Solving this differential equation IVP

Question

I'm not sure how to approach this problem. Please help with this.

dy/dx +y = xy^2 - 1/x^2

The 1/x^2 is really throwing me off.

Answer 1

We revrite equation like dy/dx + 1/x² = xy² - y;

d/dx(y - 1/x) = yx(y - 1/x);

Now new function y = v + 1/x; we ger dv/dx = (v + 1/x)xv, dv/dx - v = xv²;

We get Bernully equation: u = v ^{- 1}, du/dx = - 1/v²dv/dx;

du/dx = - v ^{- 2}dv/dx; dv/dx = - v²du/dx; - v²du/dx - v = xv²; du/dx + u = - x.

We get linear nongomogenius equation for function u: Integrated factor μ= e^∫1dx = e^x;

e^x(du/dx + u) = - xe^x; d/dx(e^xu) = - xe^x; e^xu = - ∫xe^xdx; e^xu = - (xe^x - e^x) + C; u = 1 - x + Ce^{- x};

y = 1/u + 1/x = 1/x + 1/(1 - x + Ce^-x)