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` ).

Explanation

Given differential equation is :
`(x-y)(dx+dy)=dx-dy`
`impliesdx+dy=(dx-dy)/(x-y)`
On integration , `int(dx+dy)=int(dx-dy)/(x-y)+C`
`implies x+y-log|x-y|+C`……….`(1)`
Given that, when `x=0` , then `y=-1`
`:. 0+(-1)=log(0+1)+CimpliesC=-1`
put this value in equation `(1)`,
`x+y=log|x-y|-1`
`implies log|x-y|=x+y+1`
which is the required particular solution.