y=x+ 2 tan-¹(p), it is also written as

=> y–x=2tan-¹p, [because p=dy/dx ]

(y–x)/2=tan-¹(dy/dx),given differential equation

Let,y–x=2t,then differentiate w.r to x we get

=> dy/dx –1=2dt/dx

dy/dx=2dt/dx + 1

Then,by putting the value of dy/dx in the given differential equation we get

=> 2t/2 =tan-¹(2dt/dx + 1)

=> t=tan-¹(2dt/dx + 1)

=> tan t =2dt/dx + 1

=> (tan t – 1)=2dt/dx

=> dx=2dt/(tan t –1)

=> dx =2dt/(tant –1)

=> dx={(1–tant) +(1+tant)}dt/–(1–tant)

=> dx=–(1–tant)dt/(1–tant) – (1+ tant)dt/(1–tant)

=> dx=–dt – (1+ tant)dt/(1– tant)

Because,[(1+ tan∅)/(1–tan∅)=tan(π/4 + ∅)]

=> dx=–dt –tan(π/4 + t) dt

Then,integrating on both side we get

=> x=–t + log|cos(π/4 + t)| + c

By putting t=(y–x)/2,then we get

=> x=–(y–x)/2 + log|π/4 + (y–x)/2| + c

=> x +(y–x)/2=log|π/4 + (y–x)/2| + c

=> (2x+y–x)/2=log|π/4 + (y–x)/2| + c

=> y +x=2log|π/4 + (y–x)/2| + 2c

So,y+x=2log|π/4 +(y–x)/2|+C