Question

(y/x^2)+1+1/x dy/dx=0

Answer 1

Let us solve the differential equation $\frac{y}{x^2}+1+\frac{1}{x} \frac{dy}{dx}=0,$ which is equivalent to $\frac{1}{x}y'+\frac{y}{x^2}=-1.$ Let us multiply both parts of the equation by $x^2:$ $xy'+y=-x^2.$ The last equation is equivalent to $(xy)'=-x^2.$ It follows that $xy=-\frac{x^3}{3}+C.$ We conclude that the general solution is $y=-\frac{x^2}{3}+\frac{C}{x}.$