The differential equation,

$\rm \frac{du}{dt}+2tu^2=1,$

is solved by employing a backward difference scheme within the finite difference framework. The value of 𝑢 at the (𝑛 − 1) th time-step, for some 𝑛, is 1.75. The corresponding time (t) is 3.14 s. Each time step is 0.01 s long. Then, the value of (un - un -1) _________ is (round off to three decimal places). 

Consider the differential equation

$${{dy} \over {dx}} = 4(x + 2) - y$$

For the initial condition y = 3 at x = 1, the value of y at x = 1.4 obtained using Euler's method with a step-size of 0.2 is ________. (round off to one decimal place)

The solution of the partial differential equation $${{\partial u} \over {\partial t}} = \alpha {{{\partial ^2}u} \over {\partial {x^2}}}$$ is of the form

Consider the following second order linear differential equation $${{{d^2}y} \over {d{x^2}}} = - 12{x^2} + 24x - 20$$
The boundary conditions are: at $$x=0, y=5$$ and at $$x=2, y=21$$
The value of $$y$$ at $$x=1$$ is