Consider the equation $\frac{d u}{d t} = 3 t^{2} + 1$ $u = 0$ at $t = 0$ . This is numeircally solved by using the forwaard Euler method with a step size, $Δ t = 2$ . The absolute error in the solution at the end of the first time step is
8

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Solution

The correct option is A 8
$\frac{d u}{d t} = 3 t^{2} + 1 = f (t, u)$ (let)
Forward Eular Method $u_{1} = u_{o} + h f (t_{o}, u_{o})$
Given, $t_{o} = 0, u_{o} = 0, h = Δ t = 2$
$u_{1} = 0 + 2 (3 \times 0^{2} + 1) = 2$
Now for exact value,
$\frac{d u}{d t} = 3 t^{2} + 1$
$d u = \int_{o}^{2} (3 t^{2} + 1) d t$
$u_{1} = [t^{3} + t]_{0}^{2} = 10$
Absolute error $= | 10 - 2 | = 8$

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