Find a particular solution of the differential equation

Solution:

$(x-y)(d x+d y)=d x-d y$

$\Rightarrow(x-y+1) d y=(1-x+y) d x$

$\Rightarrow \frac{d y}{d x}=\frac{1-x+y}{x-y+1}$

$\Rightarrow \frac{d y}{d x}=\frac{1-(x-y)}{1+(x-y)}$                ...(1)

Let $x-y=t$.

$\Rightarrow \frac{d}{d x}(x-y)=\frac{d t}{d x}$

$\Rightarrow 1-\frac{d y}{d x}=\frac{d t}{d x}$

$\Rightarrow 1-\frac{d t}{d x}=\frac{d y}{d x}$

Substituting the values of $x-y$ and $\frac{d y}{d x}$ in equation (1), we get:

$1-\frac{d t}{d x}=\frac{1-t}{1+t}$

$\Rightarrow \frac{d t}{d x}=1-\left(\frac{1-t}{1+t}\right)$

$\Rightarrow \frac{d t}{d x}=\frac{(1+t)-(1-t)}{1+t}$

$\Rightarrow \frac{d t}{d x}=\frac{2 t}{1+t}$

$\Rightarrow\left(\frac{1+t}{t}\right) d t=2 d x$

$\Rightarrow\left(1+\frac{1}{t}\right) d t=2 d x$        ...(2)

Integrating both sides, we get:

$t+\log |t|=2 x+\mathrm{C}$

$\Rightarrow(x-y)+\log |x-y|=2 x+\mathrm{C}$

$\Rightarrow \log |x-y|=x+y+\mathrm{C}$                ...(3)

Now, $y=-1$ at $x=0$

Therefore, equation (3) becomes:

$\log 1=0-1+C$

$\Rightarrow C=1$

Substituting C = 1 in equation (3) we get:

This is the required particular solution of the given differential equation.