∵dy/dx=sin(x-y)
==>dy=sin(x-y)dx
==>dx-dy=dx-sin(x-y)dx
==>d(x-y)=(1-sin(x-y))dx
==>d(x-y)/(1-sin(x-y))=dx
==>d(x-y)/(sin((x-y)/2)-cos((x-y)/2))^2=dx (应用(sin((x-y)/2)^2+(cos((x-y)/2)^2=1)
==>(sec((x-y)/2))^2d(x-y)/(tan((x-y)/2)-1)^2=dx (分子分母同除(cos((x-y)/2))^2)
==>2d(tan((x-y)/2)-1)/(tan((x-y)/2)-1)^2=dx
==>2/(1-tan((x-y)/2))=x+C (C是常数)
==>(x+C)(1-tan((x-y)/2))=2
∴原方程的通解是(x+C)(1-tan((x-y)/2))=2.
再问： 追问
这个您可以做吗？（3xy+2y^2+y）dx+(x^2+2xy+x+2y)dy=0求方程的一个积分因子>
再答： 此方程的积分因子是1/√(xy+y^2)。
∵(3xy+2y^2+y)dx+(x^2+2xy+x+2y)dy=0
==>(3xy+2y^2)dx+(x^2+2xy)dy+ydx+xdy+2ydy=0
==>(3xy+2y^2)dx+(x^2+2xy)dy+d(xy+y^2)=0
==>[(3xy+2y^2)dx+(x^2+2xy)dy]/√(xy+y^2)+d(xy+y^2)/√(xy+y^2)=0 (等式两端同乘1/√(xy+y^2))
==>2d[x√(xy+y^2)]+2d[√(xy+y^2)]=0
==>x√(xy+y^2)+√(xy+y^2)=C (C是常数)
==>(x+1)√(xy+y^2)=C
∴原方程的通解是(x+1)√(xy+y^2)=C。
再问： 谢谢了
