Solve dy/dx - (1+1/x)y=y^2

Question

It is in the form dy/dx +Py=Qy^n where P = -(1+1/x) and Q=y^2

Answer 1

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 e^x+ln(x)=e^xe^ln(x)=xe^x as the multiplier:

xe^xdv/dx+e^x(1+x)v=-xe^x,

d(xe^xv)/dx=-xe^x,

xe^xv=-∫xe^xdx.

Let u=x, du=dx; dv=e^xdx, v=e^x; xe^xv=-(xe^x-∫e^xdx)=-xe^x+e^x+C=e^x(1-x)+C.

Substituting v=1/y:

xe^x/y=e^x(1-x)+C, x/y=1-x+Ce^-x, y=x/(1-x+Ce^-x).