Let y=y(x) be a differentiable function in th | vedclass

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(B) Given the limit: $\lim_{t \rightarrow x} \frac{t^{2}y(x)-x^{2}y(t)}{x-t} = 3$. Applying $L$'$H$ôpital's rule with respect to $t$: $\lim_{t \righta

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$23$

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(B) Given the limit: $\lim_{t \rightarrow x} \frac{t^{2}y(x)-x^{2}y(t)}{x-t} = 3$. Applying $L$'$H$ôpital's rule with respect to $t$: $\lim_{t \rightarrow x} \frac{2ty(x)-x^{2}y'(t)}{-1} = 3$. Substituting $t=x$: $2xy(x) - x^{2}y'(x) = -3$, which simplifies to $x^{2}y'(x) - 2xy(x) = 3$. Dividing by $x^{4}$ (for $x>0$): $\frac{x^{2}y'(x) - 2xy(x)}{x^{4}} = \frac{3}{x^{4}} \Rightarrow \frac{d}{dx} \left( \frac{y(x)}{x^{2}} \right) = 3x^{-4}$. Integrating both sides: $\frac{y(x)}{x^{2}} = \int 3x^{-4} dx = -x^{-3} + C = -\frac{1}{x^{3}} + C$. So, $y(x) = Cx^{2} - \frac{1}{x}$. Given $y(1)=2$: $2 = C(1)^{2} - \frac{1}{1} \Rightarrow 2 = C - 1 \Rightarrow C = 3$. Thus, $y(x) = 3x^{2} - \frac{1}{x}$. Finally, $2y(2) = 2 \left( 3(2)^{2} - \frac{1}{2} \right) = 2 \left( 12 - 0.5 \right) = 24 - 1 = 23$.

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

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