calculusderivativesordinary differential equations
Form the differential equations of the family of circles touching two given parallel lines $y=\pm a$
My Attempt:
Let the family of circles be $(x-h)^2+y^2=a^2$, with $(h,0)$ being the centre and $a$ being the radius.
Differentiating w.r.t. x, we get,
$$2(x-h)+2yy'=0\\ \implies yy'+x=h$$
Differentiating again, I get,
$$yy''+(y')^2+1=0$$
But the answer given is
$$y^2(1+(y')^2)=a^2$$
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.
Note that when we are given a differential equation containing $n$ arbitrary constants, by differentiating it successively $n$ times and eliminating the constants gives us the required differential equation of order $n$.
Here, $n=1$. Simplifying $(1)$ by taking the square root, we get: $$8y^2 = y’(4xy-y’^2) \implies y’^3=4xyy’-8y^2 = 4y(xy’-2y)$$ that is $(2)$.
Note that, simplifying $(1)$ gave us the simplest form of the differential equation. But, it is always preferred to differentiate the equation and eliminate constants.
Best Answer
Your equation, the solution of which depends on two arbitrary constants, cannot be an acceptable differential equation defining the family (one of the constants being $a$, that is determined, and the other $h$). We can use the equation for the family to get rid of the constant.
$$(x-h)^2+y^2=a^2$$
$$2(x-h)+2yy'=0\\ \implies (x-h)^2=y^2(y')^2$$
But $(x-h)^2=a^2-y^2$
$$\dfrac{a^2-y^2}{y^2}=(y')^2$$
$$\dfrac{a^2}{y^2}-1=(y')^2$$
Finally $$a^2=y^2(1+(y')^2)$$