Problem Breakdown and Solution

Part (a): Prove that $f^2 - g^2$ is constant

We are given two differentiable functions $f$ and $g$ such that: $f'(x) = g(x) \quad \text{and} \quad g'(x) = f(x)$ We need to prove that $f^2(x) - g^2(x)$ is constant. To do so, we'll differentiate $f^2(x) - g^2(x)$.

1. Differentiate $f^2(x) - g^2(x)$ using the chain rule: $\frac{d}{dx}\left( f^2(x) - g^2(x) \right) = 2f(x)f'(x) - 2g(x)g'(x)$ Substituting $f'(x) = g(x)$ and $g'(x) = f(x)$, we get: $2f(x)f'(x) - 2g(x)g'(x) = 2f(x)g(x) - 2g(x)f(x)$ Simplifying the right-hand side: $2f(x)g(x) - 2g(x)f(x) = 0$ Therefore: $\frac{d}{dx}\left( f^2(x) - g^2(x) \right) = 0$ This means that $f^2(x) - g^2(x)$ is constant for all $x$.

Part (b): Prove that $f(x) > 0$ and $g(x) > 0$ for all $x > 0$

We are now asked to prove that, under the additional conditions $f(0) = 1$ and $g(0) = 0$, the functions $f(x)$ and $g(x)$ are positive for all $x > 0$.

Step 1: Understanding the Infimum Definitions

Let: $a = \inf \{ x > 0 \mid f(x) \leq 0 \}, \quad b = \inf \{ x > 0 \mid g(x) \leq 0 \}$ We need to show that $a = b = \infty$, meaning that neither $f(x)$ nor $g(x)$ can become non-positive for any $x > 0$.

Step 2: Argument for $a = \infty$

Assume for contradiction that $a < \infty$, i.e., there exists some $x = a$ such that $f(a) = 0$ and $f(x) > 0$ for all $x \in (0, a)$. Since $f$ is continuous, we also have $f'(a) = g(a)$. Given that $f'(x) = g(x)$, if $f(a) = 0$, then we must have: $f'(a) = g(a) = 0$ However, this contradicts the fact that $g'(a) = f(a)$, meaning that $g'(a) = 0$. Since we assumed that $a$ is the first point where $f(x) \leq 0$, and $g(x) = 0$ when $f(x) = 0$, we cannot have $g(a) \neq 0$. Thus, we conclude that $a$ must not exist.

Step 3: Conclusion on $a = b = \infty$

Similarly, assume for contradiction that $b < \infty$, i.e., there exists some $x = b$ such that $g(b) = 0$ and $g(x) > 0$ for all $x \in (0, b)$. Using similar reasoning, we deduce that this leads to a contradiction, and we conclude that both $a = \infty$ and $b = \infty$. Thus, both $f(x) > 0$ and $g(x) > 0$ for all $x > 0$.

Final Answer:

(a) $f^2(x) - g^2(x)$ is constant because its derivative is 0.
(b) $f(x) > 0$ and $g(x) > 0$ for all $x > 0$ because $a = b = \infty$.

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Would you like further details on any part of the solution? Here are some related questions you might find interesting:

1. How can we interpret the fact that $f^2(x) - g^2(x)$ is constant in terms of the relationship between $f(x)$ and $g(x)$?
2. What is the behavior of the functions $f(x)$ and $g(x)$ as $x \to \infty$?
3. How can the initial conditions $f(0) = 1$ and $g(0) = 0$ influence the behavior of the functions for all $x > 0$?
4. Can we derive a generalization for systems of differential equations of the form $f'(x) = g(x)$ and $g'(x) = f(x)$?
5. What would happen if the roles of $f$ and $g$ were reversed, i.e., $f'(x) = -g(x)$ and $g'(x) = -f(x)$?

Tip: Always consider the behavior of solutions near critical points and use the properties of the derivatives to help rule out contradictory scenarios.