#### The **method** used during the course of this study is **Adam**-**bashforth** of order 2 (AB2). 3.2.1 Second order **Adam**-**Bashforth method** (AB2) Suppose we have an ordinary differential equation y 0 = f (t, y(t)) with an initial condition y(to ) = yo and we want to solve it numerically. If we know y(t) at a time tn and want to know what y(t) is at a later. **Adams**-**Bashforth** **Methods** Like Runge-Kutta **methods**, **Adams**-**Bashforth** **methods** want to estimate the behavior of the solution curve, but instead of evaluating the derivative function at new points close to the next solution value, they look at the derivative at old solution values and use interpolation ideas, along with the current solution and. **Example** 5 Use two-step **Adams**-**Bashforth method** to nd the approximate solution of dx dt = 1+ x t; x(1) = 1; near x(1:5). Take step size h = 0:5. Solution The two-step **Adams**-**Bashforth** formula is xn+1 = xn + h 2 [3f(tn;xn) f(tn 1;xn 1)]: We note that for nding x2, we require x1 and x0. We calculate x1 with the help of second order.

**method**lead to both smaller trunca-tion and round-o↵ errors. (3) The implicit methods are typically not used by themselves, but as corrector methods for an explicit predictor

**method**. The two methods above combine to form the

**Adams**-

**Bashforth**-Moulton

**Method**as a predictor-corrector

**method**. Maple. opments based on Runge–Kutta rather than

**Adams**–

**Bashforth**formulae, for

**example**, again see work by Ascher, Ruuth, and Spiteri [3], as well as very recent work by Calvo, ... (see, for

**example**, [38]). A similar

**method**has been developed for the study of PDEs. The idea is to make a change of variable that allows us to solve for the linear part. The second-order

**Adams**-

**Bashforth method**for the integration of a single first-order differential equation d.x =f(t, x) dt is Xn+1 = X₁ + 1 h[3f(t, Xn) – f(tn-1, Xn-1)] Write down the appropriate equations for applying the same

**method**to the solution of the pair of differential equations dx == f₁(t, x, y), dy= dy = f(t, x, y) dt dt Hence find the value of X(0.3) for the initial-value.