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Решите задачу:

$y'= \frac{x-y}{x+y} \\\\y'= \frac{1-\frac{y}{x}}{1+\frac{y}{x}}\\\\t= \frac{y}{x} \; ,\; \; y=tx\; ,\; \; y'=t'x+t\\\\t'x+t= \frac{1-t}{1+t} \; \; ,\; \; t'x=-t+ \frac{1-t}{1+t}\; \; ,\; \; t'x= \frac{-t-t^2+1-t}{1+t}\\\\t'x= \frac{-(t^2+2t-1)}{t+1}\; ,\; \; \frac{dt}{dx}\cdot x = -\frac{t^2+2t-1}{t+1} \; ,\\\\\int \frac{(t+1)dt}{t^2+2t-1} =- \int \frac{dx}{x}$

$\int \frac{(t+1)dt}{t^2+2t-1}=\int \frac{(t+1)dt}{(t+1)^2-2}=[\; u=t+1\; ,\; dt=du\; ]=\\\\=\int \frac{u\, du}{u^2-2} =\frac{1}{2}\cdot \int \frac{2u\; du}{u^2-2}= \frac{1}{2}\cdot ln|u^2-2|+C_1=$

$=\frac{1}{2}\cdot ln|(\frac{y}{x}+1)^2-2|+C_1\\\\\int \frac{dx}{x} =ln|x|+C_2\\\\ \frac{1}{2}\cdot ln|( \frac{y}{x}+1)^2-2|=-ln|x|+lnC\\\\\sqrt{( \frac{y}{x} +1)^2-2}=\frac{C}{x}$