Question

Find a particular solution of the differential equation $(x-y)(dx+dy)=dx-dy$ given that y = -1, when x = 0.

Answer 1

Given, differential equation is $(x-y)(dx+dy)=dx-dy$
$\Rightarrow dx+dy=\frac{dx-dy}{x-y}$
On integrating both sides, we get $\int (dx+dy)=\int \frac{dx-dy}{x-y}+C$
Let $x-y=t\Rightarrow dx-dy=dt$
$\therefore \int dx+dy-C=\int \frac{dx-dy}{x-y}=\int \frac{dt}{t}=logt=log|x-y|\Rightarrow x+y=log|x-y|+C\dots \dots \left(i\right)$
It is given that when x = 0, y = -1
$\therefore 0+(-1)=log(0+1)+C\Rightarrow C=-1$
On substituting this value in Eq. (i), we get the required particular solution as $x+y=log|x-y|-1\Rightarrow log|x-y|=x+y+1$