Given, differential equation dxdy=2y−x+32x+y−3 Let x=X+h,y=Y+k, then dXdY=2(Y+k)−(X+h)+32(X+h)+(Y+k)−3=2Y−X+2k−h+32X+Y+2h+k−3 Let 2h+k−3=0 and −h+2k+3=0 Solving these two equations, we get h=59 and k=5−3 So, Eq. (i) becomes, dXdY=2Y−X2X+Y Let Y=vX⇒dXdY=v+XdX′dv Substituting this into Eq. (ii), we get ⇒xdxdv=2v−12+v−v⇒2v−12+v−2v2+v=2v−12v−2v2+2⇒−2v2+2v+22v−1dv=2v−12+v Integrating both sides, we get ∫−2v2+2v+22v−1dv=∫XdX Let u=−2v2+2v+2 Then du=(−4v+2)dv∴2−1∫udu=∫XdX⇒2−1ln∣u∣=ln∣X∣+c1, where c1= constant ⇒2−1ln∣−2v2+2v+2∣=ln∣X∣+c1⇒ln∣−2v2+2v+2∣=−2ln∣X∣+c2where c2=−2c1⇒ln∣X2(−2v2+2v+2)∣=c2⇒X2(−2v2+2v+2)=c, where c=ec2⇒X2(X2−2Y2+X2Y+2)=c⇒(−2Y2+2XY+2X2)=cSubstitute X and Y, we get−2(y+53)2+2(x−59)(y+53)+2(x−59)2=c⇒−2(y2+56y+259)+2(xy+53x−59y−2527)+2(x2−518x+2581)=c⇒−2y2−512y−2518+2xy+56x−518y−2554+2x2−536x+25162=c⇒−2y2+2xy+2x2−530y−530x+2590=c⇒2x2+2xy−2y2−6x−6y=c⇒x2+xy−y2−3x−3y=cThus, the general solution isx2+xy−y2−3x−3y+c=0