Starts with this implicit equation:
(x - a)^2 + y^2 == 1
(1) ( circles on x -axis ) $(x-a)^2+y^2=1$
Is the set of equations of the given circles. This set contains one parameter namely a
. So, it is the solution set of a differential equation of the first order. ( ?) Differentiating both members of (1) yields
(x - a) + y*Derivative[1][y] =0
(2) $(x-a)+y y'=0$
I must try to get a differential equation out of the implicit differentiation of equation(2) The problem is: how to do that?
-There is a total differential command.
Out of equations (1) and (2) can a be eliminated a, with should give this differential equation:
$y^2+y^2 \left(y'\right)^2=1$
Doing by hand equation(1) : $2 (x-a) (x-a)'+2 y y'=0$
Note: There are 2 methods to get y' out of a implicit equation
- differentiate both members of equation to x (note: beware of y)
- take from both members the differential ( to be done yet )