Differential Equation

Differential Equation: $(1-xy)^{-2} dx + \left[ y^2 + x^2 (1-xy)^{-2} \right] dy = 0$

Hello, can anyone solve this equation?

I can't figure it out,

$(1-xy)^{-2} dx + \left[ y^2 + x^2 (1-xy)^{-2} \right] dy = 0$

Thanks.

DE: $(x²+4) y' + 3 xy = x$

(x²+4) y' + 3 xy = x

bernoulli: $(y^4 - 2xy) dx + 3 x^2 dy= 0$

(y^4 - 2xy) dx + 3 x^2 dy= 0

Differential Equation $2y \, dx+x(x^2 \ln y -1) \, dy = 0$

2ydx+x(x2lny -1)dy=0

yexydx+xexydy=0

DE: $x \, dx + [ sin^2 (y/x) ](y \, dx - x \, dy) = 0$

xdx + sin^2 ( y/x ) [ ydx - xdy ] = 0

differential equation: Determine whether a member of the family can be found that satisfies the initial conditions

The given two-parameter family is a solution of the indicated differential equation on the
interval (−∞,∞). Determine whether a member of the family can be found that satisfies the
initial conditions.
y = c1e^x cos x + c2e^x sin x; y" − 2y' + 2y = 0,
i. y(0) = 1, y'(π) = 0
ii. y(0) = 1, y(π) = −1