Please help me solve this differential equation: dy/dx-(1+1/x)y=y^2

Answer 1

This is just a special case of the Bernoulli Equation

y' + p(x) y = q(x) y^n

where q(x) = 1

Make the substitution v = y^(1-n)
This will give you

-v' + (1+1/x)v = 1

Now you can multiply by an integrating factor

e^[∫(1+1/x) dx] = xe^x to get

xe^x v = -e^x (x-1)

and you're almost done

Answer 2

Hi..
I'm not sure how to get the final answer