Solve `(x-y)(dx+dy)=dx-dy`, given that `y=-1`, where `x=0`.

Answer

`log_(e)|x-y|=x+y+I`

Explanation

`(x-y)(dx+dy)=dx-dy`
or `(x-y+1)dy=(1-x+y)dx`
or `(dy)/(dx) = (1-x+y)/(x-y+1)`
Let `x-y=t`.
`therefore d/(dx)(x-y)=(1-t)/(1+t)`
or `(1+t)dt=2dx`
Integrating both sides, we get
`t+log|t|=2x+C`
or `(x-y)+log|x-y|=2x+C`
or `log|x-y|=x+y+C`
Now, `y=-1` at `x=0`.
Therefore, equation (2) becomes
`log1 =0-1+C`
`therefore C=1`
Substituting C=1 in equation (2), we get, `log|x-y|=x+y+1`. This is the required particular solution of the given differential equation.