Question

Нужно решить 2 диф. уравнения (не система, отдельно) e^(-y) * (1 + dy/dx) = 1
y' = (x+y)/(x-y)

Answer 1

1.

${e}^{ - y} (1 + y') = 1 \\ 1 + y' = \frac{1}{ {e}^{ - y} } \\ 1 + y'= e {}^{y} \\ \frac{dy}{dx} = {e}^{y} - 1 \\ \int\limits \frac{dy}{e {}^{y} - 1 } = \int\limits \: dx \\ \\ \\ \int\limits \frac{dy}{e {}^{y} - 1 } \\ \\ {e}^{y} - 1 = t \\ e {}^{y} dy = dt \\ dy = \frac{dt}{e {}^{y} } = \frac{dt}{t + 1} \\ \\ \int\limits \frac{dt}{t(t + 1)} \\ \\ \frac{1}{t(t + 1)} = \frac{a}{t} + \frac{b}{t + 1} \\ 1 = a(t + 1) + bt \\ 1 = at + a + bt \\ \\ 1 = a \\ 0 = a + b \\ \\ a = 1\\ b = - 1 \\ \\ \\ \int\limits \frac{dt}{t} - \int\limits \frac{dt}{t + 1} = \\ = ln( |t| ) - ln( |t + 1| ) + C= \\ = ln( | {e}^{y} - 1| - ln( | {e}^{y} | ) + C= \\ = ln( | {e}^{y} - 1 | ) - 1 + C \\ \\ \\ \\ ln( | {e}^{y} - 1 | ) - 1 = x + C$

общее решение

2.

$y' = \frac{x + y}{x - y} \\ y' = \frac{1 + \frac{y}{x} }{1 - \frac{y}{x} }$

- однородное ДУ

Замена:

$\frac{y}{x} = u \\ y' = u'x + u$

$u'x + u = \frac{1 + u}{1 - u} \\ \frac{du}{dx} x = \frac{1 + u - u(1 - u)}{1 - u} \\ \frac{du}{dx} x = \frac{1 + u - u + u {}^{2} }{1 - u} \\ \int\limits \frac{1 - u}{u {}^{2} + 1 } du = \int\limits \frac{dx}{x} \\ \int\limits \frac{du}{u {}^{2} + 1 } - \frac{1}{2} \int\limits \frac{2udu}{u {}^{2} + 1} = ln( |x| ) + C \\ arctgu - \frac{1}{2} \int\limits \frac{d( {u}^{2} + 1) }{u {}^{2} + 1 } = ln( |x| ) + C \\ arctgu - \frac{1}{2} ln( | {u}^{2} + 1| ) = ln( |x| ) + C \\ arctg \frac{y}{x} - \frac{1}{2} ln( | \frac{ {y}^{2} }{ {x}^{2} } + 1 | ) = ln( |x| ) + C$

общее решение