A positive real valued continuously differentiable functions $f$ on the real line such that for all $x$ $f^2(x) = \int_0^x \left[(f(t))^2 + (f'(t))^2\right] dt + e^2$ then $f(-1) = $ ____.
Step-by-Step Solution
Key Concept: Differentiate the given functional equation to obtain a separable differential equation $f'(x) = f(x)$.
Differentiating both sides of the given relation with respect to $x$ yields $2f(x)f'(x) = [f(x)]^2 + [f'(x)]^2$, which simplifies to $(f(x) - f'(x))^2 = 0$. Therefore $f'(x) = f(x)$. From the initial condition $f(0)^2 = e^2$, we have $f(0) = e$ (rejecting the negative solution). Solving $\frac{f'(x)}{f(x)} = 1$ gives $\ln(f(x)) = x + C$. With $f(0) = e$, we get $C = 1$, so $f(x) = e^{x+1}$.
Correct Answer: 1