ordinary differential equations
How do I find the DE of all circles passing through origin? I tried something like this
The family of circles passing through the origin is given by
$$
(x- r \cos \theta )^2 + (y- r\sin \theta)^2 = r^2
$$Differentiating once, we get
$$
2 ( (x- r \cos \theta) + (y – r \sin \theta) y') = 0
$$
Differentiating again, we get
$$
1 + y'^2 + (y – r \sin \theta) y'' = 0
$$
How to get rid of $r$ and $\theta$ algebraically? Is there any other approach?
A hint:
Eliminate $C_1$ from the first and the second equation.
Note that exactly two circles pass through each point in the strip $-1<x<1$. Therefore we should expect a differential equation of the form $y'^2=\ldots\ $. A second order differential equation has infinitely many solution curves through each point $(x_0,y_0)$ in its domain.
The equation of a circle with centre at $(h, k)$ and radius equal to $a$, is $$(x - h)^2 + (y - k)^2 = a^2~.$$
When the circle passes through the origin and centre lies on $x-$axis, then the abscissa will be equal to the radius of the circle and the $y-$co-ordinate of the centre will be zero. i.e., $h = a$ and $k = 0$.
Then the equation of the circle satisfied the given condition is$$(x - a)^2 + y^2 = a^2\tag1$$
Differentiating both side of $(1)$ with respect to $x$ , $$2(x-a)+2yy'=0\implies x-a=-yy'\implies a=x+yy'$$where $~y'\equiv\dfrac{dy}{dx}~$.
Putting the value of $~a~$, in equation $(1)$ , $$y^2~y'^2 + y^2 = (x+yy')^2$$ $$\implies y^2~y'^2 + y^2 = x^2+2xyy'+y^2y'^2$$ $$\implies y^2 = x^2+2xyy'$$ which is the required differential equation of the given family of circles.
Best Answer
For a circle polar coordinates is a natural choice. It avoids differential relations between $ (x,\theta) $. It is convenient to find a pure polar equation of a circle of arbitrary radius (or diameter D ) whose inclination $ \alpha$ to a reference direction like x-axis is arbitrary. Curvature is a second order differential and hence these arbitrary constants $ ( D, \alpha) $ need to be involved and later on eliminated in differentiation:
From the sketch it is seen they are relatively positioned and geometrically related as:
$$ r = D \cos(\theta + \alpha) $$
whose differential equation is the well known:
$$ \boxed {\frac{d^2 r}{d \theta^{2}} + r =0.} $$
Rectangular coordinates to be computed after numerical integration.
EDIT 1:
In rectangular coordinates we can derive as follows. $\alpha$ can be taken wlog as zero as seen above, anyhow it disappears.
$$ x = D \cos ^2\theta=\frac{D\,x^2}{x^2+y^2} $$
$$ x^2 + y^2 = D\,x $$
Differentiating once
$$ x + y\,y^{\prime} = D/2 $$ Differentiate again
$$\boxed {y\, y^{\prime \,\prime} + 1 + y^{\prime\, 2 }=0.}$$
One disadvantage with rectangular coordinates numerical integration is that it cannot proceed beyond point of the vertical tangent.. unless other tricks are used.