Let y=y(x) be a differentiable function in the interval (0, ∞) such...

Given the limit $\lim_{t \to x} \frac{t^2 y(x) - x^2 y(t)}{x - t} = 3$.
The limit is in $\frac{0}{0}$ form. Applying L'Hopital's Rule with respect to $t$:
$\lim_{t \to x} \frac{\frac{d}{dt}(t^2 y(x) - x^2 y(t))}{\frac{d}{dt}(x - t)} = 3$
$\lim_{t \to x} \frac{2t y(x) - x^2 y'(t)}{-1} = 3$
Substituting $t = x$:
$\frac{2x y(x) - x^2 y'(x)}{-1} = 3$
$x^2 y'(x) - 2x y(x) = 3$
Dividing by $x^4$ to make it a linear differential equation or recognizing the quotient rule form:
$\frac{x^2 y'(x) - 2x y(x)}{x^4} = \frac{3}{x^4}$
$\frac{d}{dx} \left( \frac{y(x)}{x^2} \right) = \frac{3}{x^4}$
Integrating both sides with respect to $x$:
$\frac{y(x)}{x^2} = \int 3x^{-4} dx = \frac{3x^{-3}}{-3} + C = -\frac{1}{x^3} + C$
Given $y(1) = 2$:
$\frac{2}{1^2} = -\frac{1}{1^3} + C \implies 2 = -1 + C \implies C = 3$
So, $\frac{y(x)}{x^2} = -\frac{1}{x^3} + 3$
$y(x) = -\frac{1}{x} + 3x^2$
We need to find $2y(2)$:
$y(2) = -\frac{1}{2} + 3(2^2) = -\frac{1}{2} + 12 = \frac{23}{2}$
$2y(2) = 2 \times \frac{23}{2} = 23$