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 )

1 Answer 1


The first equation is:

eqn1 = (x - a)^2 + y^2 == 1;

ReplaceAll y to create a function y[x] and do the derivative w.r.t x to get the second equation:

eqn2 = D[eqn1 /. y -> y[x], x]

Eliminate the parameter a

Eliminate[{eqn1 /. y -> y[x], eqn2}, a]


  • $\begingroup$ BMF ,thanks. That is the right differential equation! Understanding implicit differentation in MMA is not that easy $\endgroup$
    – janhardo
    Apr 28 at 20:26
  • $\begingroup$ @janhardo don't mention. Glad I was able to help. You can DSolve the last result just so you know :) $\endgroup$
    – bmf
    Apr 28 at 20:28
  • $\begingroup$ After spending some time working with the y[x] notation in MMA, I think it is more convenient to just use y as books do. Makes it more complicated the y[x] notation and more work , agree? $\endgroup$
    – janhardo
    May 2 at 19:56

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