If $f(x)$ is 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 $f\left(\frac{3}{2}\right)$ is equal to

$\frac{31}{18}$

We have,

$\lim\limits _{t \rightarrow x} \frac{t^2 f(x)-x^2 f(t)}{t-x}=1$

$\Rightarrow \lim\limits _{t \rightarrow x} \frac{t^2\{f(x)-f(t)\}-\left(x^2-t^2\right) f(t)}{t-x}=1$

$\Rightarrow \lim\limits _{t \rightarrow x}\left\{\frac{f(x)-f(t)}{t-x}\right\} t^2+\lim\limits _{t \rightarrow x}(x+t) f(t)=1$

$\Rightarrow -x^2 f'(x)+2 x f(x)=1$                [∵ f is differentiable and so continuous also]

$\Rightarrow f'(x)-\frac{2}{x} f(x)=-\frac{1}{x^2}$             .....(i)

This is a linear differential equation with integrating factor $e^{-\int \frac{2}{x} d x}=\frac{1}{x^2}$

Multiplying (i) by integrating factor $=\frac{1}{x^2}$ integrating, we get

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

It is given that $f(1)=1$. Putting $x=1$ in (ii), we get

$1=\frac{1}{3}+C \Rightarrow C=\frac{2}{3}$

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

$\Rightarrow f(x)=\frac{1}{3 x}+\frac{2 x^2}{3} \Rightarrow f\left(\frac{3}{2}\right)=\frac{2}{9}+\frac{3}{2}=\frac{31}{18}$