Problem source

Show that $\sin(k\sin^{-1} x)$, where $k$ is a constant, satisfies the  differential equation
$$(1-x^{2})\frac {\d^2 y}{\d x^2} -x\frac{\d y}{\d x} +k^{2}y=0. \tag{*}$$
In the particular case when $k=3$, find the  solution of equation $(*)$ of the form
\[
y=Ax^{3}+Bx^{2}+Cx+D,
\]
that satisfies $y=0$ and $\displaystyle \frac{\d y}{\d x}=3$ at $x=0$.
Use this result to express $\sin 3\theta$ in terms of powers of $\sin\theta$.

Solution source

\begin{align*}
&& y &= \sin(k \sin^{-1} x ) \\
&&y' &= \cos (k \sin^{-1} x) \cdot k \frac{1}{\sqrt{1-x^2}} \\
&& y'' &= -\sin (k \sin^{-1} x) \cdot k^2 \frac{1}{(1-x^2)} - \cos(k \sin^{-1} x)  \cdot k \frac{x}{(1-x^2)\sqrt{1-x^2}} \\
&& (1-x^2)y'' &= -k^2y -xy' \\
\Rightarrow && 0 &= (1-x^2)y''+xy' + k^2y
\end{align*}

\begin{align*}
&& y &= Ax^3 + Bx^2 + Cx + D \\
&& y' &= 3Ax^2 + 2Bx + C \\
&& y'' &= 6Ax+2B \\
&& 0 &= (1-x^2)(6Ax+2B) - x( 3Ax^2 + 2Bx + C) + 9(Ax^3 + Bx^2 + Cx + D ) \\
&&&= x^3(-6A-3A+9A) + x^2(-2B-2B+9B) + x(6A-C+9C) + (2B +9D) \\
\Rightarrow && B &= 0 \\
\Rightarrow && D &= 0 \\
\Rightarrow && C &= -\frac34 A \\
\\
x = 0, y = 0, y' = 0: && y &= 3x-4x^3 \\
\end{align*}

And so $\sin 3 x  = 3 \sin x - 4\sin^3 x$