Дифференциальное уравнение:1)y'+x/(x-2)=0. 2)2ydx+(y^2-6x)dy=0.

Решите задачу:

$1)\quad y'+\frac{x}{x-2}=0\\\\\frac{dy}{dx}=-\frac{x}{x+2}\\\\\int dy=-\int \frac{x}{x+2}\, dx\\\\\int dy=-\int (1-\frac{2}{x+2})dx\\\\y=-x+2\cdot ln|x+2|+C$

$2)\quad 2y\cdot dx+(y^2-6x)\cdot dy=0\\\\2y\cdot dx=-(y^2-6x)\cdot dy\\\\2y\cdot \frac{dx}{dy}=6x-y^2\\\\\frac{dx}{xy}=x'\; \; \; \to \; \; \; x'=\frac{6x-y^2}{2y}\; ,\; \; x'=3\cdot \frac{x}{y}-\frac{1}{2}\cdot y\\\\x'-3\cdot \frac{x}{y}=-\frac{y}{2}\quad linejnoe\; \; otnositelno\; \; x=x(y)\\\\x=uv\; ,\; \; x'=u'v+uv'\\\\u'v+uv'-3\cdot \frac{uv}{y}=-\frac{y}{2}\\\\u'v+u\cdot (v'-3\cdot \frac{v}{y})=-\frac{y}{2}$

$1)\; \; v'-\frac{3v}{y}=0\; ,\; \; \frac{dv}{dy}=\frac{3v}{y}\\\\\int\frac{dv}{v}=3\cdot \int \frac{dy}{y}\\\\ln|v|=3\cdot ln|y|\; \; \; \to \; \; \; v=y^3$

$2)\; \; u'\cdot y^3=-\frac{y}{2}\\\\\frac{du}{dy}=-\frac{1}{2y^2}\\\\\int du=-\frac{1}{2}\cdot \int \frac{dy}{y^2}\\\\u=-\frac{1}{2}\cdot (-\frac{1}{y})+C\\\\u=\frac{1}{2y}+C\\\\3)\; \; x=y^3\cdot (\frac{1}{2y}+C)\\\\x=\frac{y^2}{2}+Cy^3$