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Find a second degree differential equation describing family of circles: $$(x-a)^2 + (y-b)^2 = b^2.$$ I differentiated once and obtained: $$x'(t) (x(t) - a) + y'(t) (y(t) - b) = 0,$$ then differentiated for the second time and got: $$(x'(t))^2 + x(t)x''(t) - ax''(t) + (y'(t))^2 + y(t)y'(t) - by''(t) =0.$$ And...what? It doesn't make my work any easier, it doesn't simplify my first equation really... How to solve it? I would really appreciate any help.

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Are you supposed to find a differential equation of the form $F(x'(t),y'(t),t)=0$? Or of the form $F(y'(x),x)=0$? – Avi Steiner Mar 3 '13 at 19:01
You want to get rid of the constants $a$ and $b$ in your resulting second order ODE by combining the derivatives. – vonbrand Mar 3 '13 at 19:01
up vote 1 down vote accepted

Why don't you treat $y$ as a function of $x$ and differentiate three times with respect to $x$? Differentiating once gives $(x - a) + y'(y - b) = 0$. Differentiating again gives $$ 1 + y''(y - b) + (y')^2 = 0 \tag{1} \\ $$ $$ y - b = -\frac{1 + (y')^2}{y''}. \tag{2} $$ Differentiating $(1)$ gives the final equation we need $$ y'''(y - b) + y''y' + 2y'y'' = 0 $$ $$ y - b = -\frac{3y'y''}{y'''}. \tag{3} $$ We can equate $(2)$ and $(3)$ to get $$ \frac{1 + (y')^2}{y''} = \frac{3y'y''}{y'''} \\ y'''(1 + (y')^2) = 3y'(y'')^2 \\ y'''(1 + (y')^2) - 3y'(y'')^2 = 0 $$ whichever form you prefer.

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Is this a second degree equation? – Did Mar 5 '13 at 8:49
@Did Of course not. But if you want the differential equation for the family of all circles (which I did) then there are three constant: $a, b, r$, where $r$ is the radius (I am assuming that the equation written in the question has a typo for the radius squared). Thus, you can't have a second order equation. If the question did not have a typo, then it represents a smaller family of circles and to find a differential equation which are only satisfied by those circles would require a different approach. – Pratyush Sarkar Mar 6 '13 at 1:00
Sorry but there IS a second degree equation. Anyway, I guess my comment was rather directed at the OP for having accepted (pretty quickly) an answer (interesting but) in contradiction with (the title and the body of) their question. – Did Mar 6 '13 at 7:46
@Did Sorry that I was mistaken. Can you give me this second degree equation? I am interested. – Pratyush Sarkar Mar 6 '13 at 14:09
Let me suggest to have a look at my answer. – Did Mar 6 '13 at 14:10

$$x''(t)=-y'(t)\qquad y''(t)=x'(t)$$ $$\text{or, equivalently,}$$ $$ (x''(t)+y'(t))^2+(y''(t)-x'(t))^2=0$$

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