9. $[\boldsymbol{7}] g$ is a twice differentiable function over the positive reals such that
$$
\begin{aligned}
g(x)+2 x^{3} g^{\prime}(x)+x^{4} g^{\prime \prime}(x) & =0 \quad \text { for all positive reals } x . \\
\lim _{x \rightarrow \infty} x g(x) & =1
\end{aligned}
$$

Find the real number $\alpha>1$ such that $g(\alpha)=1 / 2$.
Answer: $\frac{6}{\pi}$. In the first equation, we can convert the expression $2 x^{3} g^{\prime}(x)+x^{4} g^{\prime \prime}(x)$ into the derivative of a product, and in fact a second derivative, by writing $y=1 / x$. Specifically,
$$
\begin{aligned}
0=g(x)+2 x^{3} g^{\prime}(x)+x^{4} g^{\prime \prime}(x) & =g\left(\frac{1}{y}\right)+2 y^{-3} g^{\prime}\left(\frac{1}{y}\right)+y^{-4} g^{\prime \prime}\left(\frac{1}{y}\right) \\
& =g\left(\frac{1}{y}\right)+\frac{\mathrm{d}}{\mathrm{d} y}\left[-y^{-2} g^{\prime}\left(\frac{1}{y}\right)\right] \\
& =g\left(\frac{1}{y}\right)+\frac{\mathrm{d}^{2}}{\mathrm{~d} y^{2}}\left[g\left(\frac{1}{y}\right)\right]
\end{aligned}
$$

Thus $g\left(\frac{1}{y}\right)=c_{1} \cos (y)+c_{2} \sin (y)$ or $g(x)=c_{1} \cos (1 / x)+c_{2} \sin (1 / x)$. Now the second condition gives
$$
1=\lim _{x \rightarrow \infty} c_{1} x+c_{2} \cdot \frac{\sin (1 / x)}{1 / x}=c_{2}+\lim _{x \rightarrow \infty} c_{1} x
$$

It must be that $c_{1}=0, c_{2}=1$. Now since $0<1 / \alpha<1$, the value of $\alpha$ such that $g(\alpha)=\sin (1 / \alpha)=1 / 2$ is given by $1 / \alpha=\pi / 6$ and so $\alpha=6 / \pi$.