dy/dx-(1+1/x)y=y2.
Let P(x)=-(1+1/x)=-(x+1)/x, and Q(x)=1 (not y2, because Qy2=y2), so we have dy/dx+Py=Qy2 (Bernouilli DE).
Rewriting the DE: y-2dy/dx+Py-1=Q.
Let v=y-1, dv/dx=-y-2dy/dx, so y-2dy/dx=-dv/dx, and:
-dv/dx+Py-1=Q, dv/dx-Pv=-Q, which can be solved by multiplying through by e∫-Pdx.
-∫Pdx=∫(1+1/x)dx=x+ln(x), so we have ex+ln(x)=exeln(x)=xex as the multiplier:
xexdv/dx+ex(1+x)v=-xex,
d(xexv)/dx=-xex,
xexv=-∫xexdx.
Let u=x, du=dx; dv=exdx, v=ex; xexv=-(xex-∫exdx)=-xex+ex+C=ex(1-x)+C.
Substituting v=1/y:
xex/y=ex(1-x)+C, x/y=1-x+Ce-x, y=x/(1-x+Ce-x).