Question

Решить дифференциальное уравнение (x+1)(y' + y^2) = -y

Answer 1

$(x + 1)(y' + {y}^{2} ) = - y \\ y' + {y}^{2} = - \frac{y}{x + 1} \\ y'+ \frac{y}{x + 1} = - {y}^{2}$

Уравнение Бернулли

$|\div {y}^{2} \\ \\ \frac{ y'}{ {y}^{2} } + \frac{1}{y(x + 1)} = - 1 \\ \\ \frac{1}{y} = z \\ z' = - {y}^{ - 1} \times y. = - \frac{y'}{ {y}^{2} } \\ \frac{y'}{y {}^{2} } = - z' \\ \\ - z'+ \frac{z}{x + 1} = - 1 \\ z'- \frac{z}{x + 1} = 1$

Линейное ДУ

$z = uv \\ z' = u'v + v'u \\ \\ u'v + v'u - \frac{uv}{x + 1} = 1 \\ u'v + u(v' - \frac{z}{x + 1} ) = 1 \\ \\ 1)v' - \frac{v}{x + 1} \\ \frac{dv}{dx} = \frac{v}{x + 1} \\ \int\limits \frac{dv}{v} = \int\limits \frac{dx}{x + 1} \\ ln(v) = ln(x + 1) \\ v = x + 1 \\ \\ 2)u'v = 1 \\ \frac{du}{dx} \times (x + 1) = 1 \\\int\limits \: du = \int\limits \frac{dx}{x +1 } \\ u = ln(x + 1) + C\\ \\ z = (x + 1) \times ( ln(x +1 ) + C) \\ z = C(x + 1) + (x + 1) ln( x+ 1) \\ \frac{1}{y} = C(x + 1) + (x + 1) ln(x + 1)$

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