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
