Question

Let $f$ be a differentiable function in the interval $(0, \infty)$ such that $f(1) = 1$ and $\lim_{t \to x} \frac{t^2 f(x) - x^2 f(t)}{t - x} = 1$ for each $x > 0$.
Then $2f(2) + 3f(3)$ is equal to _________.

Answer 1

Applying L'Hôpital's rule, we differentiate the numerator and the denominator:

$\lim_{t \to x} \frac{t^2 f(x) - x^2 f(t)}{t - x} = 1$ $\lim_{t \to x} \frac{2t f(x) - x^2 f'(x)}{1} = 1$

Simplifying, we get:

$2x f(x) - x^2 f'(x) = 1$

Rearranging terms:

$\frac{dy}{dx} = \frac{2}{x} \quad \text{where } y = -\frac{1}{x^2}$

The differential equation becomes:

$\frac{dy}{dx} + \frac{y}{x} = \frac{2}{x^2}$

Solving this differential equation, we assume:

$y = \frac{1}{3x} + \frac{2x^2}{3}, \quad \text{leading to: } y = \frac{2x^3 + 1}{3x}$

Calculating specific values:

$f(2) = \frac{17}{6}, \quad f(3) = \frac{55}{9}$

Thus: $2f(2) + 3f(3) = \frac{17}{3} + \frac{55}{3} = \frac{72}{3} = 24$