求微分方程xdy-2[y+xy^2(1+lnx)]dx=0的通解
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求微分方程xdy-2[y+xy^2(1+lnx)]dx=0的通解
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)
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)
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