2y-x^4y^3)dy=0

    ==>(ydx+xdy)+(2xy^2dx+x+2x^2y)-x^4y^3dy=0

    ==>d(xy)+d((xy)^2)-x^4y^3dy=0

    ==>-d(xy)-d((xy)^2)+x^4y^3dy=0

    ==>-d(xy)/(xy)^4-d((xy)^2)/((xy)^2)^2+dy/y=0 (等式两端同除(xy)^4)

    ==>(1/3)/(xy)^3+1/(xy)^2+ln│y│=ln│C│ (等式两端积分,C是常数)

    ==>(xy+1/3)/(xy)^3+ln│y│=ln│C│

    ==>ye^((xy+1/3)/(xy)^3)=C

    ∴原方程的通解是ye^((xy+1/3)/(xy)^3)=C。