Question

1. y²dx = (1 + x²) dy
2. y²dx + (x-2)dy=0
3. (1+y)dx=(x-1)dy

Помогите решить

Answer 1

Ответ:

$1 ){y}^{2} dx = (1 + {x}^{2} )dy \\ \\ \int\limits \frac{dy}{ {y}^{2} } = \int\limits \frac{dx}{1 + {x}^{2} } \\ \frac{ {y}^{ - 1} }{ - 1} = arctg(x) + C \\ - \frac{1}{y} = arctg(x) + C \\ \frac{1}{y} = - arctg(x) + C$

$2) {y}^{2} dx + (x - 2)dy = 0 \\ (x - 2)dy = - {y}^{2} dx \\ \int\limits \frac{dy}{ {y}^{2} } = - \int\limits \frac{dx}{x - 2} \\ - \frac{1}{y} = - \int\limits \frac{d(x - 2)}{x - 2} \\ \frac{1}{y} = ln(x - 2) + C$

$3)(1 + y)dx = (x - 1)dy \\ \int\limits \frac{d(y + 1)}{y + 1} = \int\limits \frac{d(x - 1)}{x - 1} \\ ln(y + 1) = ln(x + 1) + ln(C) \\ ln(y + 1) = ln(C(x - 1)) \\ y + 1 = C(x - 1)$