Differential transform method for the solutions to some initial value problems in chemistry


Under investigation in the present work are such problems in chemistry governed by differential equations. We show that, exact analytical solutions for some chemical problems could be sought by means of the Differential transform method. Actually in the calculation process, based on the differential transform, the original differential equations is first converted to a series of algebraic equations, which could be easily solved via symbolic computing software, then with the aid of the differential inverse transform one would properly seek the exact analytical solutions to these chemical problems. In particular, four typical chemical problems are studied as examples to illustrate the effectiveness of the method.


Differential equations (DEs), whether linear or nonlinear, both play important roles in chemistry as critical mathematical models, see e.g. [1,2,3,4]. These equations could describe various phenomena and processes for chemical problems, or generally in sciences, and they are usually used to reveal natural laws.

However, as we all know, one of the most important thing is to solve these DEs since their solutions could describe the properties and characteristics for the processes of such problems [5, 6]. There are abundant analytical methods to find or construct the exact or approximate analytical solutions of DEs, such as parametric solution methods [7,8,9,10], Variable separation method [11], Taylor series method [12], Adomian decomposition method [13], Homotopy analysis method [14], etc. In particular, the Differential transform method (DTM), suggested by Zhou for analyzing electric circuit problems in Ref. [15], is one of the famous analytical methods for solving DEs. Actually, this method can be seen as an extension of the Taylor series method, and nowadays DTM and generalized DTM-based approach have been successfully developed to study many type of equations, including ordinary differential equations (ODEs) [16], partial differential equations (PDEs) [17], fractional differential equations (FDEs) [18], differential-difference equation (DDEs) [19], integro-differential equations (IDEs) [20] and so on.

The main aim of this work is to show that, the DTM could be used to sought the analytical solutions, even the exact ones to some DEs arising in chemistry, which provides an extremely simple but efficient analytical mathematical idea to seek solutions for such chemical problems. The rest of this work is organised as follows. In Sect. 2, a brief description of DTM is made. Section 3 is devoted to use the DTM to seek the exact analytical solutions of many chemical DEs as examples. Conclusions are made in Sect. 4.

Main approach

This section is devoted to provide a brief description of the DTM. Consider an analytic function u(x) in domain \(\Omega\), then u(x) could be represented by a series whose center is located at the point \(x_0\in \Omega\).

The differential transform of the function u(x) takes the form

$$\begin{aligned} U(j)=\left. \frac{1}{j!}\left[ \frac{d^j u(x)}{dx^j}\right] \right| _{x=x_0}, \end{aligned}$$

where u(x) and U(j) are called the original function and the transformed function, respectively.

The differential inverse transform reads

$$\begin{aligned} u(x)=\sum _{j=0}^\infty U(j)(x-x_0)^j. \end{aligned}$$

In particular, when \(x_0=0\), the differential transform and the differential inverse transform corresponds to the analytic function u(x) becomes

$$\begin{aligned} U(j)=\left. \frac{1}{j!}\left[ \frac{d^j u(x)}{dx^j}\right] \right| _{x=0}, \end{aligned}$$


$$\begin{aligned} u(x)=\sum _{j=0}^\infty U(j) x^j. \end{aligned}$$

For instance, some fundamental mathematical operations of DTM are given in Table 1, see Refs. [16,17,18,19,20] for reference and the references therein.

Table 1 Basic operations for the DTM

Some examples

In this section, we will apply the DTM to seek the exact solutions to some chemical problems, including linear and nonlinear examples composed of single/couple differential equation(s).

Example 1

The first considered example is the following simple linear differential equation [11]

$$\begin{aligned} a_1 \frac{dy}{dx}+a_2 y=0 \end{aligned}$$

with the initial condition

$$\begin{aligned} y(0)=y_0. \end{aligned}$$

It should be note that Eq. (5) is a classical mathematical equation that can model abundant processes and phenomena in chemistry, see the applications in Table 2 for reference, more details see Ref. [11] and the references therein. On the other hand, Eq. (5) could be easily solved by utilizing separation of variables, and the exact solution to the initial value problem (IVP) (5) and (6) is

$$\begin{aligned} y(x)=y_0 e^{-\frac{a_2}{a_1}x}. \end{aligned}$$
Table 2 Applications in chemistry of the first-order differential equation \(a_1 \frac{dy}{dx}+a_2 y=0\) [11]

Below we would like to show that the DTM could also be used to find the exact solution to the IVP (5) and (6). Particularly, from the \(1\text{st}\)\(4\text{th}\) rows in Table 1, the differential transforms of IVP (5) and (6) yield

$$\begin{aligned} a_1 (j+1) Y(j+1) + a_2 Y(j)=0 \end{aligned}$$


$$\begin{aligned} Y(0)=y_0. \end{aligned}$$

By substituting \(j=0,1,2,\ldots\) to Eqs. (8) and (9) one after another, it is not difficult to achieve that

$$\begin{aligned} Y(1) = \,& y_{0} \frac{{( - a_{2} /a_{1} )^{1} }}{1},\quad Y(2) = y_{0} \frac{{( - a_{2} /a_{1} )^{2} }}{{2 \cdots 1}},\quad Y(3) = y_{0} \frac{{( - a_{2} /a_{1} )^{3} }}{{3 \cdots 2 \cdots 1}}, \\ & \cdots ,\quad Y(j) = y_{0} \frac{{( - a_{2} /a_{1} )^{j} }}{{j!}},\quad \cdots . \\ \end{aligned}$$

Ultimately with the help of the the differential inverse transform as presented in Eq. (4), one directly obtain the exact solution to IVP (5) and (6), i.e.,

$$\begin{aligned} y(x)=y_0 \sum \limits _{j=0}^\infty \frac{(-a_2/a_1)^j}{j!} x^j=y_0 e^{-\frac{a_2}{a_1}x}, \end{aligned}$$

as also shown in Eq. (7).

Example 2

Let us investigate the stiff chemical problem governed by [4]

$$\begin{aligned} \left( \begin{array}{c} y_1' \\ y_2' \\ \end{array} \right) = \left( \begin{array}{c} -1002 y_1 + 1000y_2^2 \\ y_1 - y_2 (1+y_2) \\ \end{array} \right) \end{aligned}$$

with the initial conditions

$$\begin{aligned} \left( \begin{array}{c} y_1(0) \\ y_2(0) \\ \end{array} \right) = \left( \begin{array}{c} 1 \\ 1 \\ \end{array} \right) . \end{aligned}$$

Notice that, Eq. (12) is different to Eq. (5) in Example 1 since it is nonlinear and could not be directly solved by separation of variables. In what follows, we will show the DTM is also valid to construct the exact solution to IVP (12) and (13).

According to the operations of DTM in Table 1, the differential transforms of IVP (12) and (13) at the point \(x=0\) would be translated by

$$\begin{aligned} \left( \begin{array}{c} (j+1) Y_1(j+1) \\ (j+1) Y_2(j+1) \\ \end{array} \right) = \left( \begin{array}{c} -1002 Y_1(j) + 1000 \sum \limits _{i=0}^j Y_2(i) Y_2(j-i) \\ Y_1(j) - Y_2(j) - \sum \limits _{i=0}^j Y_2(i) Y_2(j-i) \\ \end{array} \right) , \end{aligned}$$


$$\begin{aligned} \left( \begin{array}{c} Y_1(0) \\ Y_2(0) \\ \end{array} \right) = \left( \begin{array}{c} 1 \\ 1 \\ \end{array} \right) . \end{aligned}$$

It is obvious that (14) and (15) contain a closed algebra system for any given j, which can be easily calculated through symbolic computing software, e.g. Maple, Mathematica and so on. Thus, by substituting \(j=0,1,2,\ldots\) one after another into (14) and (15), one could successively obtain

$$\begin{gathered} \left( {\begin{array}{*{20}c} {Y_{1} (1)} \\ {Y_{2} (1)} \\ \end{array} } \right) = \left( {\begin{array}{*{20}c} { - 2} \\ { - 1} \\ \end{array} } \right),\quad \left( {\begin{array}{*{20}c} {Y_{1} (2)} \\ {Y_{2} (2)} \\ \end{array} } \right) = \left( {\begin{array}{*{20}c} 2 \\ {\frac{1}{2}} \\ \end{array} } \right),\quad \left( {\begin{array}{*{20}c} {Y_{1} (3)} \\ {Y_{2} (3)} \\ \end{array} } \right) = \left( {\begin{array}{*{20}c} { - \frac{4}{3}} \\ { - \frac{1}{6}} \\ \end{array} } \right),\quad \left( {\begin{array}{*{20}c} {Y_{1} (4)} \\ {Y_{2} (4)} \\ \end{array} } \right) = \left( {\begin{array}{*{20}c} {\frac{2}{3}} \\ {\frac{1}{{24}}} \\ \end{array} } \right), \hfill \\ \left( {\begin{array}{*{20}c} {Y_{1} (5)} \\ {Y_{2} (5)} \\ \end{array} } \right) = \left( {\begin{array}{*{20}c} { - \frac{4}{{15}}} \\ { - \frac{1}{{120}}} \\ \end{array} } \right),\quad \left( {\begin{array}{*{20}c} {Y_{1} (6)} \\ {Y_{2} (6)} \\ \end{array} } \right) = \left( {\begin{array}{*{20}c} {\frac{4}{{45}}} \\ {\frac{1}{{720}}} \\ \end{array} } \right),\quad \cdots ,\quad \left( {\begin{array}{*{20}c} {Y_{1} (j)} \\ {Y_{2} (j)} \\ \end{array} } \right) = \left( {\begin{array}{*{20}c} {\frac{{( - 2)^{j} }}{{j!}}} \\ {\frac{{( - 1)^{j} }}{{j!}}} \\ \end{array} } \right),\quad \cdots . \hfill \\ \end{gathered}$$

Finally, using the differential inverse transform as given in Eq. (4), we get

$$\begin{aligned} \left( \begin{array}{c} y_1(x) \\ y_2(x) \\ \end{array} \right)&= \left( \begin{array}{c} \sum \limits _{j=0}^\infty \frac{(-2)^j}{j!} x^j \\ \sum \limits _{j=0}^\infty \frac{(-1)^j}{j!} x^j \\ \end{array} \right) = \left( \begin{array}{c} e^{-2x} \\ e^{-x} \\ \end{array} \right) , \end{aligned}$$

which is the exact analytical solution to the stiff chemical problem (12) and (13).

Example 3

Consider the following nonlinear stiff chemical problem [4]

$$\begin{aligned} \left( \begin{array}{c} y_1' \\ y_2' \\ \end{array} \right) = \left( \begin{array}{c} \lambda y_1 + y_2^2 \\ - y_2 \\ \end{array} \right) , \end{aligned}$$

with the initial conditions

$$\begin{aligned} \left( \begin{array}{c} y_1(0) \\ y_2(0) \\ \end{array} \right) = \left( \begin{array}{c} -\frac{1}{\lambda +2}\\ 1 \\ \end{array} \right) . \end{aligned}$$

Similar to Examples 1 and 2, it follows from Table 1 that the differential transforms to problem (18) and (19) yield

$$\left( {\begin{array}{*{20}c} {(j + 1)Y_{1} (j + 1)} \\ {(j + 1)Y_{2} (j + 1)} \\ \end{array} } \right) = \left( {\begin{array}{*{20}c} {\lambda Y_{1} (j) + \sum\limits_{{i = 0}}^{j} {Y_{2} } (i)Y_{2} (j - i)} \\ { - Y_{2} (j)} \\ \end{array} } \right),\quad \left( {\begin{array}{*{20}c} {Y_{1} (0)} \\ {Y_{2} (0)} \\ \end{array} } \right){\text{ }}\quad = \left( {\begin{array}{*{20}c} { - \frac{1}{{\lambda + 2}}} \\ 1 \\ \end{array} } \right).{\text{ }}$$

By symbolic computing software Maple, the closed algebra system (20) would be easily solved and the calculated results are

$$\begin{aligned} \left( \begin{array}{c} Y_1(1) \\ Y_2(1) \\ \end{array} \right) = \left( \begin{array}{c} \frac{2}{\lambda +2} \\ -1 \\ \end{array} \right) ,\quad&\left( \begin{array}{c} Y_1(2) \\ Y_2(2) \\ \end{array} \right) = \left( \begin{array}{c} -\frac{2}{\lambda +2} \\ \frac{1}{2} \\ \end{array} \right) ,\quad \left( \begin{array}{c} Y_1(3) \\ Y_2(3) \\ \end{array} \right) = \left( \begin{array}{c} \frac{4}{3(\lambda +2)} \\ -\frac{1}{6} \\ \end{array} \right) ,\nonumber \\ \left( \begin{array}{c} Y_1(4) \\ Y_2(4) \\ \end{array} \right) = \left( \begin{array}{c} -\frac{2}{3(\lambda +2)} \\ \frac{1}{24} \\ \end{array} \right) ,&\left( \begin{array}{c} Y_1(5) \\ Y_2(5) \\ \end{array} \right) = \left( \begin{array}{c} \frac{4}{15(\lambda +2)} \\ -\frac{1}{120} \\ \end{array} \right) ,\quad \left( \begin{array}{c} Y_1(6) \\ Y_2(6) \\ \end{array} \right) = \left( \begin{array}{c} -\frac{4}{45(\lambda +2)} \\ \frac{1}{720} \\ \end{array} \right) ,\nonumber \\ \cdots ,\quad&\left( \begin{array}{c} Y_1(j) \\ Y_2(j) \\ \end{array} \right) = \left( \begin{array}{c} -\frac{(-2)^j}{(\lambda +2)j!} \\ \frac{(-1)^j}{j!} \\ \end{array} \right) ,\quad \cdots . \end{aligned}$$

As a consequence, combining the results in Eq. (21) with the differential inverse transform given in Eq. (4), the exact solution to the stiff chemical problem (18) and (19) would be gained as follows:

$$\begin{aligned} \left( \begin{array}{c} y_1(x) \\ y_2(x) \\ \end{array} \right)&= \left( \begin{array}{c} \sum \limits _{j=0}^\infty -\frac{(-2)^j}{(\lambda +2)j!} x^j \\ \sum \limits _{j=0}^\infty \frac{(-1)^j}{j!} x^j \\ \end{array} \right) = \left( \begin{array}{c} -\frac{e^{-2x}}{\lambda +2} \\ e^{-x} \\ \end{array} \right) . \end{aligned}$$

Example 4

Let us finally study the stiff chemical IVP [4]

$$\begin{aligned} \left( \begin{array}{c} y_1' \\ y_2' \\ \end{array} \right) = \left( \begin{array}{c} -a y_1 - b y_2 + (a+b-1) e^{-x}\\ b y_1 - a y_2 + (a-b-1) e^{-x}\\ \end{array} \right) , \end{aligned}$$

with the initial conditions

$$\begin{aligned} \left( \begin{array}{c} y_1(0) \\ y_2(0) \\ \end{array} \right) = \left( \begin{array}{c} 1 \\ 1 \\ \end{array} \right) . \end{aligned}$$

Due to Table 1, the differential transforms at \(x=0\) to IVP (23) and (24) are

$$\left( {\begin{array}{*{20}c} {(j + 1)Y_{1} (j + 1)} \\ {(j + 1)Y_{2} (j + 1)} \\ \end{array} } \right) = \left( {\begin{array}{*{20}c} { - aY_{1} (j) - bY_{2} (j) + (a + b - 1)\frac{{( - 1)^{j} }}{{j!}}} \\ {bY_{1} (j) - aY_{2} (j) + (a - b - 1)\frac{{( - 1)^{j} }}{{j!}}} \\ \end{array} } \right),\quad \left( {\begin{array}{*{20}c} {Y_{1} (0)} \\ {Y_{2} (0)} \\ \end{array} } \right){\text{ }} = \left( {\begin{array}{*{20}c} 1 \\ 1 \\ \end{array} } \right).{\text{ }}$$

Solving the closed algebra system (25) one by one implies the following results

$$\left( {\begin{array}{*{20}c} {Y_{1} (1)} \\ {Y_{2} (1)} \\ \end{array} } \right) = \left( {\begin{array}{*{20}c} { - 1} \\ { - 1} \\ \end{array} } \right),\quad \left( {\begin{array}{*{20}c} {Y_{1} (2)} \\ {Y_{2} (2)} \\ \end{array} } \right) = \left( {\begin{array}{*{20}c} {\frac{1}{2}} \\ {\frac{1}{2}} \\ \end{array} } \right), \cdots ,\left( {\begin{array}{*{20}c} {Y_{1} (j)} \\ {Y_{2} (j)} \\ \end{array} } \right) = \left( {\begin{array}{*{20}c} {\frac{{( - 1)^{j} }}{{j!}}} \\ {\frac{{( - 1)^{j} }}{{j!}}} \\ \end{array} } \right),\quad \cdots .{\text{ }}$$

Hence, with the aid of the differential inverse transform (4) and the results in Eq. (26), we could easily obtain the exact solution to the stiff chemical IVP (23) and (24), namely,

$$\begin{aligned} \left( \begin{array}{c} y_1(x) \\ y_2(x) \\ \end{array} \right)&= \left( \begin{array}{c} \sum \limits _{j=0}^\infty \frac{(-1)^j}{j!} x^j \\ \sum \limits _{j=0}^\infty \frac{(-1)^j}{j!} x^j \\ \end{array} \right) = \left( \begin{array}{c} e^{-x} \\ e^{-x} \\ \end{array} \right) . \end{aligned}$$

It should be note that, although Examples 2-4 have been investigated in Ref. [4] by means of a new class of multistep multiderivative hybrid methods to obtain solutions numerically, the present method (DTM) provides another different route to seek the solutions even the exact analytical ones to those IVPs arising in chemistry, which reveals that the DTM not only simple but also efficient.


We have successfully sought the exact analytical solutions for some linear and nonlinear differential equations arising in chemistry, in the framework of the DTM. The example results show that, no redundant terms are produced during the calculation processes, which can be implemented straightforwardly through symbolic computing software, and the obtained series solution converges to the relevant exact solution. It indicates that the DTM provides one simple but efficient analytical mathematical idea to find and analyze the solutions of such chemical problems.


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This work is supported by the National Natural Science Foundation of China (12001233), the Key Scientific Research Projects of the Higher Education Institutions of Henan Province (20B410001), and the Doctoral Fund of Henan Institute of Technology (KQ1860).

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Qin, Y., Lou, Q. Differential transform method for the solutions to some initial value problems in chemistry. J Math Chem (2021). https://doi.org/10.1007/s10910-021-01225-7

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  • Chemical problems
  • Differential equations
  • Exact solutions
  • Differential transform method