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 ______

Given $$f(1) = 1$$ and $$\lim_{t \to x} \frac{t^2 f(x) - x^2 f(t)}{t - x} = 1$$.

The limit is of the form $$\frac{0}{0}$$ when $$t = x$$. Using L'Hôpital's rule (differentiating with respect to $$t$$):

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

So: $$x^2f'(x) - 2xf(x) = -1$$.

Dividing by $$x^2$$: $$f'(x) - \frac{2}{x}f(x) = -\frac{1}{x^2}$$.

I.F. = $$e^{\int -2/x \, dx} = e^{-2\ln x} = x^{-2}$$.

$$ \frac{f(x)}{x^2} = \int \frac{-1}{x^2} \cdot \frac{1}{x^2} dx = \int -x^{-4} dx = \frac{1}{3x^3} + C $$

$$ f(x) = \frac{x^2}{3x^3} + Cx^2 = \frac{1}{3x} + Cx^2 $$

Using $$f(1) = 1$$: $$\frac{1}{3} + C = 1 \Rightarrow C = \frac{2}{3}$$.

$$ f(x) = \frac{1}{3x} + \frac{2x^2}{3} $$

$$ f(2) = \frac{1}{6} + \frac{8}{3} = \frac{1 + 16}{6} = \frac{17}{6} $$

$$ f(3) = \frac{1}{9} + \frac{18}{3} = \frac{1}{9} + 6 = \frac{55}{9} $$

$$ 2f(2) + 3f(3) = 2 \times \frac{17}{6} + 3 \times \frac{55}{9} = \frac{17}{3} + \frac{55}{3} = \frac{72}{3} = 24 $$

The answer is 24.