Problem source

This question concerns solutions of 
 the differential equation
\[
(1-x^2) \left(\frac{\d y}{\d x}\right)^2 + k^2 y^2 = k^2\,
\tag{$*$}
\]
where $k$ is a positive integer. 

For each value of $k$, let $y_k(x)$  be
the solution of $(*)$ that satisfies $y_k(1)=1$; 
you may 
assume that
there is only one such solution for each value of $k$.



\begin{questionparts}
\item
Write down the differential equation satisfied by $y_1(x)$ and
verify that $y_1(x) = x\,$. 
\item 
Write down the differential equation satisfied by $y_2(x)$ and
verify that  $y_2(x) = 2x^2-1\,$.
\item 
Let $z(x) = 2\big(y_n(x)\big)^2 -1$. Show that
\[
 (1-x^2) \left(\frac{\d z}{\d x}\right)^2 +4n^2 z^2 = 4n^2\,
\]
and hence obtain an expression for $y_{2n}(x)$ in terms of $y_n(x)$.
\item 
Let $v(x) = y_n\big(y_m(x)\big)\,$. 
Show that $v(x) = y_{mn}(x)\,$.
\end{questionparts}

Solution source

\begin{questionparts}
\item When $k =1$, we have $(1-x^2)(y')^2 + y^2 = 1$. Notice that if $y_1 = x$ we have $y_1' = 1$ and $(1-x^2) \cdot 1 + x^2 = 1$ so $y_1$ is a solution, and we are allowed to assume this is the only solution. And notice that $y_1(1) = 1$.

\item When $k = 2$ we have $(1-x^2)(y')^2 + 4y^2 = 4$. Trying $y_2 = 2x^2-1$ we see that $y_2' = 4x$ and $(1-x^2)(4x)^2 + 4(2x^2-1)^2 = 16x^2-16x^4+16x^4-16x^2+4 = 4$. We can also check that $y(1) = 2 \cdot 1^2 - 1 = 1$

\item Let $z(x) = 2(y_n(x))^2-1$, then

\begin{align*}
&& \frac{\d z}{\d x} &= 4y'_n(x)y_n(x) \\
\Rightarrow && LHS &= (1-x^2)\left ( \frac{\d z}{\d x} \right)^2 + 4n^2 z^2  \\
&&&= (1-x^2)16(y'_n(x))^2(y_n(x))^2 + 4n^2(2(y_n(x))^2-1)^2 \\
&&&= 16y_n^2(1-x^2) \left [\frac{n^2-n^2y_n^2}{(1-x^2)} \right] + 16n^2y_n^4-16n^2y_n^2+4n^2 \\
&&&= 4n^2 = RHS
\end{align*}

Therefore $y_{2n}(x) = 2(y_n(x))^2-1 = y_2(y_n(x))$ (notice also that $z(1) = 2(y_n(1))^2-1 = 2-1 = 1$).

\item Let $v(x) = y_n(y_m(x))$ so

\begin{align*}
&& (y_m')^2 &= \frac{m^2(1-y_m^2)}{1-x^2} \\
&& (y_n')^2 &= \frac{n^2(1-y_n^2)}{1-x^2} \\
\\
&& \frac{\d v}{\d x} &= y_n'(y_m(x)) \cdot y_m'(x) \\
&& (1-x^2)(v')^2 &= (1-x^2) \cdot (y_n'(y_m(x)))^2 (y_m')^2 \\
&&&=  (1-x^2) \cdot (y_n'(y_m(x)))^2 \left ( \frac{m^2(1-y_m^2)}{1-x^2} \right)  \\
&&&= (y_n'(y_m(x)))^2 m^2(1-y_m^2) \\
&&&= \frac{n^2(1-(y_n(y_m(x)))^2)}{1-y_m^2}m^2(1-y_m^2) \\
&&&= n^2m^2(1-v^2)
\end{align*}

Therefore $v$ satisfies our differential equation and $v(1) = y_n(y_m(1)) = y_n(1) = 1$ so it must be our desired solution.

\end{questionparts}


[Note: this is another question about Chebyshev polynomials, and we have proven that we can compose them nicely. This might be more easily proven as $T_n(x) = \cos(n \cos^{-1} x)$ and so $T_n(T_m(x)) = \cos (n \cos^{-1}( \cos (m \cos^{-1} x))) = \cos(nm \cos^{-1}x) = T_{nm}(x)$]