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.