Question

xdy+(x+y)dx=0 помогите пожалуйста срочно!!!

Answer 1

Ответ:

$x\, dy+(x+y)\, dx=0\\\\\dfrac{dy}{dx}=-\dfrac{x+y}{x}\ \ ,\ \ \ y'=-1-\dfrac{y}{x}\\\\\\\dfrac{y}{x}=t\ \ ,\ \ y=tx\ \ ,\ \ y'=t'x+t\\\\\\t'x+t=-1-t\ \ ,\ \ t'x=-1-2t\ \ ,\ \ \ \dfrac{dt}{dx}=\dfrac{-1-2t}{x}\\\\\\\int \dfrac{dt}{-1-2t}=\int \dfrac{dx}{x}\\\\\\-\dfrac{1}{2}\, ln|-1-2t|=ln|x|+lnC\\\\\\\dfrac{1}{\sqrt{-1-2t}}=Cx\ \ ,\ \ \sqrt{-1-2t}=\dfrac{1}{Cx}\ \ ,\ \ -1-2t=\dfrac{1}{C^2x^2}\ \ ,\ \ t=-\dfrac{1}{2}-\dfrac{1}{2C^2x^2}\ ,\\\\\\\dfrac{y}{x}=-\dfrac{1}{2}-\dfrac{1}{C^2x^2}$

$y=-\dfrac{x}{2}-\dfrac{1}{2C^2x}\ \ \ \to \ \ \ \ \boxed {\ y=-\dfrac{x}{2}-\dfrac{1}{C_{1}\, x}\ }\ \ ,\ C_1=2C^2$