You should come up with the substitution such that: `x+y+2=t =gt x+y=t-2` `(dy)/(dx) = (t-2)/t`

You need to differentiate t with respect to x such that:

`(dt)/(dx) = 1 + (dy)/(dx) =gt (dy)/(dx) = (dt)/(dx)- 1`

You need to substitute `(dt)/(dx) - 1` for`(dy)/(dx) ` in equation `(dy)/(dx) = (t-2)/t` such that:

`(dt)/(dx) - 1 = (t-2)/t =gt (dt)/(dx) = (t-2)/t + 1`

`(dt)/(dx) = (t - 2 + t)/t =gt (dt)/(dx) = (2t-2)/t`

You need to separate the variables, hence, you should divide by `(2t-2)/t ` both sides such that:

`(tdt)/(2t-2) = dx`

You need to integrate both sides such that:

`(1/2)int (tdt)/(t-1) = int dx`

`(1/2)int ((t-1)dt)/(t-1) + (1/2)int (dt)/(t-1)= int dx`

`t/2 + (1/2)ln|t-1| = x + c`

You need to substitute x+y+2 for t in equation above such that:

`(x+y+2)/2 + (1/2)ln|x+y+2| = x + c `

`x + y + 2 + ln|x+y+2| = 2x + c`

`y + ln|x+y+2| = x - 2 + c`

`y = x - 2 - ln|x+y+2| + c`

**Hence, evaluting the solution to equation yields `y = x - 2 - ln|x+y+2| + c.` **