The differential equation,
$$\frac{\mathrm{d} u}{\mathrm{~d} t}+2 t u^2=1,$$
is solved by employing a backward difference scheme within the finite difference framework. The value of $u$ at the $(n-1)^{\text {th }}$ time-step, for some $n$, is $1.75.$ The corresponding time $(t)$ is $3.14 \mathrm{~s}$. Each time step is $0.01 \mathrm{~s}$ long. Then, the value of $\left(u_n-u_{n-1}\right)$ is__________ (round off to three decimal places).