第1章 大問1

(1)
与式の2次導関数までを求めれば良い．

$$ y=\sin \left(a \sin ^{-1} x\right)$$

$$y^{\prime}=\cos \left(a \sin ^{-1} x\right) \cdot \frac{a}{\sqrt{1-x^2}}$$

$$y^{\prime \prime}=a \frac{-a\sin ^{-1}\left(\sin ^{-1} x\right) \cdot \frac{a}{\sqrt{1-x^2}} \cdot \sqrt{1-x^2}-\cos \left(a \sin ^{-1} x\right) \cdot \frac{a x}{\sqrt{1-x^2}}}{1-x^2}$$

$$=-\sin \left(a \sin ^{-1} x\right) \cdot \frac{a^2}{1-x^2}-a\cos \left(\sin ^{-1} x\right) \frac{a x}{\left(1-x^2\right)^{\frac{3}{2}}}$$

$$=-\frac{a^2}{1-x^2} y-\frac{x}{1-x^2} y^{\prime}$$

$$∴\left(1-x^2\right) y^{\prime \prime}-x y^{\prime}+a^2 y=0$$

(2)
$$
\left(1-x^2\right) y^{\prime \prime}-x y^{\prime}+a^2 z=0
$$
両辺をn回微分すると，
$$
\left\{\left(1-x^2\right) y^{\prime \prime}\right\}^{(n)}-\left(x y^{\prime}\right)^{(n)}+a^2 y^{(n)}=0
$$
ライプニッツの定理より，上式各項の微分は，
$$
\begin{aligned}
& \left\{\left(1-x^2\right) y^{\prime \prime}\right\}=\left(1-x^2\right) z^{(n+2)}+n(-2 x) y^{(n+1)}+\frac{n(n-1)}{2}(-2) y^{(n)} \
\\
& \left(x f^{\prime}\right)^{(n)}=\sum_{k=0}^n n C_k x^{(k)} z^{(n+1-k)}=x z^{(n+1)}+n y^{(n)}
\end{aligned}
$$
となるので，
$$
\left\{\left(1-x^2\right) y^{(n+2)}-2 n x y^{(n+1)}-n(n-1) y^{(n)}\right\}-\left\{x y^{(n+1)}+n z^{(n)}\right\}+a^2 y^{(n)}=0
$$
$$
\left(1-x^2\right) y^{(n+2)}-(2 n+1) x y^{(n+1)}+\left(-n^2+a^2\right) z^{(n)}=0
$$
とかける．x=0の時，
$$
y^{(n+2)}+\left(a^2-n^2\right) y^{(n)}=0
$$
$$
∴ y^{(n)}=\left(n^2-a^2\right) y^{(n-2)}
$$
と漸化式が求まった．
Case(1) $n$ が偶数
$$
\begin{aligned}
& y^{(n)}(0)=\left(n^2-a^2\right)^{\frac{n}{2}} y(0) \\
∴ & y^{(n)}(0)=0
\end{aligned}
$$
Case(ii) $n$ が奇数
$$
\begin{aligned}
y^{(n)}(0) & =\left(n^2-a^2\right)^{\frac{n-1}{2}} y{\prime}(0) \\
∴ y^{(n)}(0) & =a\left(n^2-a^2\right)^{\frac{n-1}{2}}
\end{aligned}
$$