The correct option is C ln((y+3)2+(x+2)2)+2tan−1y+3x+2=C
The intersection of y−x+1=0 and y+x+5=0 is (−2, −3). Put x=X−2, y=Y−3.
The given equation reduces to dYdX=Y−XY+X.
Putting Y=vX, we get
XdvdX=−v2+1v+1
⇒(−vv2+1−1v2+1)dv=dXX
⇒−12ln∣∣v2+1∣∣−tan−1v=ln|X|+k,kis constant of integration
⇒ln∣∣Y2+X2∣∣+2tan−1YX=C,Here C=−2k
⇒ln∣∣(y+3)2+(x+2)2∣∣+2tan−1y+3x+2=C