Order of Differential Equation using number of arbitrary constants

Question

Consider
$$y=C_1 \sin ^{-1} x+C_2 \cos ^{-1} x+C_3 \tan ^{-1} x+C_4 \cot ^{-1} x$$ [ where $C _1, C _2, C _3$ and $C _4$ are arbitrary constants ]

It is a family of curves. When we convert it to Differential Equation form , then the Constants will be eliminated.

What is the Order of that Ordinary Differential Equation of that family of curves ?

My attempts :

$ \begin{aligned} & \text { Here, } y=C_1 \sin ^{-1} x+C_2 \cos ^{-1} x+C_3 \tan ^{-1} x+C_4 \cot ^{-1} x \\ & \Rightarrow y=C_1 \sin ^{-1} x+C_2\left(\frac{\pi}{2}-\sin ^{-1} x\right)+C_3 \tan ^{-1} x+C_4\left(\frac{\pi}{2}-\tan ^{-1} x\right) \\ & =\left(C_1-C_2\right) \sin ^{-1} x+\left(C_3-C_4\right) \tan ^{-1} x+\left(C_2+C_4\right) \frac{\pi}{2} \end{aligned} $

Unlike the Original Case , where it would seem to have 4 Constants , we now have only 3 Constants.
More-over , the Constants are inter-related & might eventually reduce to 2 Constants.

Hence , I am not reach to conclusion whether the Order will be 2 or 3 ?

Answer 1

Hints are available though the comments , thanks to Ninad Munshi & Sean Roberson , that the Order is $3$ , through the considerations of the linearly dependent solutions & the sums of constants.

Here is one alternate way to look at it :

When we have $y=f(x,C_1,C_2,C_3,C_4)$ & we want to eliminate the Constants to make the Ordinary Differential Equation , we can consider making it $y \equiv 0$ & count the Parameters which are free for us to control.

Here , we can eliminate the $\sin^{-1}()$ term by making $C_2=C_1$
We can then eliminate the Constant term by making $C_4=-C_2=-C_1$
Next , we can then eliminate the $\tan^{-1}()$ by making $C_3=C_4=-C_1$

Hence , when we choose some $C_1$ arbitrarily , we can make it $y \equiv 0$ where the other Constants are restricted by this $C_1$ , which is free to vary.

Hence the Order is $4-1=3$.

OBSERVATION :

When making it $y \equiv 0$ , if we can not even vary $C_1$ , then Order will be $4-0=4$.

In Case , we could vary 2 Constants while making it $y \equiv 0$ , then Order will be $4-2=2$.

Basically , we have to ignore the "Dependent Parameters" & count the "Constants" which are free to vary.

Answer 2

You've already found that the solution space consists of the members of the family $$y(x) = A \arcsin x + B \arctan x + C, \quad A, B, C \in \Bbb R .$$ The Wronskian determinant of the solutions $\arcsin x$, $\arctan x$, $1$ is nonzero in an open, dense subset of $(-1, 1)$, which shows that they form a linearly independent set there, hence the equation has order $3$.

Remark A standard computation gives that a linear, homogeneous differential equation with the specified solution space (on $(-1, 1)$) is $$(x^7 - 3 x^5 - x^3 + 3 x) y'''(x) + (4 x^6 - 19 x^4 + 6 x^2 - 3) y''(x) + (2 x^5 - 14 x^3) y'(x) = 0 .$$