解：∵y'=sin²(x-y+1)
==>1-d(x-y+1)/dx=sin²(x-y+1) (∵d(x-y+1)/dx=1-dy/dx)
==>d(x-y+1)/dx=1-sin²(x-y+1)
==>d(x-y+1)/dx=cos²(x-y+1)
==>d(x-y+1)/cos²(x-y+1)=dx
==>tan(x-y+1)=x+C (对等式两端积分，C是常数)
∴原方程的通解是tan(x-y+1)=x+C。
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