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Obtain the D.E. having a solution as the equation representing all circles whose radius 1 and center's on the line $y=x$.

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closed as off-topic by Danny Cheuk, Dan Rust, Shuhao Cao, Start wearing purple, Potato Jul 1 '13 at 4:56

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I don't understand -- these circles overlap each other so can't be common solutions to one (reasonable) ODE. – user29743 Jun 30 '13 at 14:46
@countinghaus Of course they can. If you denote by $(a,a)$ the position of the center, you can think of $a$ as being 1st integral of the equation. – Start wearing purple Jun 30 '13 at 14:50

From the geometrical description it follows that the domain $\Omega$ of this differential equation is given by $\Omega:=\{(x,y)\>|\>|x-y|\leq\sqrt{2}\}$.

Let a point $(u,v)\in\Omega$ be given, and assume that it lies on the circle $$\gamma_a:\quad(x-a)^2+(y-a)^2=1\ .\tag{1}$$ In the neighborhood of $(u,v)$ this circle can be viewed as the graph of a function $$x\mapsto y(x),\qquad y(u)=v\ ,$$ and in view of $(1)$ we have $$(x-a)^2+\bigl(y(x)-a\bigr)^2\equiv1\ .$$ This implies $$2(x-a)+2\bigl(y(x)-a\bigr)y'(x)\equiv0\ .$$ Putting $x:=u$ here we find that our circle $\gamma_a$ at $(u,v)$ has the slope $$y'=-{u-a\over v-a}\ .\tag{2}$$ But note that there are actually two circles $\gamma_a$ passing through the given point $(u,v)\in\Omega$. Their "addresses" $a_1$, $a_2$ are the solutions of the equation $$(u-a)^2+(v-a)^2-1=0\ ,$$ or $$a_{1,2}={1\over2}\bigl(u+v\pm\sqrt{2-(u-v)^2} \bigr)\ .$$ Plugging this into $(2)$ we obtain $$y'(u,v)=-{u-v\mp\sqrt{2-(u-v)^2}\over v-u\mp\sqrt{2-(u-v)^2}}\ .$$ Writing now $(x,y)$ instead of $(u,v)$ we finally have the DE $$y'={y-x\pm\sqrt{2-(y-x)^2}\over y-x\mp\sqrt{2-(y-x)^2}}\ .$$

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So, the equation of the family of circles can be $$(x-a)^2+(y-a)^2=1\ \ \ \ (1)$$

Any point on the circle can be $P(a+\cos t,a+\sin t)$

$x=a+\cos t,y= a+\sin t$

$\implies dx=-\sin t dt,dy =\cos tdt$ and $\frac{dy}{dx}=\frac{dy}{dt}/\frac{dx}{dt}=-\cot t$

Now, $x-\cos t=y-\sin t\implies \sin t-\cos t=y-x$

$\implies (y-x)^2=1-2\sin t\cos t$

$$\text{Now, }\sin t\cos t=\frac{\frac{\sin t\cos t}{\sin^2t}}{\frac1{\sin^2t}}=\frac{\cot t}{1+\cot^2t}=\frac{-\frac{dy}{dx}}{1+\left(-\frac{dy}{dx}\right)^2} $$

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