(1)dy/(1-y²)=tanxdx
两边积分,1/2*ln|(y+1)/(y-1)|=ln|secx|+c1
(y+1)/(y-1)=csec²x
(2)对应的齐次方程为dy/dx=-2tanx*y
dy/y=-2tanxdx
ln|y|=2ln|cosx|+c1
y=ccos²x
把c换成u=u(x),则y'=u'cos²x+u*2cosx*(-sinx)
代入原方程得u'cos²x-2usinxcosx+2ucos²xtanx=sinx
u'=sinx/cos²x
u=1/cosx+c
∴y=cosx+ccos²x
