求微分方程xdy-2[y+xy^2(1+lnx)]dx=0的通解

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  • x dy - 2[y + xy²(1 + lnx)] dx = 0

    x·dy/dx - 2y = 2xy²(1 + lnx)、两边除以xy²

    (1/y²)(dy/dx) - 2/(xy) = 2(1 + lnx)

    令z = 1/y、dz/dx = dz/dy·dy/dx = (- 1/y²)(dy/dx)、代入原式得

    - dz/dx - 2z/x = 2(1 + lnx)

    dz/dx + (2/x)z = - 2(1 + lnx)、e^[∫ 2/x dx] = e^(2lnx) = x²

    x²·dz/dx + 2xz = - 2x²(1 + lnx)

    d(x²z)/dx = - 2x²(1 + lnx)

    x²z = - 2∫ (x² + x²lnx) = - 2[x³/3 + (1/3)x³lnx - x³/9] + C

    x²z = - 4x³/9 - (2/3)x³lnx + C

    z = - 4x/9 - (2/3)xlnx + C/x²

    1/y = - 4x/9 - (2/3)xlnx + C/x²

    y = 9x²/(9C - 4x³ - 6x³lnx)