Lety = y (x) be a differentiable function in the interval $$(0, \infty)$$ such that y(l) = 2, and $$\lim_{t \rightarrow x} \left( \frac{t^{2}y(x)-x^{2}y(t)}{x-t} \right) = 3$$ for each x > 0. Then 2){2) is equal to

We need to find 2y(2) given that y(1) = 2 and $$\lim_{t \rightarrow x} \frac{t^2y(x) - x^2y(t)}{x - t} = 3$$.

The limit $$\lim_{t \rightarrow x} \frac{t^2 y(x) - x^2 y(t)}{x - t}$$ is of the form $$\frac{0}{0}$$ when $$t \to x$$. Applying L'Hopital's rule by differentiating with respect to t yields $$= \lim_{t \rightarrow x} \frac{2t \cdot y(x) - x^2 y'(t)}{-1} = -(2x \cdot y(x) - x^2 y'(x)) = x^2 y'(x) - 2xy(x).$$

Setting this expression equal to 3 gives $$x^2 y'(x) - 2xy(x) = 3$$.

This can be rewritten as the linear differential equation $$y' - \frac{2}{x}y = \frac{3}{x^2}$$.

The integrating factor is $$IF = e^{\int -2/x \, dx} = e^{-2\ln x} = \frac{1}{x^2}$$.

Multiplying both sides by this integrating factor gives $$\frac{d}{dx}\left(\frac{y}{x^2}\right) = \frac{3}{x^4}$$, and integrating leads to $$\frac{y}{x^2} = \int \frac{3}{x^4} dx = -\frac{1}{x^3} + C$$, so that $$y = -\frac{x^2}{x^3} + Cx^2 = -\frac{1}{x} + Cx^2$$.

Applying the initial condition $$y(1) = 2$$ gives $$2 = -1 + C \implies C = 3$$, so $$y(x) = 3x^2 - \frac{1}{x}$$.

Finally, $$y(2) = 3(4) - \frac{1}{2} = 12 - 0.5 = 11.5$$ and $$2y(2) = 2 \times 11.5 = 23$$. Therefore, 2y(2) = Option 3: 23.