Differentiating both sides of we get or equivalently (this is the solution)
Differentiating w.r.t.x we have
2x-2+2y dy/dx-2dy/dx =0 So,x - 1+ydy/dx-dy/dx =0
So x-1+(y-1)dy/dx=0
dy/dx=1-x/y-1
differentiating again w.r.t.x,we get
1+(y-1)d²y/dx² +(dy/dx)^2=0
Now how to calculate required differential equation.
If my calculations of 1st and 2nd order derivative is wrong, reply me accordingly.
Sir,
You are correct. Original problem is as follows:
Find the differential equation of all circles having a fixed radius a.
Answer given in the study material is as below:
{1+ (dy/dx)^2}^3=a^2{d²y/dx²}^2
Hence give me a hint to arrive at the above answer.
Regards
Sir,
I found the differential equation of all circles having fixed radius a.First derivative of equation of these circles is
Second derivative of equation of these circles is
d²y/dx²=[-(dy/dx)^2-1]/(y-k)
Differential equation of all circles having fixed radius a is given in post number #5 to this thread.
However thanks for your reply.