SOLUTION OF AN EXTRAORDINARY DIFFERENTIAL EQUATION BY ADOMIAN DECOMPOSITION METHOD

Large classes of linear and nonlinear differential equations, both ordinary as well as partial, can be solved by the Adomian decomposition method [3, 4, 6]. This method is much more simpler in computation and quicker in convergence than any other method available in the open literature. The application of the fractional differential equation in physical problems is available in the book of Bracewell [11]. Recently, the solution of the fractional differential equation has been obtained through the Adomian decomposition method by the researchers in [7, 14]. In this paper, we solve a differential equation containing a fractional derivative of order half along with an ordinary first-order derivative using the Adomian decomposition method. Then the solution obtained by this method is verified with that of the transformed ordinary differential equation derived from the original fractional differential equation. For the sake of convenience, we first of all give definitions of fractional integral and fractional derivative introduced by Riemann-Liouville as discussed in [18, 20, 21].


Introduction
Large classes of linear and nonlinear differential equations, both ordinary as well as partial, can be solved by the Adomian decomposition method [3,4,6]. This method is much more simpler in computation and quicker in convergence than any other method available in the open literature.
The application of the fractional differential equation in physical problems is available in the book of Bracewell [11]. Recently, the solution of the fractional differential equation has been obtained through the Adomian decomposition method by the researchers in [7,14].
In this paper, we solve a differential equation containing a fractional derivative of order half along with an ordinary first-order derivative using the Adomian decomposition method. Then the solution obtained by this method is verified with that of the transformed ordinary differential equation derived from the original fractional differential equation.
For the sake of convenience, we first of all give definitions of fractional integral and fractional derivative introduced by Riemann-Liouville as discussed in [18,20,21]. Definition 1.1 (fractional integral). Let q > 0 denote a real number. Assuming f (x) to be a function of class C (n) (the class of functions with continuous nth derivatives), the fractional integral of a function f of order −q is given by (1.1) Definition 1.2 (fractional derivative). Let q > 0 denote a real number and n the smallest integer exceeding q such that n − q > 0 (n = 0 if q < 0). Assuming f (x) to be a function of class C (n) (the class of functions with continuous nth derivatives), the fractional derivative of a function f of order q is given by

The decomposition method. We consider an equation in the form
Lu where L is an easily or trivially invertible linear operator, R is the remaining linear part, and N represents a nonlinear operator. The general solution of the given equation is decomposed into the sum where u 0 is the solution of the linear part. Our approach will be to write any nonlinear term in terms of the Adomian A n polynomials. It has been derived by Adomian that Nu = ∞ n=0 A n , where the A n are special polynomials obtained for the particular nonlinearity Nu = f (u) and generated by Adomian [3,4,5]. These A n polynomials depend, of course, on the particular nonlinearity.
The A n are given as S. S. Ray and R. K. Bera 333 and can be found from the formula (for n ≥ 1) where the c(ν,n) are products (or sums of products) of ν components of u whose subscripts sum to n, divided by the factorial of the number of repeated subscripts [5]. Therefore, the general solution becomes where u 0 = φ + L −1 g and Lφ = 0.
To identify the terms in ∞ n=1 u n , it has been derived by Adomian that From (1.9), we can write u 1 = −L −1 Ru 0 − L −1 A 0 . Thus u 1 can be calculated in terms of the known u 0 . Now,

Solution of an extraordinary differential equation
A relationship involving one or more derivatives of an unknown function f with respect to its independent variable x is known as an ordinary differential equation. Similar relationship involving at least one differintegral of noninteger order may be termed as an extraordinary differential equation. Such an equation is solved when an explicit expression for f is exhibited. As with ordinary differential equations, the solutions of extraordinary differential equations often involve integrals and contain arbitrary constants as discussed in [20]. These types of equations are also known as fractional differential equations. The application of extraordinary differential equation is now available in many physical and 334 Solution of an extraordinary differential equation by ADM technical areas [21]. It can be mentioned here that the simplified fractional-order differential equation appearing in applied problems is of the form where n − 1 < α ≤ n. For 0 < α ≤ 2, this equation is called the relaxation-oscillation equation [21]. Moreover, the Bagley-Torvik equation [10] A d 2 y(t) (where A = 0 and B,C ∈ R) arises in the modelling of the motion of a rigid plate immersed in a Newtonian fluid [21]. Another equation in the form of (where A = 0 and B,C ∈ R) introduced by Caputo [12] arises in the modelling of the motion of a single degree-of-freedom oscillator with damping system [22]. Phenomena in electromagnetics, acoustics, viscoelasticity, electrochemistry, and materials science are described by differential equations of fractional order (see [8,9,13,15,16,17,19]).
Here we apply the Adomian decomposition method for solving an extraordinary differential equation and then compare this solution with that obtained by an alternative method.
In the present paper, we consider the Adomian decomposition method for solving the following extraordinary differential equation: We suppose that L ≡ d/dx. Therefore, by the Adomian decomposition method, we can write where c is an arbitrary constant. This implies that S. S. Ray and R. K. Bera 335 In the light of the Adomian decomposition method, we assume y(x) = y 0 (x) + y 1 (x) + y 2 (x) + ··· to be the solution of (2.4), where y 0 (x) = c, Similarly, x k/2 Γ(k/2 + 1)   . (2.10) As in [2,21], the explicit formula for E 1/2,1 (z) is z k Γ(k/2 + 1) = e z 2 1 + erf(z) .