Differential Equations - Arbitrary and fixed constants

Question

I am having a problem understanding,

1. Whether to keep a constant (arbitrary or fixed) in the solution of a differential equation
2. Figuring which is an arbitrary constant and which a fixed constant
3. How many times to differentiate the equation (equal to the number of constants?)

So in the above given questions, in the first one, that is 5th question. We take the original equation $ (x-h) ^2+(y-k) ^2 = a^2 $ and differentiate it twice (not thrice) and get the solution, keeping the constant a in the answer. So I assumed that a must not be an arbitrary constant.

But then in the next one, we take a as an arbitrary constant. Why? (as we differentiate the equation thrice) to ultimately get $ y3=0 $

And if in the last question, r is given so it cannot be an arbitrary constant. Then, the equation doesn't have any arbitrary constants at all. It would become $ x^2 + y^2 - 2ry =0 $ or $ x^2 + y^2 + 2ry =0 $ so it would be a fixed equation and not having any arbitrary constants. So how would we have a differential equation for it? And by my understanding it itself would be the differential equation with degree = 0

I want to know where all am I going wrong.

Thanks for the patience to understand. Let me know if there is some confusion

Answer 1

Using curvature formula for $y$ as a function of $x$:

$$\kappa = \frac{y''}{(1+y'^{2})^{3/2}}$$

For a sphere of radius $a$,

$$\kappa=\frac{1}{a}$$

Therefore

$$ \left[ 1+\left( \frac{dy}{dx} \right)^2 \right]^3 =a^{2} \left( \frac{d^{2} y}{dx^{2}} \right)^2$$

Alternatively, we may verify this by implicit differentiation.

\begin{align} 2(x-h)+2(y-k) \frac{dy}{dx} = 0 \\ \frac{dy}{dx} &= -\frac{x-h}{y-k} \\ 1+\left( \frac{dy}{dx} \right)^2+(y-k) \frac{d^2y}{dx^2} = 0 \\ \frac{d^2y}{dx^2} &= \frac{(x-h)^2+(y-k)^2}{(y-k)^3} \\ &= \frac{a^2}{(y-k)^3} \end{align}

What next is simply plug and play to check which option is the case.

Answer 2

5th Qn

The question must specify which constant is to be retained into the last differential equation. It should at least say as a given constant or capitalize it etc. to distinguish between them.

(I for one like) An equation of all circles of same radii $A$ should be

$$(x-h)^2+ (y-k)^2 = A^2$$

which has a DE second order. Caps means don't differentiate $A$ as an arbitrary constant

$$\kappa = \frac{y''}{(1+y'^{2})^{3/2}} =\frac{1}{A} $$

But DE of all circles in $x-y$ plane

$$(x-h)^2+ (y-k)^2 = a^2$$

would be

$$ y^{\prime \prime \prime}= \dfrac{3{ y^ {\prime\prime}}^2}{(y^{'}+1/y^{'})}$$

6th Qn

Second order with two differentiations gives a constant and once more a zero.

7th Qn

Your result is correct but $r$ is an arbitrary constant! Imagine a lot of circles touching x-axis at origin.

$$ \frac{x^2+y^2}{y} = const. $$

$$ \frac{dy}{dx} =\frac{2xy}{x^2-y^2}$$

So it is of first degree.