Solve the following differential equations: `(x+y)(dx-dy)=dx+dy`

Explanation

Given differential equation is
`" "(x+y)(dx-dy)=dx+dy`
`rArr" "(x+y)(1-(dy)/(dx))=1+(dy)/(dx)" "...(i)`
Put `" "x+y=z`
`rArr" "1+(dy)/(dx)=(dz)/(dx)`
On substituting these values in Eq. (i), we get
`" "z(1-(dz)/(dx)+1)=(dz)/(dx)`
`rArr" "z(2-(dz)/(dx))=(dz)/(dx)`
`rArr" "2z-z(dz)/(dx)-(dz)/(dx)=0`
`rArr" "2z-(z+1)(dz)/(dx)=0`
`rArr" "(dz)/(dx)=(2z)/(z+1)`
`rArr" "((z+1)/(z))dz=2dx`
On integrating both sides, we get
`" "int(1+(1)/(z))dz=2intdx`
`rArr" "z+logz=2x-logC`
`rArr" "(x+y)+log(x+y)=2x-logC" "[because z=x+y]`
`rArr" "2x-x-y=logC+log(x+y)`
`rArr" "x-y=log|C(x+y)|`
`rArr" "e^(x-y)=C(x+y)`
`rArr" "(x+y)=(1)/(C)e^(x-y)`
`rArr" "x+y=Ke^(x-y)" "[becauseK=(1)/(2)]`