求微分方程y'-ytanx=1∕cosx的通解

e^∫-tanxdx=e^∫-sinx/cosxdx=e^∫dcosx/cosx=e^lncosx=cosx
y'－ytanx＝cosx
cosxy'－ytanxcosx＝cos²x
cosxy'－ysinx＝cos²x
(cosxy)`=cos²x
cosxy=∫cos²xdx+C
=1/2∫(1+cos2x)dx+C
=1/2(x+1/2∫cos2xd2x)+C
=1/2x+1/4∫cos2xd2x+C
=1/2x+1/4sin2x+C
y=(1/2x+1/4sin2x+C)/cosx