先求通解y'+ysinx=0y'/y=-sinxdxlny=cosx+C1y=Ce^cosx运用常数变异法y=C(x)e^cosxy'=C'(x)e^cosx-C(x)e^cosx*sinx代入原方程得C'(x)e^cosx-C(x)e^cosx*sinx+C(x)e^cosx*sinx=sin^3xC'(x)e^cosx=sin^3xC'(x)=sin^3x*e...