The particular solution of differential equation \((\mathrm{x}+\mathrm{y}) \mathrm{dy}+(\mathrm{x}-\mathrm{y}) \mathrm{d} \mathrm{x}=0\) at \(\mathrm{x}=\mathrm{y}=1\) is

\(
(x+y) d y+(x-y) d x=0
\)
We have \(\frac{d y}{d x}=-\frac{(x-y)}{x+y}\)
Put \(y=v x \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}\)
\(
\therefore v+x \frac{d v}{d x}=\frac{-(x-v x)}{x+v x}=\frac{-x(1-v)}{x(1+v)}=\frac{-(1-v)}{1+v}
\)
\(\therefore x \frac{d v}{d x}=\frac{-1+v}{1+v}-v=\frac{-1+v-v-v^2}{1+v} \Rightarrow x \frac{d v}{d x}=\frac{-\left(1+v^2\right)}{1+v}\)
\(
\therefore \int \frac{(1+v)}{1+v^2} d v=-\int \frac{d x}{x} \Rightarrow \int \frac{d v}{1+v^2}+\frac{1}{2} \int \frac{2 v}{1+v^2} d v=-\) \(\int \frac{d x}{x}
\)
\( \therefore \tan ^{-1}(\mathrm{v})+\frac{1}{2} \log \left|1+\mathrm{v}^2\right|=-\log \mathrm{x}+\log \mathrm{c}_1 \)
\(\Rightarrow \tan ^{-1}\left(\frac{\mathrm{y}}{\mathrm{x}}\right)+\frac{1}{2} \log \left|1+\frac{\mathrm{y}^2}{\mathrm{x}^2}\right|=-\log \mathrm{x}+\log \mathrm{c}_1\)
\(
2 \tan ^{-1}\left(\frac{y}{x}\right)+\log \left|\frac{x^2+y^2}{x^2}\right|+2 \log x=\log c\) \(\ldots\left[\because \log c=2 \log c_1\right]
\)
\(
\therefore 2 \tan ^{-1}\left(\frac{y}{x}\right)+\log \left|\frac{x^2+y^2}{x^2} \times x^2\right|=\log c
\)