Question

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

Answer 1

The correct answer is: $log|x-y|=x+y+1$
$(x-y)(dx+dy)=dx-dy$
$⇒(x-y+1)dy=(1-x+y)dx$
$⇒\frac{dx}{dy}=\frac{1-x+y}{x-y+1}$
$⇒\frac{dy}{dx}=\frac{1-(x-y)}{1+(x-y)}...(1)$
Let $x-y=t.$
$⇒\frac{d}{dx}(x-y)=\frac{dt}{dx}$
$⇒1-\frac{dy}{dx}=\frac{dt}{dx}$
$⇒1-\frac{dt}{dx}=\frac{dy}{dx}$
Substituting the values of $x-y$ and $\frac{dy}{dx}$ in equation(1),we get:
$1-\frac{dt}{dx}=\frac{1-t}{1+t}$
$⇒\frac{dt}{dx}=1-(\frac{1-t}{1+t})$
$⇒\frac{dt}{dx}=\frac{(1+t)-(1-t)}{1+t}$
$⇒\frac{dt}{dx}=\frac{2t}{1+t}$
$⇒(\frac{1+t}{t})dt=2dx$
$⇒(1+\frac{1}{t})dt=2dx....(2)$
Integrating both sides,we get:
$t+log|t|=2x+C$
$⇒(x-y)+log|x-y|=2x+C$
$⇒log|x-y|=x+y+C...(3)$
Now,$⇒log|x-y|=x+y+C...(3)$
Therefore,equation(3)becomes:
$log\,1=0-1+C$
$⇒C=1$
Substituting $C=1$ in equation(3),we get:
$log|x-y|=x+y+1$
This is the required particular solution of the given differential equation.