Question
Thu April 07, 2011 By: Rohit Varshney
 

Obtain he differential equation of the family of circles x^2+y^2+2gx+2fy+c=0; where g,f and c are arbitary constants

Expert Reply
Mon April 11, 2011
The given equation is
x2+y2+2gx+2fy+c=0       (1)
 
Differentiating once with respect to x, we get
2x+2yy'+2g+2fy'=0       (2)
 
Differentiating again, we get
2+2yy''+2(y')2+2fy''=0      
or, 1+yy''+(y')2+fy''=0      (3)
 
Differentiating once again, we get:
y'y''+yy'''+2y'y''+fy'''=0
or, yy'''+3y'y''+fy'''=0        (4)
 
Substituting the value of f from equation (2) in equation (4), we get the required differential equation as:
y''' (1+y'2) - 3y'y''2 = 0
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