令y=ux
则y‘=u+xu’
带入原式
u+xu'=u+cosu
xu'=cosu
x* du/dx=cosu
du/cosu=dx/x
cosu*du/cosu ^2 = dx/x
dsinu/(1-sin^2)=dx/x
令sinu=v
1/2 ln [(v-1)/(v+1)]= lnx+C
v-1/v+1=Cx^2
v=sinu=sin (y/x)
则:(sin(y/x)-1)/(sin(y/x+1)=Cx^2