Question

Помогите пожалуйста

Answer 1

$x^2dy + (y - 1)dx = 0\\x^2dy = (1 - y)dx\\\frac{dy}{1 - y} = \frac{dx}{x^2}\\\int \frac{dy}{1 - y} = \int \frac{dx}{x^2}\\-\ln(1 - y) = -\frac{1}{x} + c_1\\y - 1 = c_2e^{\frac{1}{x}}, \quad (c_2 = e^{c_1})\\y = 1 + c_2e^{\frac{1}{x}}$