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Can someone please post a detailed step-by-step procedure. Given the circle with a radius a, what is the differential equation of the circle.

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quoting you, "what is the differential equation of the circle"? – Ilya Dec 3 '11 at 18:40
Does this other account belong to you as well? If so, we can flag the moderators to merge these two accounts; it'll make it convenient for you to keep track of your questions. Thanks, – Srivatsan Dec 3 '11 at 23:45
Nope, that account does not belong to me Srivatsan. – Nikhil Mulley Dec 4 '11 at 5:12

2 Answers 2

up vote 1 down vote accepted

From the implicit equation of the circle $(x-u)^2+(y-v)^2=a^2$, you get $$x'(x-u)+y'(y-v)=0$$ by implicit differentiation. Add the initial condition $$x(0)=u+a, \quad y(0)=v$$

You can write the differential equations as $$ x'=-y+v, \quad y' = x-u $$ which is especially nice for circles centered at the origin.

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Thanks Norbert and lhf. Appreciate your answers. I am however surprised to see clean mathematical notations here. Do you use any software as such to paste these equations here? (Pardon me, I am using math.stackexchange for the first time. – Nikhil Mulley Dec 3 '11 at 19:14
@Nikhil, this site understands TeX. View the source of the web pages to see how it's done. See also Welcome to MSE. – lhf Dec 3 '11 at 19:18
Thanks :-) BTW, I was also wondering if there is any kind of math application, which would give me sort of help to find the steps as in for example: Evaluate the integral x^2 (sqrt(a pow3 + x pow3)) dx. – Nikhil Mulley Dec 13 '11 at 12:46

Circle equation $$ (x-C_1)^2+(y-C_2)^2=a^2\quad (0) $$ Differentiate twice by $x$ $$ (x-C_1)+y'(y-C_2)=0\quad (1) $$ $$ 1+(y-C_2)y''+(y')^2=0\quad (2) $$ From $(2)$ we obtain $$ C_2=y+\frac{(y')^2+1}{y''} $$ Then substitute in $(1)$ and $(0)$ $$ (x-C_1)-y'\frac{(y')^2+1}{y''}=0\quad(3) $$ $$ (x-C_1)^2+\left(\frac{(y')^2+1}{y''}\right)^2=a^2\quad(4) $$ From $(3)$ we obtain $$ x-C_1=y'\frac{(y')^2+1}{y''} $$ Then substitute in $(4)$ $$ \left(y'\frac{(y')^2+1}{y''}\right)^2+\left(\frac{(y')^2+1}{y''}\right)^2=a^2 $$ After some simplifications we get $$ ((y')^2+1)^3=(ay'')^2 $$

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