On the Generalized Simplest Equations: Toward the Solution of Nonlinear Differential Equations with Variable Coefficients

Abstract

The simplest equations with variable coefficients are considered in this research. The purpose of this study is to extend the procedure for solving the nonlinear differential equation with variable coefficients. In this case, the generalized Riccati equation is solved and becomes a basis to tackle the nonlinear differential equations with variable coefficients. The method shows that Jacobi and Weierstrass equations can be rearranged to become Riccati equation. It is also important to highlight that the solving procedure also involves the reduction of higher order polynomials with examples of Korteweg de Vries and elliptic-like equations. The generalization of the method is also explained for the case of first order polynomial differential equation.

1. Introduction

Despite the advent of supercomputers in numerical methods, increasing activities are devoted to solving nonlinear differential equations by analytical method in recent years [1, 2, 3]. Analytical solutions have their own importance concerning the physical phenomena as they are often pave the way to the construction of right theory [4]. Many methods have been proposed concerning this important problem and generally, for the problems with constant coefficients. One of the useful methods is the method of simplest Equation [5] or for some authors, the auxiliary Equations [6]. The method is built by the utilization of the first integral of simplest nonlinear differential equations, such as Bernoulli and Riccati Equations [7]. The method had produced many new solutions of the considered nonlinear differential equations, generally with constant coefficients [8, 9].

For the more general cases, we have found that the method can be extended such as involving the solution of the nonlinear differential equations with variable coefficients. The nature of variable coefficients often arises in the equation describing the heterogenous media and composites [10] or in other cases are produced by the coordinate transformation of the partial differential Equations [11]. Those two categories are developed rapidly in recent years with the capacity of high-speed super computers which sufficient for computing nonlinear problem with complex geometries [12, 13], as sometimes desired by engineering design activities. The role of analytical solutions is as a benchmark to validate the computer algorithm with simpler geometries as it is usually performed [14].

In this chapter, the solution method of the simplest equations is different from the cases of constant coefficients except, on Bernoulli equation. Hence, we will start from Riccati equation instead of Bernoulli equation as the simplest equation to highlight the novelty of the procedure. The method is then followed by examples and conclusion.

2. The first integral of the simplest equations with variable coefficients

2.1 Riccati equation

Consider the Riccati equation with variable coefficients as follows,

$$A_{\xi }=a_1\left(\xi \right)A^2+a_2\left(\xi \right)A+a_3\left(\xi \right)$$E1

Let $A={\beta }_1{\beta }_2$ and the above equation can be rearranged as,

$$\begin{array}{l}{\beta }_2{\beta }_{1\xi }+{\beta }_1{\beta }_{2\xi }=a_1{\beta }_1^2{\beta }_2^2+a_2{\beta }_1{\beta }_2+a_3\text{or}\\ {\beta }_2{\beta }_{1\xi }-a_1{\beta }_1^2{\beta }_2^2-a_2{\beta }_1{\beta }_2=-{\beta }_1{\beta }_{2\xi }+a_3={\mathit{\gamma \beta }}_1{\beta }_2.\end{array}$$

and is separated as

$${\beta }_{1\xi }-a_1{\beta }_1^2{\beta }_2-\left(a_2+\gamma \right){\beta }_1=0\text{and}{\beta }_{2\xi }+{\mathit{\gamma \beta }}_2-\frac{a_3}{{\beta }_1}=0$$E2

The solutions for ${\beta }_1$ and ${\beta }_2$ are

$${\beta }_2=e^{{\int }_{\xi }\left(a_2+\gamma \right)\mathit{d\xi }}{\left[{\int }_{\xi }{e}^{{\int }_{\xi }\left(a_2+\gamma \right)\mathit{d\xi }}{a}_1{\beta }_2\mathit{d\xi }+C_1\right]}^{-1}$$

and

$${\beta }_3=e^{-{\int }_{\xi }\mathit{\gamma d\xi }}\left({\int }_{\xi }{e}^{{\int }_{\xi }\mathit{\gamma d\xi }}\frac{a_3}{{\beta }_1}\mathit{d\xi }+C_2\right)$$E3

The relation for $A={\beta }_1{\beta }_2$ is thus,

$$A={\beta }_1{\beta }_2=e^{{\int }_{\xi }{a}_2\mathit{d\xi }}{\left[{\int }_{\xi }{e}^{{\int }_{\xi }\left(a_2+\gamma \right)\mathit{d\xi }}{a}_1{\beta }_2\mathit{d\xi }+C_1\right]}^{-1}\left({\int }_{\xi }{e}^{{\int }_{\xi }\mathit{\gamma d\xi }}\frac{a_3}{{\beta }_1}\mathit{d\xi }+C_2\right)$$E4

Without loss of generality, suppose that ${\beta }_1=e^{{\int }_{\xi }\mathit{\gamma d\xi }}$ and thus the above relation is performed as,

$${\beta }_1{\beta }_2=e^{{\int }_{\xi }{a}_2\mathit{d\xi }}{\left[{\int }_{\xi }{e}^{{\int }_{\xi }{a}_2\mathit{d\xi }}{a}_1{\beta }_1{\beta }_2\mathit{d\xi }+C_1\right]}^{-1}\left({\int }_{\xi }{a}_3\mathit{d\xi }+C_2\right)$$E5

Rearrange Eq. (5) and integrate once,

$$a_1{e}^{{\int }_{\xi }{a}_2\mathit{d\xi }}{\beta }_1{\beta }_2\left[{\int }_{\xi }{e}^{{\int }_{\xi }{a}_2\mathit{d\xi }}{a}_1{\beta }_1{\beta }_2\mathit{d\xi }+C_1\right]=a_1{e}^{2{\int }_{\xi }{a}_2\mathit{d\xi }}\left({\int }_{\xi }{a}_3\mathit{d\xi }+C_2\right)$$

or

$${\left[{\int }_{\xi }{e}^{{\int }_{\xi }{a}_2\mathit{d\xi }}{a}_1{\beta }_1{\beta }_2\mathit{d\xi }+C_1\right]}^2=2{\int }_{\xi }{a}_1{e}^{2{\int }_{\xi }{a}_2\mathit{d\xi }}\left({\int }_{\xi }{a}_3\mathit{d\xi }+C_2\right)\mathit{d\xi }+C_3$$

The solution for $A$ is then,

$$A={\beta }_1{\beta }_2=\frac{\sqrt{2}}{2}{e}^{{\int }_{\xi }{a}_2\mathit{d\xi }}\left({\int }_{\xi }{a}_3\mathit{d\xi }+C_2\right){\left[{\int }_{\xi }{a}_1{e}^{2{\int }_{\xi }{a}_2\mathit{d\xi }}\left({\int }_{\xi }{a}_3\mathit{d\xi }+C_2\right)\mathit{d\xi }+C_3\right]}^{-\frac{1}{2}}$$E6

where the coefficients $a_i$ will be determined later from the substitution into the considered nonlinear differential equations.

2.2 The Jacobi and Weierstrass equations

It is interesting to know that other simplest equations can also be rearranged into the Riccati-type equations. The famous examples are Jacobi [15] and Weierstrass Equations [16], which can solve a large class of nonlinear differential equations. Let us consider Jacobi type equation with variable coefficients,

$${\varphi }_{\xi }^2=b_1\left(\xi \right){\varphi }^4+b_2\left(\xi \right){\varphi }^3+b_3\left(\xi \right){\varphi }^2+b_4\left(\xi \right)\varphi +b_5\left(\xi \right)$$E7

and the Weierstrass equation as follows,

$${\varphi }_{\xi }^2=b_1\left(\xi \right){\varphi }^3+b_2\left(\xi \right){\varphi }^2+b_3\left(\xi \right)\varphi +b_4\left(\xi \right)$$E8

Here, the reader should not be confused by the coefficients which represent different functions with the same index. Take $\varphi =\frac{1}{\nu }+a\left(\xi \right)$ and the Weierstrass equation becomes Jacobi equation which admits the similar method of solution.

Concerning the search for obtaining solution of (7) and (8), the balancing principle suggests the substitution of the first order series $\varphi =b_6+b_{7}A$ as in the following,

$$\begin{array}{l}{\left(b_{6\xi }+b_{7\xi }A+b_7{A}_{\xi }\right)}^2=b_1{b}_7^4{A}^4+\left(4b_1{b}_6{b}_7^3+b_2{b}_7^3\right)A^3+\left(6b_1{b}_6^2{b}_7^2+3b_2{b}_6{b}_7^2+b_3{b}_7^2\right)A^2\\ +\left(4b_1{b}_6^3{b}_7+3b_2{b}_6^2{b}_7+2b_3{b}_6{b}_7+b_4{b}_7\right)A+b_1{b}_6^4+b_2{b}_6^3+b_3{b}_6^2+b_4{b}_6+b_5\end{array}$$E9

Performing the Riccati equation $A_{\xi }=a_1{b}_7{A}^2+a_{2}A+a_3$ into (9) and we generate the following expression,

$$\begin{array}{l}{\left(b_{6\xi }+b_{7\xi }A+b_7{A}_{\xi }\right)}^2={\left[a_1{b}_7^2{A}^2+\left(a_2{b}_7+b_{7\xi }\right)A+a_3{b}_7+b_{6\xi }\right]}^2=a_1^2{b}_7^4{A}^4+2a_1{b}_7\left(a_2{b}_7+b_{7\xi }\right)A^3\\ +\left[2a_1{b}_7\left(a_3{b}_7+b_{6\xi }\right)+{\left(a_2{b}_7+b_{7\xi }\right)}^2\right]A^2+2\left(a_2{b}_7+b_{7\xi }\right)\left(a_3{b}_7+b_{6\xi }\right)A+{\left(a_3{b}_7+b_{6\xi }\right)}^2\end{array}$$

The coefficients of polynomial are then related with the coefficients in (9) in order to determine $a_1,a_2,a_3,b_6,b_7$ as functions of the known $b_1,b_2,b_3,b_4$ and $b_5$ as follows,

$$\begin{array}{l}{b}_1{b}_7^4=a_1^2{b}_7^4\\ 4b_1{b}_6{b}_7^3+b_2{b}_7^3=2a_1{b}_7\left(a_2{b}_7+b_{7\xi }\right)\\ 6b_1{b}_6^2{b}_7^2+3b_2{b}_6{b}_7^2+b_3{b}_7^2=2a_1{b}_7\left(a_3{b}_7+b_{6\xi }\right)+{\left(a_2{b}_7+b_{7\xi }\right)}^2\\ 4b_1{b}_6^3{b}_7+3b_2{b}_6^2{b}_7+2b_3{b}_6{b}_7+b_4{b}_7=2\left(a_2{b}_7+b_{7\xi }\right)\left(a_3{b}_7+b_{6\xi }\right)\\ b_1{b}_6^4+b_2{b}_6^3+b_3{b}_6^2+b_4{b}_6+b_5={\left(a_3{b}_7+b_{6\xi }\right)}^2\end{array}$$E10

Hence, the first equation gives,

$$a_1=f_0\left(b_1\right)$$E11

and the second equation is then,

$$a_2{b}_7+b_{7\xi }=\frac{4b_1{b}_6{b}_7^2+b_2{b}_7^2}{2f_0}$$E12

The next relation produces,

$$a_3{b}_7+b_{6\xi }=\frac{6b_1{b}_6^2{b}_7+3b_2{b}_6{b}_7+b_3{b}_7}{2f_0}-\frac{1}{8f_0^3}{\left(4b_1{b}_6+b_2\right)}^2{b}_7^3$$E13

Eqs. (12) and (13) are thus substituted into the fourth relation of (12) to form the third order polynomial equation in term of $b_6$ as follows,

$$\begin{array}{l}\left(32f_0^4{b}_1+64b_1^3{b}_7^4-192f_0^2{b}_1^2{b}_7^2\right)b_6^3+\left(24f_0^4{b}_2+32b_1^2{b}_2{b}_7^4-16b_1^2{b}_2{b}_7^4-96f_0^2{b}_1{b}_2{b}_7^2-48f_0^2{b}_1{b}_2{b}_7^2\right)b_6^2\\ +\left(16f_0^4{b}_3+4b_1{b}_2^2{b}_7^4+8b_1{b}_2^2{b}_7^2-32f_0^2{b}_1{b}_3{b}_7^2-24f_0^2{b}_2^2{b}_7^2\right)b_6+8f_0^4{b}_4+b_2^3{b}_7^4-8f_0^2{b}_2{b}_3{b}_7^2=0\end{array}$$E14

which the roots will determine the solution for $b_6$ as functions of $b_1,b_2,b_3,b_4,b_7$, or $b_6=f_1\left(b_7\right)$ in simple unknown variable. The step now is to find the polynomial expression for $b_7$ from the last relation of (10) as,

$$\begin{array}{l}{b}_1{f}_1^4\left(b_7\right)+b_2{f}_1^3\left(b_7\right)+b_3{f}_1^2\left(b_7\right)+b_4{f}_1\left(b_7\right)+b_5=\\ {\left[\frac{6b_1{f}_1^2\left(b_7\right)b_7+3b_2{f}_1\left(b_7\right)b_7+b_3{b}_7}{2f_0}-\frac{1}{8f_0^3}{\left(4b_1{f}_1\left(b_7\right)+b_2\right)}^2{b}_7^3\right]}^2\end{array}$$E15

Therefore, the last equation gives the expression for $b_7$ as polynomial equation of higher order, and the generated polynomial is,

$$a_n{b}_7^n+a_{n-1}{b}_7^{n-1}+a_{n-2}{b}_7^{n-2}+a_{n-3}{b}_7^{n-3}+..\dots \dots \dots +a_2{b}_7^2+a_1{b}_7+a_0=0$$E16

In this case the higher order polynomial will be solved by reducing the order.

3. Reduction of higher order polynomial

Consider the sixth order polynomial equation as in the following,

$$b_7^6+a_4{b}_7^5+a_5{b}_7^4+a_6{b}_7^3+a_7{b}_7^2+a_8{b}_7+a_9=0$$

First, multiply the above equation with the function$\alpha$ and rearranged as,

$$B^6+a_4\alpha B^5+a_5{\alpha }^2{B}^4+a_6{\alpha }^3{B}^3+a_7{\alpha }^4{B}^2+a_8{\alpha }^{5}B+a_9{\alpha }^6+\phi =\phi$$E17

where $B=\alpha b_7$. The polynomial equation is cut as in the following,

$$\left(B^4+b_1{B}^3+b_2{B}^2+b_{3}B+b_4\right)B^2+b_5{B}^2-\left(B^4+b_1{B}^3+b_2{B}^2+b_{3}B+b_4\right)b_6=\phi$$

Note that, the coefficients $b_i$ in this section is different from the previous section. Expanding for the new coefficients,

$$B^6+b_1{B}^5+\left(b_2-b_5\right)B^4+\left(b_3-b_1{b}_5\right)B^3+\left(b_4+b_5-b_2{b}_5\right)B^2-b_3{b}_{5}B-b_4{b}_6=\phi$$E18

Hence, the relation for coefficients is,

$$\begin{array}{l}{b}_1=a_1\alpha \\ b_2-b_6=a_2{\alpha }^2\\ b_3-b_1{b}_6=a_3{\alpha }^3\\ b_4+b_5-b_2{b}_6=a_4{\alpha }^4\text{or}\\ -b_3{b}_6=a_5{\alpha }^5\\ -b_4{b}_6=a_6{\alpha }^6+\phi \\ b_6=\frac{\phi }{b_5}\\ b_1=a_1\alpha \\ b_2-b_6=a_2{\alpha }^2\\ b_3-a_1\alpha \left(b_2-a_2{\alpha }^2\right)=a_3{\alpha }^3\\ b_4+b_5-{\left(b_2-a_2{\alpha }^2\right)}^2=a_4{\alpha }^4+a_2{\alpha }^2\left(b_2-a_2{\alpha }^2\right)\\ -\left[a_3{\alpha }^3+a_1\alpha \left(b_2-a_2{\alpha }^2\right)\right]\left(b_2-a_2{\alpha }^2\right)=a_5{\alpha }^5\\ -\left[a_4{\alpha }^4+a_2{\alpha }^2\left(b_2-a_2{\alpha }^2\right)+{\left(b_2-a_2{\alpha }^2\right)}^2-b_5\right]\left(b_2-a_2{\alpha }^2\right)=a_6{\alpha }^6+\phi \\ \left(b_2-a_2{\alpha }^2\right)=-\frac{\left[a_4{\alpha }^4+a_2{\alpha }^2\left(b_2-a_2{\alpha }^2\right)+{\left(b_2-a_2{\alpha }^2\right)}^2-b_5\right]\left(b_2-a_2{\alpha }^2\right)+a_6{\alpha }^6}{b_5}\end{array}.$$

The fifth coefficient relation is rearranged as,

$${\left(b_2-a_2{\alpha }^2\right)}^2+\frac{a_3}{a_1}{\alpha }^2\left(b_2-a_2{\alpha }^2\right)+\frac{a_5}{a_1}{\alpha }^4=0$$E19

and the roots are,

$$\left(b_2-a_2{\alpha }^2\right)=\frac{1}{2}{\alpha }^2\left[-\frac{a_3}{a_1}\pm {\left(\frac{a_3^2}{a_1^2}-4\frac{a_5}{a_1}\right)}^{\frac{1}{2}}\right]=f_0{\alpha }^2$$E20

Also, the last relation is rewritten as,

$${\left(b_2-a_2{\alpha }^2\right)}^3+a_2{\alpha }^2{\left(b_2-a_2{\alpha }^2\right)}^2+a_4{\alpha }^4\left(b_2-a_2{\alpha }^2\right)+a_6{\alpha }^6=0$$E21

Note that performing (19) into (21) will remove $b_5$ and $\alpha$. Thus, it is necessary to take other relation, i.e. $a_6{\alpha }^6={\alpha }^{12}+b_5$, which will produce the cubic equation as follows,

$${\alpha }^{12}+\left(f_0^3+a_2{f}_0^2+a_4{f}_0\right){\alpha }^6+b_5=0$$E22

which has the roots as,

$${\alpha }^6=-\frac{1}{2}\left(f_0^3+a_2{f}_0^2+a_4{f}_0\right)\pm \frac{1}{2}{\left[{\left(f_0^3+a_2{f}_0^2+a_4{f}_0\right)}^2-4b_5\right]}^{\frac{1}{2}}$$E23

Substituting back into $a_6{\alpha }^6={\alpha }^{12}+b_5$ to get,

$$\begin{array}{l}{\left\{-\frac{1}{2}\left(f_0^3+a_2{f}_0^2+a_4{f}_0\right)\pm \frac{1}{2}{\left[{\left(f_0^3+a_2{f}_0^2+a_4{f}_0\right)}^2-4b_5\right]}^{\frac{1}{2}}\right\}}^2\\ +b_5=-\frac{1}{2}{a}_6\left(f_0^3+a_2{f}_0^2+a_4{f}_0\right)\pm \frac{1}{2}{a}_6{\left[{\left(f_0^3+a_2{f}_0^2+a_4{f}_0\right)}^2-4b_5\right]}^{\frac{1}{2}}\end{array}\text{or}$$

$$\begin{array}{l}\frac{1}{4}\left[{\left(f_0^3+a_2{f}_0^2+a_4{f}_0\right)}^2-4b_5\right]-\frac{1}{2}\left(f_0^3+a_2{f}_0^2+a_4{f}_0\right){\left[{\left(f_0^3+a_2{f}_0^2+a_4{f}_0\right)}^2-4b_5\right]}^{\frac{1}{2}}\\ +\frac{1}{4}{\left(f_0^3+a_2{f}_0^2+a_4{f}_0\right)}^2+b_5=-\frac{1}{2}{a}_6\left(f_0^3+a_2{f}_0^2+a_4{f}_0\right)\pm \frac{1}{2}{a}_6{\left[{\left(f_0^3+a_2{f}_0^2+a_4{f}_0\right)}^2-4b_5\right]}^{\frac{1}{2}}\\ \text{or}\end{array}$$

$$b_5=\frac{1}{4}{\left(f_0^3+a_2{f}_0^2+a_4{f}_0\right)}^2-\frac{1}{4}{\left[\frac{{\left(f_0^3+a_2{f}_0^2+a_4{f}_0\right)}^2+a_6\left(f_0^3+a_2{f}_0^2+a_4{f}_0\right)}{\left(f_0^3+a_2{f}_0^2+a_4{f}_0+a_6\right)}\right]}^2$$E24

Therefore,$\alpha$ is also determined by (24) and so all the coefficients, $b_1,b_2,b_3,b_4,b_6,\phi$. The polynomial equation of sixth order is then re-expressed as,

$$\left(B^4+b_1{B}^3+b_2{B}^2+b_{3}B+b_4\right)\left(B^2-b_6\right)=-b_5\left(B^2-\frac{\phi }{b_5}\right)$$E25

as reduced into the quartic equation the roots can be obtained by radical solution.

The procedure described by (17–25) can be applied and iterated into (16) until the polynomial equation of $b_7$ is reduced into quartic equation. Hence, all the coefficients for Riccati equation the first order series, i.e. $a_1,a_2,a_3,b_6,b_7$ are determined and produce the solution as,

$$\begin{array}{l}\varphi =b_6+b_{7}A\text{or}\\ \varphi =b_6+\frac{\sqrt{2}}{2}{b}_7{e}^{{\int }_{\xi }{a}_2\mathit{d\xi }}\left({\int }_{\xi }{a}_3\mathit{d\xi }+C_2\right){\left[{\int }_{\xi }{a}_1{b}_7{e}^{2{\int }_{\xi }{a}_2\mathit{d\xi }}\left({\int }_{\xi }{a}_3\mathit{d\xi }+C_2\right)\mathit{d\xi }+C_3\right]}^{-\frac{1}{2}}\end{array}$$E26

Thus, following the method explained by (2–6) and (17–25), we have arrived at the solution of Jacobi and Weierstrass equations with variable coefficients.

4. Solution examples

4.1 The elliptic-like equation

As an application, consider the elliptic-like equation with forcing function,

$${\varphi }_{\mathit{\xi \xi }}+b_1\left(\xi \right){\varphi }^3+b_2\left(\xi \right)\varphi =b_3\left(\xi \right)$$E27

The balance principle suggests that the solution should be in the form,

$$\varphi =b_4+b_{5}A$$E28

Substituting into (27) will reproduce the following expression,

$$b_5{A}_{\mathit{\xi \xi }}+2b_{5\xi }{A}_{\xi }+b_1{b}_5^3{A}^3+3b_1{b}_4{b}_5^2{A}^2+\left(b_{5\mathit{\xi \xi }}+3b_1{b}_4^2{b}_5+b_2{b}_5\right)A+b_{4\mathit{\xi \xi }}+b_1{b}_4^3+b_2{b}_4=b_3$$E29

The next step is to differentiate the Riccati equation once,

$$A_{\mathit{\xi \xi }}=2a_1^2{A}^3+\left(3a_1{a}_2+a_{1\xi }\right)A^2+\left(2a_1{a}_3+a_2^2+a_{2\xi }\right)A+a_2{a}_3+a_{3\xi }$$

Substituting into Eq. (29) and it will produce the polynomial equation as in the following,

$$\begin{array}{l}\left(2a_1^2{b}_5+b_1{b}_5^3\right)A^3+\left(3a_1{a}_2{b}_5+a_{1\xi }{b}_5+2a_1{b}_{5\xi }+3b_1{b}_4{b}_5^2\right)A^2\\ +\left(2a_1{a}_3{b}_5+a_2^2{b}_5+a_{2\xi }{b}_5+2a_2{b}_{5\xi }+b_{5\mathit{\xi \xi }}+3b_1{b}_4^2{b}_5+b_2{b}_5\right)A\\ +a_2{a}_3{b}_5+a_{3\xi }{b}_5+2a_3{b}_{5\xi }+b_{4\mathit{\xi \xi }}+b_1{b}_4^3+b_2{b}_4=b_3\end{array}$$

or the next step is to relate the coefficients as,

$$\begin{array}{l}2a_1^2{b}_5+b_1{b}_5^3=0\\ 3a_1{a}_2{b}_5+a_{1\xi }{b}_5+2a_1{b}_{5\xi }+3b_1{b}_4{b}_5^2=0\\ 2a_1{a}_3{b}_5+a_2^2{b}_5+a_{2\xi }{b}_5+2a_2{b}_{5\xi }+b_{5\mathit{\xi \xi }}+3b_1{b}_4^2{b}_5+b_2{b}_5=0\\ a_2{a}_3{b}_5+a_{3\xi }{b}_5+2a_3{b}_{5\xi }+b_{4\mathit{\xi \xi }}+b_1{b}_4^3+b_2{b}_4=b_3\end{array}$$

In this case, the first equation gives,

$$a_1=f\left(b_1\right)b_5$$E30

For the second equation,

$$3{\mathit{fa}}_2{b}_5+f_{\xi }{b}_5+3{\mathit{fb}}_{5\xi }+3b_1{b}_4{b}_5=0$$

Thus, provide the expression for $b_5$ as,

$$b_5=C_4{f}^{-\frac{1}{3}}{e}^{-{\int }_{\xi }\left(a_2+\frac{b_1{b}_4}{f}\right)\mathit{d\xi }}$$E31

The third and fourth equations produce,

$$a_3{b}_5=b_5^{-1}{e}^{-{\int }_{\xi }{a}_2\mathit{d\xi }}\left[{\int }_{\xi }{b}_5{e}^{{\int }_{\xi }{a}_2\mathit{d\xi }}\left(b_3-b_{4\mathit{\xi \xi }}-b_1{b}_4^3-b_2{b}_4\right)\mathit{d\xi }+C_4\right]$$E32

Substituting (32) into (31),

$$-a_2^2-a_{2\xi }-2a_2\frac{b_{5\xi }}{b_5}-\frac{b_{5\mathit{\xi \xi }}}{b_5}-3b_1{b}_4^2-b_2=2{\mathit{fb}}_5^{-1}{e}^{-{\int }_{\xi }{a}_2\mathit{d\xi }}\left[{\int }_{\xi }{b}_5{e}^{{\int }_{\xi }{a}_2\mathit{d\xi }}\left(b_3-b_{4\mathit{\xi \xi }}-b_1{b}_4^3-b_2{b}_4\right)\mathit{d\xi }+C_4\right]$$

Replace $b_5$ with (10),

$$\begin{array}{l}\frac{2}{3}\frac{f_{\xi }}{f}{a}_2+\frac{1}{3}\frac{f_{\mathit{\xi \xi }}}{f}-\frac{4}{9}{\left(\frac{f_{\xi }}{f}\right)}^2-{\left(\frac{b_1{b}_4}{f}\right)}^2+{\left(\frac{b_1{b}_4}{f}\right)}_{\xi }-3b_1{b}_4^2-b_2=\\ 2{\mathit{fe}}^{{\int }_{\xi }\frac{b_1{b}_4}{f}\mathit{d\xi }}\left[{\int }_{\xi }{e}^{-{\int }_{\xi }\frac{b_1{b}_4}{f}\mathit{d\xi }}\left(b_3-b_{4\mathit{\xi \xi }}-b_1{b}_4^3-b_2{b}_4\right)\mathit{d\xi }+C_4\right]\end{array}$$E33

which then solves $a_2$ regardless of $b_4$. In this case, we take $b_4$ as the chosen fundamental variable and the resulted coefficients, $b_5,a_1,a_2,a_3$ depend on $b_4$ and with the known coefficients $b_1,b_2,b_3$. Therefore, the solution of (27) is generated as,

$$\begin{array}{l}\varphi \left(\xi \right)=b_5\left(b_4\right)\left\{\frac{\sqrt{2}}{2}{e}^{{\int }_{\xi }{a}_2\left(b_4\right)\mathit{d\xi }}\left({\int }_{\xi }{a}_3\left(b_4\right)\mathit{d\xi }+C_2\right){\left[{\int }_{\xi }{a}_1\left(b_4\right)e^{2{\int }_{\xi }{a}_2\left(b_4\right)\mathit{d\xi }}\left({\int }_{\xi }{a}_3\left(b_4\right)\mathit{d\xi }+C_2\right)\mathit{d\xi }+C_3\right]}^{-\frac{1}{2}}\right\}\\ +b_4\end{array}$$E34

4.2 Korteweg de Vries equation

The next example is for the Korteweg de Vries type equation,

$${\varphi }_{\mathit{\xi \xi \xi }}+b_1\left(\xi \right){\mathit{\varphi \varphi }}_{\xi }+b_2\left(\xi \right){\varphi }_{\xi }+b_3\left(\xi \right)\varphi +b_4\left(\xi \right)=0$$E35

The balancing principle with application of Riccati equation will determined the ansatz,

$$\varphi =b_5+b_{6}A+b_7{A}^2$$E36

Performing into (35) will produce,

$$\begin{array}{l}2b_7{\mathit{AA}}_{\mathit{\xi \xi \xi }}+b_6{A}_{\mathit{\xi \xi \xi }}+6b_7{A}_{\xi }{A}_{\mathit{\xi \xi }}+6b_{7\xi }{A}_{\xi }^2+6b_{7\xi }{\mathit{AA}}_{\mathit{\xi \xi }}+2b_1{b}_7^2{A}^3{A}_{\xi }+3b_1{b}_6{b}_7{A}^2{A}_{\xi }\\ +\left(6b_{7\mathit{\xi \xi }}+b_1{b}_6^2+2b_1{b}_5{b}_7+2b_2{b}_7\right){\mathit{AA}}_{\xi }+3b_{6\xi }{A}_{\mathit{\xi \xi }}+\left(3b_{6\mathit{\xi \xi }}+b_1{b}_5{b}_6+b_2{b}_6\right)A_{\xi }+b_1{b}_7{b}_{7\xi }{A}^4\\ +\left(b_1{b}_{6\xi }{b}_7+b_1{b}_6{b}_{7\xi }\right)A^3+\left(b_{7\mathit{\xi \xi \xi }}+b_1{b}_5{b}_{7\xi }+b_1{b}_6{b}_{6\xi }+b_1{b}_{5\xi }{b}_7+b_2{b}_{7\xi }+b_3{b}_7\right)A^2\\ +\left(b_{6\mathit{\xi \xi \xi }}+b_1{b}_5{b}_{6\xi }+b_1{b}_{5\xi }{b}_6+b_2{b}_{6\xi }+b_3{b}_6\right)A+b_{5\mathit{\xi \xi \xi }}+b_1{b}_5{b}_{5\xi }+b_2{b}_{5\xi }+b_3{b}_5+b_4=0\end{array}$$E37

Performing the Riccati equation into (37) will produce the following polynomial,

$$\begin{array}{l}\left(12b_7{a}_1^3+2b_1{b}_7^2{a}_1+12a_1^3{b}_7\right)A^5+\left(\begin{array}{l}18a_1{a}_{1\xi }{b}_7+54a_1^2{a}_2{b}_7+6a_1^3{b}_6+b_1{b}_7{b}_{7\xi }+2a_2{b}_1{b}_7^2\\ +18a_1^2{b}_{7\xi }+3a_1{b}_1{b}_6{b}_7\end{array}\right)A^4\\ +\left(\begin{array}{l}40a_1^2{a}_3{b}_7+32a_1{a}_2^2{b}_7+12a_{1\xi }{a}_2{b}_7+18a_1{a}_{2\xi }{b}_7+2a_{1\mathit{\xi \xi }}{b}_7+6a_1{a}_{1\xi }{b}_6+12a_1^2{a}_2{b}_6+b_1{b}_{6\xi }{b}_7\\ +b_1{b}_6{b}_{7\xi }+2a_3{b}_1{b}_7^2+6a_2^3{b}_7+30a_1{a}_2{b}_{7\xi }+6a_{1\xi }{b}_{7\xi }+6a_1^2{b}_{6\xi }+3a_2{b}_1{b}_6{b}_7+6a_1{b}_{7\mathit{\xi \xi }}\\ +a_1{b}_1{b}_6^2+2a_1{b}_1{b}_5{b}_7+2a_1{b}_2{b}_7\end{array}\right)A^3\\ +\left(\begin{array}{l}52a_1{a}_2{a}_3{b}_7+14a_{1\xi }{a}_3{b}_7+10a_1{a}_{3\xi }{b}_7+12a_2{a}_{2\xi }{b}_7+2a_{2\mathit{\xi \xi }}{b}_7+8a_2^3{b}_7+8a_1^2{a}_3{b}_6+7a_1{a}_2^2{b}_6\\ +3a_{1\xi }{a}_2{b}_6+6a_1{a}_{2\xi }{b}_6+a_{1\mathit{\xi \xi }}{b}_6+24a_1{a}_3{b}_{7\xi }+12a_2^2{b}_{7\xi }+6a_{2\xi }{b}_{7\xi }+9a_1{a}_2{b}_{6\xi }+3a_{1\xi }{b}_{6\xi }+b_{7\mathit{\xi \xi \xi }}\\ +b_1{b}_5{b}_{7\xi }+b_1{b}_6{b}_{6\xi }+b_1{b}_{5\xi }{b}_7+b_2{b}_{7\xi }+b_3{b}_7+3a_3{b}_1{b}_6{b}_7+6a_2{b}_{7\mathit{\xi \xi }}+a_2{b}_1{b}_6^2+2a_2{b}_1{b}_5{b}_7\\ +2a_2{b}_2{b}_7+3a_1{b}_{6\mathit{\xi \xi }}+a_1{b}_1{b}_5{b}_6+a_1{b}_2{b}_6\end{array}\right)A^2\\ +\left(\begin{array}{l}16a_1{a}_3^2{b}_7+14a_2^2{a}_3{b}_7+10a_{2\xi }{a}_3{b}_7+8a_2{a}_{3\xi }{b}_7+2a_{3\mathit{\xi \xi }}{b}_7+8a_1{a}_2{a}_3{b}_6+4a_{1\xi }{a}_3{b}_6+2a_1{a}_{3\xi }{b}_6\\ +3a_2{a}_{2\xi }{b}_6+a_{2\mathit{\xi \xi }}{b}_6+a_2^3{b}_6+18a_2{a}_3{b}_{7\xi }+6a_{3\xi }{b}_{7\xi }+6a_1{a}_3{b}_{6\xi }+3a_2^2{b}_{6\xi }+3a_{2\xi }{b}_{6\xi }+b_{6\mathit{\xi \xi \xi }}\\ +b_1{b}_5{b}_{6\xi }+b_1{b}_{5\xi }{b}_6+b_2{b}_{6\xi }+b_3{b}_6+6a_3{b}_{7\mathit{\xi \xi }}+a_3{b}_1{b}_6^2+2a_3{b}_1{b}_5{b}_7+2a_3{b}_2{b}_7+3a_2{b}_{6\mathit{\xi \xi }}\\ +a_2{b}_1{b}_5{b}_6+a_2{b}_2{b}_6\end{array}\right)A\\ +\left(\begin{array}{l}2a_1{a}_3^2{b}_6+a_2^2{a}_3{b}_6+2a_{2\xi }{a}_3{b}_6+a_2{a}_{3\xi }{b}_6+a_{3\mathit{\xi \xi }}{b}_6+6a_2{a}_3^2{b}_7+6a_3{a}_{3\xi }{b}_7+3a_2{a}_3{b}_{6\xi }\\ +3a_{3\xi }{b}_{6\xi }+b_{5\mathit{\xi \xi \xi }}+b_1{b}_5{b}_{5\xi }+b_2{b}_{5\xi }+6a_3^2{b}_{7\xi }+3a_3{b}_{6\mathit{\xi \xi }}+a_3{b}_1{b}_5{b}_6+a_3{b}_2{b}_6+b_3{b}_5+b_4\end{array}\right)=0\end{array}$$E38

From this step on, there is a little hope to solve all the coefficients as they are equal to zero. As it has also to be reduced, it is important to note that the problem of reduction here is different from the case of Jacobi equation since all the coefficients are in principle solvable in algebraic form. In this case, it is not practical to reduce the fifth order polynomial as the even highest power, i.e., as a tenth order polynomial equation. The calculation will become too tedious as the detail expression is needed in the reduced polynomial equation. The next sub section will illustrate the reduction of an odd highest power polynomial equation.

4.3 Reduction of fifth order polynomial

Consider Eq. (38) as follows,

$$d_1{A}^5+d_2{A}^4+d_3{A}^3+d_4{A}^2+d_{5}A+d_6=0$$

Multiply by the function, $\beta$ and rearrange,

$$d_1{B}^5+d_2\beta B^4+d_3{\beta }^2{B}^3+d_4{\beta }^3{B}^2+d_5{\beta }^{4}B+d_6{\beta }^5+\phi =\phi$$E39

where, $B=\mathit{\beta A}$. Rearranged Eq. (39) as given by,

$$\left(b_1{B}^3+b_2{B}^2+b_{3}B+b_4\right)B^2+b_5{B}^2-\left(b_1{B}^3+b_2{B}^2+b_{3}B+b_4\right)b_6=\phi$$E40

Expanding the all the coefficients as,

$$b_1{B}^5+b_2{B}^4+\left(b_3-b_1{b}_6\right)B^3+\left(b_4+b_5-b_2{b}_6\right)B^2-b_3{b}_{6}B-b_4{b}_6=\phi$$

Relate the coefficients as in the following,

$$\begin{array}{l}{b}_1=d_1\\ b_2=d_2\beta \\ b_3-b_1{b}_6=d_3{\beta }^2\\ b_4+b_5-d_2\beta \frac{1}{d_1}\left(b_3-d_3{\beta }^2\right)=d_4{\beta }^3\\ -\frac{1}{d_1^2}{\left(b_3-d_3{\beta }^2\right)}^2=d_5{\beta }^4+d_3{\beta }^2\frac{1}{d_1}\left(b_3-d_3{\beta }^2\right)\\ -\left[d_4{\beta }^3+d_2\beta \frac{1}{d_1}\left(b_3-d_3{\beta }^2\right)-b_5\right]\frac{1}{d_1}\left(b_3-d_3{\beta }^2\right)=d_6{\beta }^5+\phi \\ \frac{1}{d_1}\left(b_3-d_3{\beta }^2\right)=-\left\{\frac{\left[d_4{\beta }^3+d_2\beta \frac{1}{d_1}\left(b_3-d_3{\beta }^2\right)-b_5\right]\frac{1}{d_1}\left(b_3-d_3{\beta }^2\right)+d_5{\beta }^5}{b_5}\right\}\end{array}$$E41

The fifth equation of (41) gives the roots as,

$$\frac{1}{d_1}\left(b_3-d_3{\beta }^2\right)=\frac{1}{2}{\beta }^2\left[d_3\pm {\left(d_3^2-4d_5\right)}^{\frac{1}{2}}\right]={\beta }^2{f}_0$$E42

Moving to the last equation, the functions $b_5$ and $\beta$ disappear from the operation. In this case we will consider the test function, $b_5+{\beta }^{10}=d_6{\beta }^5$, and will perform as,

$${\beta }^{10}+d_4{\beta }^3\frac{1}{d_1}\left(b_3-d_3{\beta }^2\right)+d_2\beta \frac{1}{d_1^2}{\left(b_3-d_3{\beta }^2\right)}^2+b_5=0$$E43

Substituting for $b_3$, the expression for $\beta$ is,

$$\begin{array}{l}{\beta }^{10}+\left(d_4{f}_0+d_2{f}_0^2\right)d_4{\beta }^5+b_5=0\\ {\beta }^5=-\frac{1}{2}\left(d_4{f}_0+d_2{f}_0^2\right)\pm \frac{1}{2}{\left[{\left(d_4{f}_0+d_2{f}_0^2\right)}^2-4b_5\right]}^{\frac{1}{2}}\end{array}$$E44

Substitute back to, $b_5+{\beta }^{10}=d_6{\beta }^5$ as follows

$${\left\{-\frac{1}{2}\left(d_4{f}_0+d_2{f}_0^2\right)\pm \frac{1}{2}{\left[{\left(d_4{f}_0+d_2{f}_0^2\right)}^2-4b_5\right]}^{\frac{1}{2}}\right\}}^2+b_5=-\frac{1}{2}{d}_6\left(d_4{f}_0+d_2{f}_0^2\right)\pm \frac{1}{2}{d}_6{\left[{\left(d_4{f}_0+d_2{f}_0^2\right)}^2-4b_5\right]}^{\frac{1}{2}}\text{or}$$

$$\begin{array}{l}\frac{1}{4}\left[{\left(d_4{f}_0+d_2{f}_0^2\right)}^2-4b_5\right]-\frac{1}{2}\left(d_4{f}_0+d_2{f}_0^2\right){\left[{\left(d_4{f}_0+d_2{f}_0^2\right)}^2-4b_5\right]}^{\frac{1}{2}}\\ +\frac{1}{4}{\left(d_4{f}_0+d_2{f}_0^2\right)}^2+b_5=-\frac{1}{2}{d}_6\left(d_4{f}_0+d_2{f}_0^2\right)\pm \frac{1}{2}{d}_6{\left[{\left(d_4{f}_0+d_2{f}_0^2\right)}^2-4b_5\right]}^{\frac{1}{2}}\\ \text{or}\end{array}$$

$$b_5=\frac{1}{4}{\left(d_4{f}_0+d_2{f}_0^2\right)}^2-\frac{1}{4}{\left[\frac{{\left(d_4{f}_0+d_2{f}_0^2\right)}^2+d_6\left(d_4{f}_0+d_2{f}_0^2\right)}{\left(d_4{f}_0+d_2{f}_0^2+d_6\right)}\right]}^2$$E45

which then solves $b_5$, $\beta$, $\phi$ and thus generates all the coefficients of $b_i$. The polynomial is then rewritten as,

$$\left(b_1{B}^3+b_2{B}^2+b_{3}B+b_4\right)\left(B^2-b_6\right)=-b_5\left(B^2-\frac{\phi }{b_5}\right)$$

which is reduced as,

$$d_1{A}^3+d_2{A}^2+\left\{\frac{1}{2}{d}_1\left[d_3\pm {\left(d_3^2-4d_5\right)}^{\frac{1}{2}}\right]+d_3\right\}A+d_4+\frac{1}{2}{d}_2\left[d_3\pm {\left(d_3^2-4d_5\right)}^{\frac{1}{2}}\right]=0$$E46

Eq. (46) dictates that the relations, $d_1=d_2=d_3=d_4=0$ will satisfy for the solution. Hence, the coefficients are then,

$$\begin{array}{l}12b_7{a}_1^3+2b_1{b}_7^2{a}_1+12a_1^3{b}_7=0\\ 18a_1{a}_{1\xi }{b}_7+54a_1^2{a}_2{b}_7+6a_1^3{b}_6+b_1{b}_7{b}_{7\xi }+2a_2{b}_1{b}_7^2+18a_1^2{b}_{7\xi }+3a_1{b}_1{b}_6{b}_7=0\\ 40a_1^2{a}_3{b}_7+32a_1{a}_2^2{b}_7+12a_{1\xi }{a}_2{b}_7+18a_1{a}_{2\xi }{b}_7+2a_{1\mathit{\xi \xi }}{b}_7+6a_1{a}_{1\xi }{b}_6+12a_1^2{a}_2{b}_6\\ +b_1{b}_{6\xi }{b}_7+b_1{b}_6{b}_{7\xi }+2a_3{b}_1{b}_7^2+6a_2^3{b}_7+30a_1{a}_2{b}_{7\xi }+6a_{1\xi }{b}_{7\xi }+6a_1^2{b}_{6\xi }+3a_2{b}_1{b}_6{b}_7\\ +6a_1{b}_{7\mathit{\xi \xi }}+a_1{b}_1{b}_6^2+2a_1{b}_1{b}_5{b}_7+2a_1{b}_2{b}_7=0\\ 52a_1{a}_2{a}_3{b}_7+14a_{1\xi }{a}_3{b}_7+10a_1{a}_{3\xi }{b}_7+12a_2{a}_{2\xi }{b}_7+2a_{2\mathit{\xi \xi }}{b}_7+8a_2^3{b}_7+8a_1^2{a}_3{b}_6\\ +7a_1{a}_2^2{b}_6+3a_{1\xi }{a}_2{b}_6+6a_1{a}_{2\xi }{b}_6+a_{1\mathit{\xi \xi }}{b}_6+24a_1{a}_3{b}_{7\xi }+12a_2^2{b}_{7\xi }+6a_{2\xi }{b}_{7\xi }+9a_1{a}_2{b}_{6\xi }\\ +3a_{1\xi }{b}_{6\xi }+b_{7\mathit{\xi \xi \xi }}+b_1{b}_5{b}_{7\xi }+b_1{b}_6{b}_{6\xi }+b_1{b}_{5\xi }{b}_7+b_2{b}_{7\xi }+b_3{b}_7+3a_3{b}_1{b}_6{b}_7+6a_2{b}_{7\mathit{\xi \xi }}\\ +a_2{b}_1{b}_6^2+2a_2{b}_1{b}_5{b}_7+2a_2{b}_2{b}_7+3a_1{b}_{6\mathit{\xi \xi }}+a_1{b}_1{b}_5{b}_6+a_1{b}_2{b}_6=0\end{array}$$E47

The first equation gives,

$$b_7=f\left(b_1\right)a_1^2$$E48

The second equation is rewritten as,

$$18a_{1\xi }{\mathit{fa}}_1^3+54{\mathit{fa}}_1^3{a}_2+6a_1^3{b}_6+b_1{\mathit{ff}}_{\xi }{a}_1^4+2b_1{f}^2{a}_1^3{a}_{1\xi }+2a_2{b}_1{f}^2{a}_1^4+18f_{\xi }{a}_1^4+18{\mathit{fa}}_1^3{a}_{1\xi }+3b_1{b}_6{\mathit{fa}}_1^2=0\text{or}$$

$$18a_{1\xi }f+54{\mathit{fa}}_2+6b_6+b_1{\mathit{ff}}_{\xi }{a}_1+2b_1{f}^2{a}_{1\xi }+2a_2{b}_1{f}^2{a}_1+18f_{\xi }{a}_1+18{\mathit{fa}}_{1\xi }+3b_1{b}_{6}f=0\text{or}$$

$$\left(36f+2b_1{f}^2\right)a_{1\xi }+54{\mathit{fa}}_2+6b_6+\left(b_1{\mathit{ff}}_{\xi }+2a_2{b}_1{f}^2+18f_{\xi }\right)a_1+3b_1{b}_{6}f=0$$

The solution for $b_6$ is then,

$$\begin{array}{l}\left(36f+2b_1{f}^2\right)a_{1\xi }=-\left(b_1{\mathit{ff}}_{\xi }+2a_2{b}_1{f}^2+18f_{\xi }\right)a_1-54{\mathit{fa}}_2-\left(6+3{\mathit{fb}}_1\right)b_6\\ b_6=-\frac{1}{\left(6+3{\mathit{fb}}_1\right)}\left[\left(b_1{\mathit{ff}}_{\xi }+2a_2{b}_1{f}^2+18f_{\xi }\right)a_1+\left(36f+2b_1{f}^2\right)a_{1\xi }+54{\mathit{fa}}_2\right]=h_1\left(a_1,a_2\right)\end{array}$$E49

The third equation will produce,

$$\begin{array}{l}40a_1^2{a}_{3}f+32a_1^2{a}_2^{2}f+12a_{1\xi }{a}_2{\mathit{fa}}_1+18a_{2\xi }{\mathit{fa}}_1^2+2a_{1\mathit{\xi \xi }}{\mathit{fa}}_1+6a_{1\xi }{b}_6+12a_1{a}_2{b}_6+b_1{b}_{6\xi }{\mathit{fa}}_1+b_1{b}_6{f}_{\xi }{a}_1\\ +2{\mathit{fa}}_{1\xi }{b}_1{b}_6+2a_3{b}_1{f}^2{a}_1^3+6a_2^3{\mathit{fa}}_1+30f_{\xi }{a}_1^2{a}_2+30{\mathit{ff}}_{\xi }{a}_1^2{a}_2+60f^2{a}_{1\xi }{a}_1{a}_2+6a_{1\xi }{f}_{\xi }{a}_1+12{\mathit{fa}}_{1\xi }^2\\ +6a_1{b}_{6\xi }+3a_2{b}_1{b}_6{\mathit{fa}}_1+6f_{\mathit{\xi \xi }}{a}_1^2+24f_{\xi }{a}_{1\xi }{a}_1+12{\mathit{fa}}_{1\mathit{\xi \xi }}{a}_1+12{\mathit{fa}}_{1\xi }^2+b_1{b}_6^2+2b_1{b}_5{\mathit{fa}}_1^2+2b_2{\mathit{fa}}_1^2=0\end{array}$$

Take the expression for $b_5$ as,

$$b_5=-\frac{1}{2b_1{\mathit{fa}}_1^2}\left[h_2\left(a_1,a_2\right)+\left(40{\mathit{fa}}_1^2+2b_1{f}^2{a}_1^3\right)a_3\right]$$

with,

$$\begin{array}{l}{h}_2\left(a_1,a_2\right)=32a_1^2{a}_2^{2}f+12a_{1\xi }{a}_2{\mathit{fa}}_1+18a_{2\xi }{\mathit{fa}}_1^2+2a_{1\mathit{\xi \xi }}{\mathit{fa}}_1+6a_{1\xi }{h}_1+12a_1{a}_2{h}_1\\ +b_1{h}_{1\xi }{\mathit{fa}}_1+b_1{h}_1{f}_{\xi }{a}_1+24f_{\xi }{a}_{1\xi }{a}_1+6a_2^3{\mathit{fa}}_1+30f_{\xi }{a}_1^2{a}_2+30{\mathit{ff}}_{\xi }{a}_1^2{a}_2+60f^2{a}_{1\xi }{a}_1{a}_2\\ +6a_{1\xi }{f}_{\xi }{a}_1+12{\mathit{fa}}_{1\xi }^2+6a_1{h}_{1\xi }+3a_2{b}_1{h}_1{\mathit{fa}}_1+6f_{\mathit{\xi \xi }}{a}_1^2+2{\mathit{fa}}_{1\xi }{b}_1{h}_1+12{\mathit{fa}}_{1\mathit{\xi \xi }}{a}_1+12{\mathit{fa}}_{1\xi }^2\\ +b_1{h}_1^2+2b_2{\mathit{fa}}_1^2\end{array}$$E50

The fourth relation of (47) will generate,

$$\begin{array}{l}{h}_3\left(a_1,a_2\right)=\left(24f_{\xi }{a}_1^3+48{\mathit{fa}}_1^2{a}_{1\xi }+3b_1{b}_6{\mathit{fa}}_1^2\right)a_3+6{\mathit{fa}}_{1\xi }{a}_{1\mathit{\xi \xi }}12a_2{a}_{2\xi }{\mathit{fa}}_1^2+2a_{2\mathit{\xi \xi }}{\mathit{fa}}_1^2\\ +8a_2^3{\mathit{fa}}_1^2+7a_1{a}_2^2{h}_1+3a_{1\xi }{a}_2{h}_1+6a_1{a}_{2\xi }{h}_1+a_{1\mathit{\xi \xi }}{h}_1+12f_{\xi }{a}_1^2{a}_2^2+24{\mathit{fa}}_1{a}_{1\xi }{a}_2^2+6f_{\xi }{a}_1^2{a}_{2\xi }\\ +12{\mathit{fa}}_1{a}_{1\xi }{a}_{2\xi }+9a_1{a}_2{h}_{1\xi }+3a_{1\xi }{h}_{1\xi }+f_{\mathit{\xi \xi \xi }}{a}_1^2+2f_{\mathit{\xi \xi }}{a}_1{a}_{1\xi }+4f_{\mathit{\xi \xi }}{a}_1{a}_{1\xi }+6f_{\xi }{a}_{1\xi }^2+6f_{\xi }{a}_1{a}_{1\mathit{\xi \xi }}\\ +2{\mathit{fa}}_1{a}_{1\mathit{\xi \xi \xi }}+b_1{h}_1{h}_{1\xi }+f_{\xi }{a}_1^2{b}_2+2{\mathit{fa}}_1{a}_{1\xi }{b}_2+b_3{\mathit{fa}}_1^2+2a_2{b}_2{\mathit{fa}}_1^2+3a_1{h}_{1\mathit{\xi \xi }}+a_1{b}_2{h}_1\\ +6f_{\mathit{\xi \xi }}{a}_1^2{a}_2+24f_{\xi }{a}_1{a}_{1\xi }{a}_2+12{\mathit{fa}}_{1\xi }^2{a}_2+12{\mathit{fa}}_1{a}_{1\mathit{\xi \xi }}{a}_2+a_2{b}_1{h}_1^2+\frac{h_2}{2b_1{\mathit{fa}}_1^2}+b_1{\mathit{fa}}_1^2{\left(\frac{h_2}{2b_1{\mathit{fa}}_1^2}\right)}_{\xi }\end{array}$$E51

which then produce the solution of $a_3$.

Note that $a_1$ and $a_2$ are the chosen fundamental variables and according to (48–51) and with the known coefficients $b_1,b_2,b_3,b_4$, they will define, $b_5,b_6,b_7,a_3,\beta$. Therefore, the solution of Korteweg de Vries equation is generated as,

$$\begin{array}{ll}\varphi \left(\xi \right)& =b_5\left(a_1,a_2\right)+b_6\left(a_1,a_2\right)\left\{\frac{\sqrt{2}}{2}\frac{e^{{\int }_{\xi }{a}_2\mathit{d\xi }}\left({\int }_{\xi }{a}_3\left(a_1,a_2\right)\mathit{d\xi }+C_2\right)}{{\left[{\int }_{\xi }{a}_1{e}^{2{\int }_{\xi }{a}_2\mathit{d\xi }}\left({\int }_{\xi }{a}_3\left(a_1,a_2\right)\mathit{d\xi }+C_2\right)\mathit{d\xi }+C_3\right]}^{\frac{1}{2}}}\right\}\\ & +b_7\left(a_1,a_2\right)\left\{\frac{1}{2}\frac{e^{2{\int }_{\xi }{a}_2\mathit{d\xi }}{\left({\int }_{\xi }{a}_3\left(a_1,a_2\right)\mathit{d\xi }+C_2\right)}^2}{\left[{\int }_{\xi }{a}_1{e}^{2{\int }_{\xi }{a}_2\mathit{d\xi }}\left({\int }_{\xi }{a}_3\left(a_1,a_2\right)\mathit{d\xi }+C_2\right)\mathit{d\xi }+C_3\right]}\right\}\end{array}$$E52

Since only a few of the considered equation has a special polynomial to be solved by equating all the variable coefficients to zero, it is important to note that the reduction of polynomial order would be an important step. Solving all coefficients to zero often be an obstacle because the difficulty would be the same or even more than the original nonlinear ODEs. In this case, the reduction of polynomial manipulates and reduces the need for solving all coefficients.

However, it is possible not to search for the expression of variable coefficients, i.e., $b_5,b_6,b_7,a_1,a_2$ and $a_3$. First the roots of Eq. (37) are determined first as $\varphi$, and then Eq. (1) is decomposed as,

$$D_{\xi }=a_1{\mathit{BD}}^2+\left(a_2-\frac{B_{\xi }}{B}\right)D+\frac{a_3}{B}$$E53

with $A=\mathit{BD}$. The solution of (53) is then,

$$D=\frac{\sqrt{2}}{2B}{e}^{{\int }_{\xi }{a}_2\mathit{d\xi }}\left({\int }_{\xi }\frac{a_3}{B}\mathit{d\xi }+C_2\right){\left[{\int }_{\xi }\frac{a_1}{B}{e}^{2{\int }_{\xi }{a}_2\mathit{d\xi }}\left({\int }_{\xi }\frac{a_3}{B}\mathit{d\xi }+C_2\right)\mathit{d\xi }+C_3\right]}^{-\frac{1}{2}}\text{or}$$

$$A=\mathit{BD}=\frac{\sqrt{2}}{2}{e}^{{\int }_{\xi }{a}_2\mathit{d\xi }}\left({\int }_{\xi }\frac{a_3}{B}\mathit{d\xi }+C_2\right){\left[{\int }_{\xi }\frac{a_1}{B}{e}^{2{\int }_{\xi }{a}_2\mathit{d\xi }}\left({\int }_{\xi }\frac{a_3}{B}\mathit{d\xi }+C_2\right)\mathit{d\xi }+C_3\right]}^{-\frac{1}{2}}$$E54

The definition for $B$ is determined by substituting the polynomial solution, $\varphi$ into (54) as in the following,

$$A=\varphi =\frac{\sqrt{2}}{2}{e}^{{\int }_{\xi }{a}_2\mathit{d\xi }}\left({\int }_{\xi }\frac{a_3}{B}\mathit{d\xi }+C_2\right){\left[{\int }_{\xi }\frac{a_1}{B}{e}^{2{\int }_{\xi }{a}_2\mathit{d\xi }}\left({\int }_{\xi }\frac{a_3}{B}\mathit{d\xi }+C_2\right)\mathit{d\xi }+C_3\right]}^{-\frac{1}{2}}$$

Rearranging the above equation as,

$${\varphi }^2\left[{\int }_{\xi }\frac{a_1}{B}{e}^{2{\int }_{\xi }{a}_2\mathit{d\xi }}\left({\int }_{\xi }\frac{a_3}{B}\mathit{d\xi }+C_2\right)\mathit{d\xi }+C_3\right]=\frac{1}{2}{e}^{2{\int }_{\xi }{a}_2\mathit{d\xi }}{\left({\int }_{\xi }\frac{a_3}{B}\mathit{d\xi }+C_2\right)}^2$$

Differentiating once,

$$\frac{a_1}{B}{e}^{2{\int }_{\xi }{a}_2\mathit{d\xi }}\left({\int }_{\xi }\frac{a_3}{B}\mathit{d\xi }+C_2\right)=\left(\frac{a_2}{{\varphi }^2}{e}^{2{\int }_{\xi }{a}_2\mathit{d\xi }}-\frac{{\varphi }_{\xi }}{{\varphi }^3}{e}^{2{\int }_{\xi }{a}_2\mathit{d\xi }}\right){\left({\int }_{\xi }\frac{a_3}{B}\mathit{d\xi }+C_2\right)}^2+\frac{a_3}{B}\frac{1}{{\varphi }^2}{e}^{2{\int }_{\xi }{a}_2\mathit{d\xi }}\left({\int }_{\xi }\frac{a_3}{B}\mathit{d\xi }+C_2\right)\text{or}$$

$$\frac{a_1}{B}=\left(\frac{a_2}{{\varphi }^2}-\frac{{\varphi }_{\xi }}{{\varphi }^3}\right)\left({\int }_{\xi }\frac{a_3}{B}\mathit{d\xi }+C_2\right)+\frac{a_3}{B}\frac{1}{{\varphi }^2}$$E55

Eq. (55) is a first order ODE in $B$ and can be easily solved, which then prove that $A=\varphi$ without establishing the explicit expression for variable coefficients.

5. Generalized method

In this section, the method of solution to the Riccati equation is extended for the class of the first order polynomial differential equation as,

$$A_{\xi }=a_n{A}^n+a_{n-1}{A}^{n-1}+a_{n-2}{A}^{n-2}+\dots \dots \dots +a_3{A}^3+a_2{A}^2+a_{1}A+a_0$$E56

The above equation can be always re-expressed as,

$$A_{\xi }=\left(b_n{A}^{n-2}+b_{n-1}{A}^{n-3}+b_{n-2}{A}^{n-4}+\dots \dots \dots +b_{3}A+b_2\right)A^2+b_{1}A+\left(b_n{A}^{n-2}+b_{n-1}{A}^{n-3}+b_{n-2}{A}^{n-4}+\dots \dots \dots +b_{3}A+b_2\right)b_0$$

or

$$A_{\xi }=b_n{A}^n+b_{n-1}{A}^{n-1}+\left(b_{n-2}+b_n{b}_0\right)A^{n-2}+\dots \dots +\left(b_3+b_{n-1}{b}_0\right)A^3+\left(b_2+b_{n-2}{b}_0\right)A^2+\left(b_1+b_3{b}_0\right)A+b_2{b}_0$$E57

which the coefficients will be reformulated as,

$$\begin{array}{l}{b}_n=a_n,b_{n-1}=a_{n-1},b_2{b}_0=a_0\\ b_{n-2}+b_n{b}_0=a_{n-2}\\ b_3+b_{n-1}{b}_0=a_3\\ b_2+b_{n-2}{b}_0=a_2\\ b_1+b_3{b}_0=a_1\end{array}\text{or}\begin{array}{l}{b}_n=a_n,b_{n-1}=a_{n-1},b_2{b}_0=a_0\\ a_2{b}_2^2-b_2^3+a_n{a}_0=a_{n-2}{a}_0{b}_2\\ b_3+a_{n-1}\frac{a_0}{b_2}=a_3\\ b_2^2+b_{n-2}{a}_0=a_2{b}_2\\ b_1+b_3\frac{a_0}{b_2}=a_1\end{array}$$E58

In this case, we will always obtain the new coefficients $b_i$. Proceeding into the other equations andmultiply the equation by the function $\alpha$, to get,

$$\begin{array}{l}\alpha A_{\xi }=\left(b_n{A}^{n-2}+b_{n-1}{A}^{n-3}+b_{n-2}{A}^{n-4}+\dots \dots \dots +b_{3}A+b_2\right)\alpha A^2+b_1\mathit{\alpha A}\\ +\left(b_n{A}^{n-2}+b_{n-1}{A}^{n-3}+b_{n-2}{A}^{n-4}+\dots \dots \dots +b_{3}A+b_2\right)\alpha b_0\end{array}\text{or}$$

$$\begin{array}{l}{B}_{\xi }=\left(b_n{\alpha }^{2-n}{B}^{n-2}+b_{n-1}{\alpha }^{3-n}{B}^{n-3}+b_{n-2}{\alpha }^{4-n}{B}^{n-4}+\dots \dots \dots +b_3{\alpha }^{-1}B+b_2\right){\alpha }^{-1}{B}^2+\left(b_1+\frac{{\alpha }_t}{\alpha }\right)B\\ +\left(b_n{\alpha }^{2-n}{B}^{n-2}+b_{n-1}{\alpha }^{3-n}{B}^{n-3}+b_{n-2}{\alpha }^{4-n}{B}^{n-4}+\dots \dots \dots +b_3{\alpha }^{-1}B+b_2\right)\alpha b_0\end{array}$$E59

where $B=\mathit{\alpha A}$. Then, all the new coefficients in $b_i$ will be determined. The step is now to solve the Riccati equation. Let $B={\beta }_2{\beta }_3$, the equation can be rearranged as,

$$\begin{array}{l}{\beta }_3{\beta }_{2\xi }+{\beta }_2{\beta }_{3\xi }=\left(b_n{\alpha }^{2-n}{B}^{n-2}+b_{n-1}{\alpha }^{3-n}{B}^{n-3}+b_{n-2}{\alpha }^{4-n}{B}^{n-4}+\dots \dots \dots +b_3{\alpha }^{-1}B+b_2\right){\alpha }^{-1}{\beta }_2^2{\beta }_3^2\\ +\left(b_1+\frac{{\alpha }_{\xi }}{\alpha }\right){\beta }_2{\beta }_3+\left(b_n{\alpha }^{2-n}{B}^{n-2}+b_{n-1}{\alpha }^{3-n}{B}^{n-3}+b_{n-2}{\alpha }^{4-n}{B}^{n-4}+\dots \dots \dots +b_3{\alpha }^{-1}B+b_2\right)\alpha b_0\end{array}\text{or}$$

$$\begin{array}{l}{\beta }_3{\beta }_{2\xi }-\left(b_n{\alpha }^{2-n}{B}^{n-2}+b_{n-1}{\alpha }^{3-n}{B}^{n-3}+b_{n-2}{\alpha }^{4-n}{B}^{n-4}+\dots \dots \dots +b_3{\alpha }^{-1}B+b_2\right){\alpha }^{-1}{\beta }_2^2{\beta }_3^2\\ -\left(b_1+\frac{{\alpha }_{\xi }}{\alpha }\right){\beta }_2{\beta }_3=-{\beta }_2{\beta }_{3\xi }+\left(\begin{array}{l}{b}_n{\alpha }^{2-n}{B}^{n-2}+b_{n-1}{\alpha }^{3-n}{B}^{n-3}+b_{n-2}{\alpha }^{4-n}{B}^{n-4}\\ +\dots \dots \dots +b_3{\alpha }^{-1}B+b_2\end{array}\right)\alpha b_0={\mathit{\gamma \beta }}_2{\beta }_3\end{array}$$E60

and is separated as,

$$\begin{array}{l}{\beta }_{2\xi }-\left(b_n{\alpha }^{2-n}{B}^{n-2}+b_{n-1}{\alpha }^{3-n}{B}^{n-3}+b_{n-2}{\alpha }^{4-n}{B}^{n-4}+\dots \dots \dots +b_3{\alpha }^{-1}B+b_2\right){\alpha }^{-2}{\beta }_2^2{\beta }_3\\ -\left(b_1+\frac{{\alpha }_{\xi }}{\alpha }+\gamma \right){\beta }_2=0\\ \text{and}\end{array}$$

$${\beta }_{3\xi }+{\mathit{\gamma \beta }}_3-\frac{1}{{\beta }_2}\left(b_n{\alpha }^{2-n}{B}^{n-2}+b_{n-1}{\alpha }^{3-n}{B}^{n-3}+b_{n-2}{\alpha }^{4-n}{B}^{n-4}+\dots \dots \dots +b_3{\alpha }^{-1}B+b_2\right)\alpha b_0=0$$E61

The solutions for ${\beta }_2$ and ${\beta }_3$ are,

$$\begin{array}{l}{\beta }_2=-e^{{\int }_{\xi }\left(b_1+\frac{{\alpha }_{\xi }}{\alpha }+\gamma \right)\mathit{d\xi }}x\\ {\left[{\int }_{\xi }{e}^{{\int }_t\left(b_1+\frac{{\alpha }_{\xi }}{\alpha }+\gamma \right)\mathit{d\xi }}\left(\begin{array}{l}{b}_n{\alpha }^{2-n}{B}^{n-2}+b_{n-1}{\alpha }^{3-n}{B}^{n-3}+b_{n-2}{\alpha }^{4-n}{B}^{n-4}\\ +\dots \dots \dots +b_3{\alpha }^{-1}B+b_2\end{array}\right){\alpha }^{-1}{\beta }_3\mathit{d\xi }+C_1\right]}^{-1}\end{array}\text{and}$$

$${\beta }_3=e^{-{\int }_{\xi }\mathit{\gamma d\xi }}\left[{\int }_{\xi }{e}^{{\int }_{\xi }\mathit{\gamma d\xi }}\left(\begin{array}{l}{b}_n{\alpha }^{2-n}{B}^{n-2}+b_{n-1}{\alpha }^{3-n}{B}^{n-3}+b_{n-2}{\alpha }^{4-n}{B}^{n-4}\\ +\dots \dots \dots +b_3{\alpha }^{-1}B+b_2\end{array}\right)\alpha \frac{b_0}{{\beta }_2}\mathit{d\xi }+C_2\right]$$E62

The relation for $B={\beta }_2{\beta }_3$ is thus,

$$\begin{array}{l}B={\beta }_2{\beta }_3=-e^{{\int }_{\xi }\left(b_1+\frac{{\alpha }_{\xi }}{\alpha }\right)\mathit{d\xi }}{\left[\begin{array}{l}{\int }_{\xi }{e}^{{\int }_{\xi }\left(b_1+\frac{{\alpha }_{\xi }}{\alpha }+\gamma \right)\mathit{d\xi }}\left(b_n{\alpha }^{2-n}{B}^{n-2}+b_{n-1}{\alpha }^{3-n}{B}^{n-3}+b_{n-2}{\alpha }^{4-n}{B}^{n-4}+\dots \dots \dots +b_3{\alpha }^{-1}B+b_2\right){\alpha }^{-1}{\beta }_3\mathit{d\xi }\\ +C_1\end{array}\right]}^{-1}x\\ \left[{\int }_{\xi }{e}^{{\int }_{\xi }\mathit{\gamma d\xi }}\left(b_n{\alpha }^{2-n}{B}^{n-2}+b_{n-1}{\alpha }^{3-n}{B}^{n-3}+b_{n-2}{\alpha }^{4-n}{B}^{n-4}+\dots \dots \dots +b_3{\alpha }^{-1}B+b_2\right)\alpha b_0\mathit{d\xi }+C_2\right]\end{array}$$

Without loss of generality, suppose that ${\beta }_2=\phi e^{{\int }_{\xi }\mathit{\gamma d\xi }}$ and the above relation is performed as,

$$\begin{array}{l}{\beta }_2{\beta }_3=-e^{{\int }_{\xi }\left(b_1+\frac{{\alpha }_{\xi }}{\alpha }\right)\mathit{d\xi }}{\left[{\int }_{\xi }{e}^{{\int }_{\xi }\left(b_1+\frac{{\alpha }_{\xi }}{\alpha }\right)\mathit{d\xi }}\left(\begin{array}{l}{b}_n{\alpha }^{2-n}{B}^{n-2}+b_{n-1}{\alpha }^{3-n}{B}^{n-3}\\ +b_{n-2}{\alpha }^{4-n}{B}^{n-4}+\dots \dots \dots +b_3{\alpha }^{-1}B\\ +b_2\end{array}\right){\alpha }^{-1}{\mathit{\phi \beta }}_2{\beta }_3\mathit{d\xi }+C_1\right]}^{-1}\\ \left[{\int }_{\xi }\left(b_n{\alpha }^{2-n}{B}^{n-2}+b_{n-1}{\alpha }^{3-n}{B}^{n-3}+b_{n-2}{\alpha }^{4-n}{B}^{n-4}+\dots \dots \dots +b_3{\alpha }^{-1}B+b_2\right){\mathit{\alpha \phi }}^{-1}{b}_0\mathit{d\xi }+C_2\right]\end{array}\text{or}$$

$$\begin{array}{l}{\beta }_2{\beta }_3\left[{\int }_{\xi }{e}^{{\int }_{\xi }\left(b_1+\frac{{\alpha }_{\xi }}{\alpha }\right)\mathit{d\xi }}\left(\begin{array}{l}{b}_n{\alpha }^{2-n}{B}^{n-2}+b_{n-1}{\alpha }^{3-n}{B}^{n-3}+b_{n-2}{\alpha }^{4-n}{B}^{n-4}\\ +\dots \dots \dots +b_3{\alpha }^{-1}B+b_2\end{array}\right){\alpha }^{-1}{\mathit{\phi \beta }}_2{\beta }_3\mathit{d\xi }+C_1\right]=\\ -e^{{\int }_{\xi }\left(b_1+\frac{{\alpha }_{\xi }}{\alpha }\right)\mathit{d\xi }}\left[{\int }_t\left(\begin{array}{l}{b}_n{\alpha }^{2-n}{B}^{n-2}+b_{n-1}{\alpha }^{3-n}{B}^{n-3}+b_{n-2}{\alpha }^{4-n}{B}^{n-4}\\ +\dots \dots \dots +b_3{\alpha }^{-1}B+b_2\end{array}\right){\mathit{\alpha \phi }}^{-1}{b}_0\mathit{d\xi }+C_2\right]\end{array}$$

Rearrange the above equation as,

$$\begin{array}{l}{e}^{{\int }_{\xi }{b}_1\mathit{d\xi }}\left(b_n{\alpha }^{2-n}{B}^{n-2}+b_{n-1}{\alpha }^{3-n}{B}^{n-3}+b_{n-2}{\alpha }^{4-n}{B}^{n-4}+\dots \dots \dots +b_3{\alpha }^{-1}B+b_2\right){\mathit{\phi \beta }}_2{\beta }_3\\ \left[{\int }_{\xi }{e}^{{\int }_{\xi }{b}_1\mathit{d\xi }}\left(b_n{y}^{n-2}+b_{n-1}{y}^{n-3}+b_{n-2}{y}^{n-4}+\dots \dots \dots +b_{3}y+b_2\right){\mathit{\phi \beta }}_2{\beta }_3\mathit{d\xi }+C_1\right]=\\ -e^{2{\int }_{\xi }{b}_1\mathit{d\xi }}\mathit{\alpha \phi }\left(b_n{\alpha }^{2-n}{B}^{n-2}+b_{n-1}{\alpha }^{3-n}{B}^{n-3}+b_{n-2}{\alpha }^{4-n}{B}^{n-4}+\dots \dots \dots +b_3{\alpha }^{-1}B+b_2\right)\\ \left[{\int }_{\xi }\left(b_n{\alpha }^{2-n}{B}^{n-2}+b_{n-1}{\alpha }^{3-n}{B}^{n-3}+b_{n-2}{\alpha }^{4-n}{B}^{n-4}+\dots \dots \dots +b_3{\alpha }^{-1}B+b_2\right){\mathit{\alpha \phi }}^{-1}{b}_0\mathit{d\xi }+C_2\right]\end{array}$$

Let $e^{2{\int }_{\xi }{b}_1\mathit{d\xi }}\mathit{\alpha \phi }={\mathit{\alpha \phi }}^{-1}{b}_0$ and integrate the above equation to get,

$$\begin{array}{l}{\left[{\int }_{\xi }{e}^{{\int }_{\xi }{b}_1\mathit{d\xi }}\left(b_n{\alpha }^{2-n}{B}^{n-2}+b_{n-1}{\alpha }^{3-n}{B}^{n-3}+b_{n-2}{\alpha }^{4-n}{B}^{n-4}+\dots \dots \dots +b_3{\alpha }^{-1}B+b_2\right){\mathit{\phi \beta }}_2{\beta }_3\mathit{dt}+C_1\right]}^2=\\ -{\left[{\int }_{\xi }\left(b_n{\alpha }^{2-n}{B}^{n-2}+b_{n-1}{\alpha }^{3-n}{B}^{n-3}+b_{n-2}{\alpha }^{4-n}{B}^{n-4}+\dots \dots \dots +b_3{\alpha }^{-1}B+b_2\right){\mathit{\alpha \phi }}^{-1}{b}_0\mathit{d\xi }+C_2\right]}^2\\ \text{or}\end{array}$$

$$\begin{array}{l}{e}^{{\int }_{\xi }{b}_1\mathit{d\xi }}\left(b_n{\alpha }^{2-n}{B}^{n-1}+b_{n-1}{\alpha }^{3-n}{B}^{n-2}+b_{n-2}{\alpha }^{4-n}{B}^{n-3}+\dots \dots \dots +b_3{\alpha }^{-1}{B}^2+b_{2}B\right){\phi }^2=\\ -\left(b_n{\alpha }^{2-n}{B}^{n-2}+b_{n-1}{\alpha }^{3-n}{B}^{n-3}+b_{n-2}{\alpha }^{4-n}{B}^{n-4}+\dots \dots \dots +b_3{\alpha }^{-1}B+b_2\right)\alpha b_0\end{array}$$

The solution for $B$ is then reduced into the solution of the polynomial equation. Thus, let $A={\alpha }^{-1}B=\varphi$, where$\varphi$ is the expression from the solution of the resulting polynomial equation which is similar to (38). The expression for $\alpha$ can be determined by the inverse method as in (53–55) for the first order polynomial differential Eq. (56).

6. Conclusion

In this chapter, we propose the method of the simplest or the auxiliary equation to solve the nonlinear differential equation with variable coefficients. The method is based on the solution of the generalized Riccati equation as the simplest equation. It is found that the other known simplest equations, i.e., Jacobi and Weierstrass equation, are also solved by the Riccati equation. The applications with the variable coefficients elliptic-like and Korteweg de Vries equations show that the problem of solving nonlinear differential equations with variable coefficients are simplified, especially by the reduction of the resulting polynomial equation in solving the Korteweg de Vries equation. The generalization of the method is also derived in detail.

Conflict of interest

Authors declare that there is no conflict of interest.