Finding the $n$-th derivative and proving a given condition The question that is getting me confused is: if $y = (\sin^{-1} x)^2$, prove that $(1-x^2) (y_{n+2}) - (2n+1)x y_{n+1}  + n^2 y_{n} = 0$.
Could you please explain the problem solving tricks used for such questions so that I can tackle similar questions. Thank you in advance.
 A: This is closest to an answer I can get for the current form of the question. Perhaps someone else will be able to help you more.
If I did not make a mistake below, it seems that the equality you want to prove is not true.

First the easy part. Let us have a look at the equation
$$(1-x^2) y_{n+2} - (2n+1)x y_{n+1}  - n^2 y_{n} = 0$$
(Notice that I have changed sign in one place; I have here $-n^2y_n$ instead of $+n^2y_n$.)
You did not say that in your post, but I assume that $y_n$ stands for $n$-th derivative of the given function.
A natural approach to prove things like this might be induction. Let us ignore the case $n=0$ at the moment.
The inductive step is relatively easy.
If you already know that
$$(1-x^2) y_{n+2} - (2n+1)x y_{n+1}  - n^2 y_{n} = 0$$
then simply by differentiating both sides you get
$$
\begin{align}
(1-x^2) y_{n+3} - 2xy_{n+2} -  (2n+1)x y_{n+2} - (2n+1) y_{n+1} - n^2
y_{n+1} &= 0\\
(1-x^2) y_{n+3} - (2n+3)x y_{n+2} - (n^2+2n+1) y_{n+1} &= 0\\
(1-x^2) y_{n+3} - (2n+3)x y_{n+2} - (n+1)^2 y_{n+1} &= 0
\end{align}
$$
So now it remains to check whether this is true for $n=0$. Then the proof would be finished.
I.e., we want to check whether we have
$$(1-x^2)y_2-xy_1=0,$$
which is equivalent to $xy_1=(1-x^2)y_2$. (Basically, this should be the easy part, we should just differentiate twice and plug in.)

One problem with your question is that there are two common interpretations of $\sin^{-1}(x)$. It can mean either $\frac1{\sin x}$ or $\arcsin x$ (i.e., the inverse function to sine).
This is well-known notational problem, which occasionally might cause confusion, see here.
But let us try both possibilities.

$$y_0=\arcsin^2 x$$
$$y_1=y_0'=2\cdot\frac{\arcsin x}{\sqrt{1-x^2}}$$
$$y_2=y_1'=2\cdot\frac{1-\frac{\arcsin x}{\sqrt{1-x^2}}}{1-x^2}$$
You can check the results on WolframAlpha: the first derivative, the second derivative
We see that the equation $xy_1=(1-x^2)y_2$ is not fulfilled in this case.

$$y_0=\frac1{\sin^2x}$$
$$y_1=y_0'=-\frac{2\sin x\cos x}{\sin^4x} = -\frac{2\cos x}{\sin^3x}$$
$$y_2=y_1' = \frac{2\sin^4x+6\cos^2x\sin^2x}{\sin^6x} = \frac{2\sin^2x+6\cos^2x}{\sin^4x}$$
If you wish, you can again check the results on WA or elsewhere.
The equality $xy_1=(1-x^2)y_2$ is not fulfilled in this case, either.

The fact that I had to change the sign for inductive step and the recursive formula we want to prove is not fulfilled for $n=0$ leads me to the question: What is the origin of this exercise? Are you sure you have copied it correctly?
