dy+(y-(y^2)*cosx+(y^2)*sinx)dx=0y'+y-(y^2)*cosx+(y^2)*sinx=0-y'/y^2=1/y-cosx+sinx设z=1/y代入：z'=z-cosx+sinxz'=z的通z=Ce^x 令特解y=Asinx+Bcosx代入z'=z-cosx+sinx求得A=-1B=0通1/y=z=Ce^x-sinx...