Problem source

Show that $y=\sin^{2}(m\sin^{-1}x)$ satisfies the differential equation
\[
(1-x^{2})y^{(2)}=xy^{(1)}+2m^{2}(1-2y),
\]
and deduce that, for all $n\geqslant1,$ 
\[
(1-x^{2})y^{(n+2)}=(2n+1)xy^{(n+1)}+(n^{2}-4m^{2})y^{(n)},
\]
where $y^{(n)}$ denotes the $n$th derivative of $y$. 

Derive the Maclaurin series for $y$, making it clear what the general
term is.

Solution source

\begin{align*}
&& y &= \sin^2 (m \sin^{-1} x) \\
\Rightarrow && y' &= 2 \sin (m \sin^{-1} x) \cdot \cos (m \sin^{-1} x) \cdot m \cdot \frac1{\sqrt{1-x^2}} \\
\Rightarrow && y'' &= 2 \cos^2(m \sin^{-1} x) \cdot m^2 \cdot \frac{1}{1-x^2} + \\
&&&\quad\quad-2\sin^2(m \sin^{-1} x) m^2 \frac{1}{1-x^2} + \\
&&&\quad\quad\quad-\sin(m \sin^{-1} x) \cdot \cos(m \sin^{-1} x) \cdot m \cdot (1-x^2)^{-\frac32} \cdot (-2x) \\
\Rightarrow && (1-x^2)y^{(2)} &= 2m^2-4m^2y+xy' \\
&&&= xy^{(1)} + 2m^2(1-2y) \\
\\
\Rightarrow && (1-x^2)y^{(n+2)}-2nxy^{(n+1)}-2\binom{n}{2}y^{(n)} &= xy^{(n+1)}+ny^{(n)} -4m^2y^{(n)} \\
\Rightarrow && (1-x^2)y^{(n+2)} &= (2n+1)xy^{(n+1)}+(n(n-1)+n-4m^2)y^{(n)} \\
&&&= (2n+1)xy^{(n+1)}+(n^2-4m^2)y^{(n)} \\
\end{align*}

\begin{align*}
&& y(0) &= \sin^2(m \sin^{-1} 0) \\
&&&= \sin^2 0 = 0 \\
\\
&& y'(0) &= 0 \\
&& (1-0^2)y^{(2)}(0) &= 2m^2(1-2y(0)) \\
\Rightarrow && y^{(2)}(0) &= 2m^2 \\
\\
&& y^{(n+2)} (0) &= (2n+1) \cdot 0 \cdot y^{(n+1)} +(n^2-4m^2)y^{(n)}(0) \\
&&&= (n^2-4m^2)y^{(n)}(0) \\
\\
&& y^{(2)}(0) &= 2m^2 \\
&& y^{(4)}(0) &= (4-4m^2) \cdot 2m^2 \\
&&&= -8m(m+1)m(m-1) \\
&& y^{(6)}(0) &= 32m(m+2)(m+1)m(m-1)(m-2) \\
&& y^{(2k)}(0) &= (-1)^{k+1}2^{2k-1}m (m+k)\cdots(m-k) \text{ if }k < m \\
\\
&& y &= m^2x^2 -2m\binom{m+1}{3} x^4 + \frac{16}{3}m\binom{m+2}{5}x^6 - \cdots  \\
&&&+ (-1)^{k}\frac{2^{2k}}{k+1} m \binom{m+k}{2k+1}x^{2k+2}+\cdots \\
&&&= mx^2\sum_{k=0}^{m-1} \frac{(-1)^k2^{2k}}{k+1}\binom{m+k}{2k+1}x^{2k}
\end{align*}