求微分方程的通解（1+x2）ydy-x（1+y2）dx＝0

网友回答

(1 + x²)y dy - x(1 + y²) dx = 0
(1 + x²)y dy = x(1 + y²) dx
y/(1 + y²) dy = x/(1 + x²) dx
(1/2)ln(1 + y²) = (1/2)ln(1 + x²) + C
ln(1 + y²) = ln(1 + x²) + C
1 + y² = e^[ln(1 + x²) + C]
y² = [(1 + x²) * e^C] - 1