(x+y)y'+1=2e^(-y)(x+y)dy+dx=2e^(-y)dx(xe^ydy+e^ydx)=2dx-ye^ydyd(xe^y)=d2x-ye^ydy通解xe^y=2x-ye^y+e^y+Cx=1,y=0C=-1特解 xe^y=2x-ye^y+e^y-1∫ye^ydy=∫yde^y=ye^y-e^y
