The differential equation \(\dfrac{du}{dt} + 2tu^{2} = 1\) is solved by a backward difference scheme. At the \((n-1)\)-th time step, \(u_{n-1}=1.75\) and \(t_{n-1}=3.14\,\text{s}\). With \(\Delta t=0.01\,\text{s}\), find \(u_n-u_{n-1}\) (round off to three decimals).