(x-2y)dy+dx=0的通解求求怎么写啊

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∵(x-2y)dy+dx=0
==>xe^ydy-2ye^ydy+e^ydx=0 (等式两端同乘e^y)
==>xd(e^y)+e^ydx=2yd(e^y)
==>d(xe^y)=2yd(e^y)
==>∫d(xe^y)=∫2yd(e^y)
==>xe^y=2(y-1)e^y+C (C是常数)
==>x=2(y-1)+Ce^(-y)
∴原方程的通解是x=2(y-1)+Ce^(-y).