微分方程解法(1-x^2)dy+(x+xy^2)dx=0

如题所述

解：∵(1-x^2)dy+(x+xy^2)dx=0
==>x(y^2+1)dx=(x^2-1)dy
==>xdx/(x^2-1)=dy/(y^2+1)
==>ln│x^2-1│=2arctany+ln│C│ (C是常数)
==>x^2-1=Ce^(arctany)
==>x^2=1+Ce^(arctany)
∴原方程的通解是x^2=1+Ce^(arctany)。

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