1、首先求解
(x^2-1)y'+2xy=0
->dy/y+2xdx/(x^2-1)=0
->lny+ln(x^2-1)=C
->y(x^2-1)=C
然后
d[y(x^2-1)]=(x^2-1)dy+2xdx
对比左边,有
(x^2-1)dy+2xdx=cosxdx
->d[y(x^2-1)]=dsinx
->y(x^2-1)-sinx=C
->y=(C+sinx)/(x^2-1)
第二题:还是先解(x-2)y'-y=0
得到y/(x-2)=C
也就是d[y/(x-2)]=0
而d[y/(x-2)]=dy/(x-2)-y/(x-2)^2
于是原来的狮子就是
(x-2)^2d[y/(x-2)]=2(x-2)^3
也就是d[y/(x-2)]=2(x-2)=d[(x-2)^2]
因此y=(x-2)^3+C(x-2)