Question

xdy + (y-x^2y^2)dx = 0

Answer 1

$$y'+\dfrac{1}{x}y=xy^2$$

First order Bernoulli ODE

$$t=y^{1-n}=y^{1-2}=y^{-1}$$

$$y=t^{-1}, y'=-t^{-2}t'$$

$$-t^{-2}t'+\dfrac{1}{x}t^{-1}=xt^{-2}$$

$$-t'+\dfrac{1}{x}t=x$$

$$\mu=\dfrac{1}{x}$$

$$-\dfrac{1}{x}t'+\dfrac{1}{x^2}t=1$$

$$(\dfrac{1}{x}t)'=\dfrac{1}{x}t'-\dfrac{1}{x^2}$$

$$(\dfrac{1}{x}t)'=-1$$

$$\dfrac{1}{x}t=-x+c_1$$

$$t=-x^2+c_1x$$

$$y=\dfrac{1}{-x^2+c_1x}$$