11.2. Find all differentiable functions $f:(0,+\infty) \rightarrow(0,+\infty)$ that satisfy the relations $f\left(\frac{1}{2}\right)=1$ and $f^{\prime}(x)=-f^{3}(x)$.
## Solution.

We have $\left(f^{-2}(x)\right)^{\prime}=-2 f^{-3}(x) f^{\prime}(x)=2 f^{-3}(x) f^{3}(x)=2$, which means the derivative of the function $f^{-2}(x)$ is a constant. It follows that the function $f^{-2}(x)$ is a linear function, which can be written in the form $f^{-2}(x)=a x+b$. By differentiating, we get $2=\left(f^{-2}(x)\right)^{\prime}=a$. Furthermore, from the condition $f\left(\frac{1}{2}\right)=1$ we have $1=f^{-2}\left(\frac{1}{2}\right)=2 \cdot \frac{1}{2}+b=1+b$, which implies $b=0$. We obtain $f^{-2}(x)=2 x$, i.e., $f(x)=\frac{1}{\sqrt{2 x}}$.

We observe that this function is defined on the entire interval $(0,+\infty)$ and is differentiable on the same interval. We have $f\left(\frac{1}{2}\right)=\frac{1}{\sqrt{2 \cdot \frac{1}{2}}}=1$ and,

$$
f^{\prime}(x)=\left(\frac{1}{\sqrt{2 x}}\right)^{\prime}=\left(\frac{1}{\sqrt{2}} x^{-1 / 2}\right)^{\prime}=\frac{1}{\sqrt{2}} \cdot\left(-\frac{1}{2}\right) \cdot x^{-3 / 2}=-\left(\frac{1}{\sqrt{2}} \cdot x^{-1 / 2}\right)^{3}=-f^{3}(x)
$$

Thus, the function $f:(0,+\infty) \rightarrow(0,+\infty), f(x)=\frac{1}{\sqrt{2 x}}$, is the unique one that satisfies the condition of the problem.