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

<p>$$\lim _\limits{t \rightarrow x} \frac{t^2 f(x)-x^2 f(t)}{(t-x)}=1 \quad\left(\frac{0}{0} \text { form }\right)$$</p> <p>$$\begin{aligned} & \lim _{t \rightarrow x} \frac{2 \operatorname{tf}(x)-x^2 f^{\prime}(t)}{1}=1 \\ & \Rightarrow 2 x f(x)-x^2 f'(x)=1 \\ & \frac{d y}{d x}-\frac{2 x y}{x^2}=\frac{-1}{x^2} \\ & \Rightarrow \frac{d y}{d x}-\left(\frac{2}{x}\right) y=\frac{-1}{x^2} \\ & \Rightarrow \text { I.F. }=e^{\int \frac{-2}{x} d x}=e^{-2 \ln x}=x^{-2}=\frac{1}{x^2} \end{aligned}$$</p> <p>$$\begin{aligned} \Rightarrow & y\left(\frac{1}{x^2}\right)=\int\left(\frac{-1}{x^2}\right)\left(\frac{1}{x^2}\right) d x+C \\ & \frac{y}{x^2}=\frac{1}{3 x^3}+C \text { at } x=1, y=1 \\ \Rightarrow & 1=\frac{1}{3}+C \Rightarrow C=\frac{2}{3} \\ \Rightarrow & y=\frac{1}{3 x}+\frac{2}{3} x^2=f(x) \\ \Rightarrow & 2 f(2)+3 f(3)=24 \end{aligned}$$</p>