4. Let $f: \mathbf{R} \rightarrow \mathbf{R}$ be a smooth function such that $f^{\prime}(x)^{2}=f(x) f^{\prime \prime}(x)$ for all $x$. Suppose $f(0)=1$ and $f^{(4)}(0)=9$. Find all possible values of $f^{\prime}(0)$.
Solution: $\pm \sqrt{3}$
Let $f^{\prime}(0)=a$. Then the equation gives $f^{\prime \prime}(0)=a^{2}$. Differentiating the given equation gives
$$
2 f^{\prime}(x) f^{\prime \prime}(x)=f(x) f^{\prime \prime \prime}(x)+f^{\prime}(x) f^{\prime \prime}(x),
$$
or $f^{\prime}(x) f^{\prime \prime}(x)=f(x) f^{\prime \prime \prime}(x)$. Differentiating once more gives
$$
f^{\prime}(x) f^{\prime \prime \prime}(x)+f^{\prime \prime}(x)^{2}=f(x) f^{(4)}(x)+f^{\prime}(x) f^{\prime \prime \prime}(x)
$$
or $f^{\prime \prime}(x)^{2}=f(x) f^{(4)}(x)$, giving $9=f^{(4)}(0)=a^{4}$. Thus $a= \pm \sqrt{3}$. These are indeed both attainable by $f(x)=e^{ \pm x \sqrt{3}}$.
Alternative Solution: Rewrite the given equation as $\frac{f^{\prime \prime}(x)}{f^{\prime}(x)}=\frac{f^{\prime}(x)}{f(x)}$. Integrating both sides gives $\ln f^{\prime}(x)=\ln f(x)+C_{1}$, and exponentiating gives $f^{\prime}(x)=C f(x)$. This has solution $f(x)=A e^{C x}$ for constants $A$ and $C$. Since $f(0)=1, A=1$, and differentiating we find that $C^{4}=f^{(4)}(0)=9$, yielding $f^{\prime}(0)=C= \pm \sqrt{3}$.