Since its inception, the differential transform method (DTM) has established a good reputation for providing convenient solutions to differential equations, both linear and nonlinear. The DTM converges to analytical solutions in form of polynomial series rapidly and does not require any discretization, linearization or perturbation [1, 2]. The mathematical literature abounds with applications of the DTM in various fields of science and engineering; e.g. see [3–7]. In a recent effort, Babolian et al. have ingeniously exploited the Adomian decomposition method to compute Laplace transforms [8]. In a similar theme, Abbasbandy has adopted He’s homopoty perturbation method to derive Laplace transforms [9]. Our objective in this short note is to propose an alternative way to compute Laplace transforms by means of the DTM. This scheme eliminates the inherent need for integration, which sometimes is not analytically tractable, to obtain Laplace transforms.

The differential transform of a given univariate function u(x) is given by the following formula:

$$ U\left( k \right) = \frac{1}{k!}\left[ {\frac{{d^{k} u\left( x \right)}}{{dx^{k} }}} \right]_{x = 0} , $$
(1)

where U(k) is the transformed function.

Also, the relevant differential inverse transform for function U(k) is defined as.

$$ u\left( x \right) = \sum\limits_{k = 0}^{\infty } {\left\{ {U\left( k \right)x^{k} } \right\}} . $$
(2)

For brevity, authors do not discuss the n-dimensional differential transform and refer the interested reader to [3]. Some basic operations of the one-dimensional differential transform are listed in Table 1. Proofs of the given operations in Table 1 can be extracted from [2, 3].

Table 1 A list of operations for the one-dimensional differential transform
Full size table

According to the proposed method, in order to have the DTM provide Laplace transform of a desired function, say Q, it is necessary to devise a differential equation whose solution is related to L{Q}. Let us consider the very fundamental first-ordered ODE:

$$ \left\{ \begin{array}{ll} u^{\prime}\left( x \right) = su\left( x \right) + Q\left( x \right), \hfill \\ u\left( 0 \right) = 0, \hfill \\ \end{array} \right. $$
(3)

where s is a positive constant. Through the use of the integrating factor, the analytical solution to (3) is given by

$$ ue^{ - sx} = \int Q e^{ - sx} dx . $$
(4)

The left hand-side of Eq. (4) equates Laplace transform of function Q, if the integration is considered from zero to infinity. That is

$$ L\left\{ Q \right\} = \left. {\left[ {e^{ - sx} u} \right]} \right|_{x = 0}^{x = + \infty } . $$
(5)

Thus, in order to obtain L{Q}, we first employ the DTM to solve Eq. (3) and simply substitute the solution into Eq. (5).

In what follows, we give some examples, which illustrate how our approach yields Laplace transform of a desired function. For better reference, we will name Eq. (3) as the “generator equation”, henceforth.

FormalPara Example 1

Let us determine L{e ax}.

As explained above, we build the generator equation as

$$ \left\{ \begin{array}{ll} u^{\prime} = su + e^{ax} , \hfill \\ u\left( 0 \right) = 0, \hfill \\ \end{array} \right. $$
(6)

and its transformed pair yields

$$ \left\{ {\begin{array}{ll} {U\left( {k + 1} \right) = \frac{s}{k + 1}U\left( k \right) + \frac{{a^{k} }}{{\left( {k + 1} \right)!}},} \\ {U\left( 0 \right) = 0.} \\ \end{array} } \right. $$
(7)

Through Eq. (7), we calculate

$$ \left\{ \begin{gathered} U\left( 1 \right) = 1, \, U\left( 2 \right) = \frac{s}{2} + \frac{a}{2!}, \, U\left( 3 \right) = \frac{{s^{2} }}{3!} + \frac{sa}{3!} + \frac{{a^{2} }}{3!}, \hfill \\ U\left( 4 \right) = \frac{{s^{3} }}{4!} + \frac{{s^{2} a}}{4!} + \frac{{sa^{2} }}{4!} + \frac{{a^{3} }}{4!}, \, U\left( 5 \right) = \frac{{s^{4} }}{5!} + \frac{{s^{3} a}}{5!} + \frac{{s^{2} a^{2} }}{5!} + \frac{{sa^{3} }}{5!} + \frac{{a^{4} }}{5!}, \, \cdots . \hfill \\ \end{gathered} \right. $$
(8)

So, we easily revert U(k) to

$$ \begin{gathered} u = \frac{1}{s}\left( {sx + \frac{{\left( {sx} \right)^{2} }}{2!} + \frac{{\left( {sx} \right)^{3} }}{3!} + \cdots } \right) + \frac{a}{{s^{2} }}\left( {\frac{{\left( {sx} \right)^{2} }}{2!} + \frac{{\left( {sx} \right)^{3} }}{3!} + \frac{{\left( {sx} \right)^{4} }}{4!} + \cdots } \right) \hfill \\ + \frac{{a^{2} }}{{s^{3} }}\left( {\frac{{\left( {sx} \right)^{3} }}{3!} + \frac{{\left( {sx} \right)^{4} }}{4!} + \frac{{\left( {sx} \right)^{5} }}{5!} + \cdots } \right) + \frac{{a^{3} }}{{s^{4} }}\left( {\frac{{\left( {sx} \right)^{4} }}{4!} + \frac{{\left( {sx} \right)^{5} }}{5!} + \frac{{\left( {sx} \right)^{6} }}{6!} + \cdots } \right) + \cdots , \hfill \\ \end{gathered} $$
(9)

or equivalently,

$$ \begin{gathered} u = \frac{1}{s}\left( {e^{sx} - 1} \right) + \frac{a}{{s^{2} }}\left( {e^{sx} - 1 - sx} \right) \hfill \\ + \frac{{a^{2} }}{{s^{3} }}\left( {e^{sx} - 1 - sx - \frac{{\left( {sx} \right)^{2} }}{2!}} \right) + \frac{{a^{3} }}{{s^{4} }}\left( {e^{sx} - 1 - sx - \frac{{\left( {sx} \right)^{2} }}{2!} - \frac{{\left( {sx} \right)^{3} }}{3!}} \right) + \cdots . \hfill \\ \end{gathered} $$
(10)

Hence,

$$ L\left\{ {e^{ax} } \right\} = \left. {\left[ {e^{ - sx} u} \right]} \right|_{x = 0}^{x = + \infty } = \frac{1}{s} + \frac{a}{{s^{2} }} + \frac{{a^{2} }}{{s^{3} }} + \frac{{a^{3} }}{{s^{4} }} + \cdots . $$
(11)

One can readily identify that the sequence in Eq. (11) is a geometric progression with 1/s and a as its common ratio and scale factor, respectively. Thus, by increasing the number of components toward infinity, we can calculate the result of the infinite series in (11) as

$$ L\left\{ {e^{ax} } \right\} = \mathop {\lim }\limits_{n \to + \infty } \frac{1}{s}\frac{{\left( {1 - \left( \frac{a}{s} \right)^{n + 1} } \right)}}{{1 - \frac{a}{s}}} = \frac{\frac{1}{s}}{{1 - \frac{a}{s}}} = \frac{1}{s - a},\quad s > \left| a \right|.\,\;{\square } $$
(12)
FormalPara Example 2

Calculate \( L\left\{ {\frac{\sin \left( x \right)}{x}} \right\} \).

The transformed equivalent of the generator equation for this problem is

$$ \left\{ \begin{array}{ll} U\left( {k + 1} \right) = \frac{sU\left( k \right)}{k + 1} + \left\{ \begin{array}{ll} \frac{{\left( { - 1} \right)^{\frac{k}{2}} }}{{\left( {k + 1} \right)\left( {k + 1} \right)!}},\quad k{\text{ is even}} \hfill \\ 0,\quad k{\text{ is odd}} \hfill \\ \end{array} \right. \hfill \\ U\left( 0 \right) = 0. \hfill \\ \end{array} \right. $$
(13)

Recursively, it follows from Eq. (13) that

$$ \left\{ \begin{gathered} U\left( 1 \right) = 1, \, U\left( 2 \right) = \frac{s}{2!}, \, U\left( 3 \right) = \frac{{s^{2} }}{3!} - \frac{1}{3 \times 3!}, \, U\left( 4 \right) = \frac{{s^{3} }}{4!} - \frac{s}{3 \times 4!}, \hfill \\ U\left( 5 \right) = \frac{{s^{4} }}{5!} - \frac{{s^{2} }}{3 \times 5!} + \frac{1}{5 \times 5!}, \, U\left( 6 \right) = \frac{{s^{5} }}{6!} - \frac{{s^{3} }}{3 \times 6!} + \frac{s}{5 \times 6!}, \hfill \\ U\left( 7 \right) = \frac{{s^{6} }}{7!} - \frac{{s^{4} }}{3 \times 7!} + \frac{{s^{2} }}{5 \times 7!} - \frac{1}{7 \times 7!}, \, U\left( 8 \right) = \frac{{s^{7} }}{8!} - \frac{{s^{5} }}{3 \times 8!} + \frac{{s^{3} }}{5 \times 8!} - \frac{s}{7 \times 8!}, \cdots . \hfill \\ \end{gathered} \right. $$
(14)

Therefore,

$$ \begin{gathered} u = \frac{1}{s}\left( {e^{sx} - 1} \right) - \frac{1}{{3s^{3} }}\left( {e^{sx} - 1 - sx - \frac{{\left( {sx} \right)^{2} }}{2!}} \right) + \frac{1}{{5s^{5} }}\left( {e^{sx} - 1 - sx - \frac{{\left( {sx} \right)^{2} }}{2!} - \frac{{\left( {sx} \right)^{3} }}{3!} - \frac{{\left( {sx} \right)^{4} }}{4!}} \right) \hfill \\ - \frac{1}{{7s^{7} }}\left( {e^{sx} - 1 - sx - \frac{{\left( {sx} \right)^{2} }}{2!} - \frac{{\left( {sx} \right)^{3} }}{3!} - \frac{{\left( {sx} \right)^{4} }}{4!} - \frac{{\left( {sx} \right)^{5} }}{5!} - \frac{{\left( {sx} \right)^{6} }}{6!}} \right) + \cdots . \hfill \\ \end{gathered} $$
(15)

Thus,

$$ L\left\{ {\frac{\sin \left( x \right)}{x}} \right\} = \left. {\left[ {e^{ - sx} u} \right]} \right|_{x = 0}^{x = + \infty } = \frac{1}{s} - \frac{1}{{3s^{3} }} + \frac{1}{{5s^{5} }} - \frac{1}{{7s^{7} }} + \cdots . $$
(16)

Knowing that

$$ \arctan \left( x \right) = \int\limits_{0}^{x} {\frac{d\xi }{{1 + \xi^{2} }}} = \int\limits_{0}^{x} {d\xi } - \int\limits_{0}^{x} {\xi^{2} d\xi } + \int\limits_{0}^{x} {\xi^{4} d\xi } - \int\limits_{0}^{x} {\xi^{6} d\xi } + \cdots , \, 0 \le x < 1 , $$
(17)

it follows that

$$ \arctan \left( \frac{1}{s} \right) = \frac{1}{s} - \frac{1}{{3s^{3} }} + \frac{1}{{5s^{5} }} - \frac{1}{{7s^{7} }} + \cdots , \, 0 \le \frac{1}{s} < 1 . $$
(18)

In view of Eqs. (16) and (18), we conclude

$$ L\left\{ {\frac{\sin \left( x \right)}{x}} \right\} = \arctan \left( \frac{1}{s} \right),\,0 \le \frac{1}{s} < 1.\,\, {\square } $$
(19)
FormalPara Example 3

Calculate L{J 0(x)}, where J 0 denotes the Bessel function of first kind of zero order.

Similar to the previous examples, we calculate the following differential transform series components by the correct choice of the generator equation:

$$ \left\{ \begin{gathered} U\left( 0 \right) = 0, \, U\left( 1 \right) = 1, \, U\left( 2 \right) = \frac{s}{2}, \, U\left( 3 \right) = \frac{{s^{2} }}{6} - \frac{1}{12}, \hfill \\ U\left( 4 \right) = \frac{{s^{3} }}{24} - \frac{s}{48}, \, U\left( 5 \right) = \frac{{s^{4} }}{120} - \frac{{s^{2} }}{ 2 4 0} + \frac{1}{320}, \hfill \\ U\left( 6 \right) = \frac{{s^{5} }}{ 7 2 0} - \frac{{s^{3} }}{ 1 4 4 0} + \frac{s}{ 1 9 2 0}, \, U\left( 7 \right) = \frac{{s^{6} }}{ 5 0 4 0} - \frac{{s^{4} }}{ 1 0 0 8 0} + \frac{{s^{2} }}{ 1 3 4 4 0} - \frac{1}{ 1 6 1 2 8}, \cdots . \hfill \\ \end{gathered} \right. $$
(20)

Now, from definition (2), it follows that

$$ \begin{gathered} u = \frac{1}{s}\left( {e^{sx} - 1} \right) - \frac{1}{{2s^{3} }}\left( {e^{sx} - 1 - sx - \frac{{\left( {sx} \right)^{2} }}{2!}} \right) \hfill \\ + \frac{3}{{8s^{5} }}\left( {e^{sx} - 1 - sx - \frac{{\left( {sx} \right)^{2} }}{2!} - \frac{{\left( {sx} \right)^{3} }}{3!} - \frac{{\left( {sx} \right)^{4} }}{4!}} \right) \hfill \\ - \frac{5}{{16s^{7} }}\left( {e^{sx} - 1 - sx - \frac{{\left( {sx} \right)^{2} }}{2!} - \frac{{\left( {sx} \right)^{3} }}{3!} - \frac{{\left( {sx} \right)^{4} }}{4!} - \frac{{\left( {sx} \right)^{5} }}{5!} + \frac{{\left( {sx} \right)^{6} }}{6!}} \right) + \cdots . \hfill \\ \end{gathered} $$
(21)

Therefore,

$$ L\left\{ {J_{0} \left( x \right)} \right\} = \left. {\left[ {e^{ - sx} u} \right]} \right|_{x = 0}^{x = + \infty } = \frac{1}{s} - \frac{1}{{2s^{3} }} + \frac{3}{{8s^{5} }} - \frac{5}{{16s^{7} }} + \cdots .\,\,{\square } $$
(22)

By virtue of the MacLaurin expansion series, we find the closed form for the previous sequence as \( {1 \mathord{\left/ {\vphantom {1 {\sqrt {\left( {s^{2} + 1} \right)} }}} \right. \kern-0pt} {\sqrt {\left( {s^{2} + 1} \right)} }} \).

To conclude, unlike the classic routine that imposes integration, our scheme only requires simple algebraic operations plus differentiation for the derivation of Laplace transforms.