A New Analytical Method for Solving General Riccati Equation

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A New Analytical Method for Solving General

Riccati Equation

Yasar Pala, Mutlu Ozgur Ertas

*

Engineering Faculty, Uludag University, Turkey

Abstract

In this paper, the general Riccati equation is analytically solved by a new transformation. By the method developed, looking at the transformed equation, whether or not an explicit solution can be obtained is readily determined. Since the present method doesn’t require a proper solution for the general solution, it is especially suitable for equations whose proper solutions cannot be seen at a first glance. Since the transformed second order linear equation obtained by the present transformation has the simplest form it can have, it is immediately seen whether or not the original equation can be solved analytically. The present method is exemplified by several examples.

Keywords

Riccati Equation, Riccati, Ordinary

Differential Equation, Nonlinear Differential Equation, Analytical Solution, Proper Solution

1. Introduction

The generalized Riccati equation is defined by

𝑑𝑑𝑑𝑑

𝑑𝑑𝑑𝑑+𝑃𝑃(𝑥𝑥)𝑦𝑦+𝑄𝑄(𝑥𝑥)𝑦𝑦2− 𝑅𝑅(𝑥𝑥) = 0 (1)

Here, 𝑃𝑃(𝑥𝑥), 𝑄𝑄(𝑥𝑥) and 𝑅𝑅(𝑥𝑥) are arbitrary functions of

𝑥𝑥 . This equation is widely encountered in analytical mechanics, engineering and other fields. Therefore, depending on the functions 𝑃𝑃(𝑥𝑥), 𝑄𝑄(𝑥𝑥) and 𝑅𝑅(𝑥𝑥), several methods have been developed to solve various types of Riccati equations. [1-11]

If (𝑥𝑥) = 0, then the transformation 𝑦𝑦= 1⁄𝑧𝑧 reduces Eq.(1) into a first order linear equation. For the case of

(𝑥𝑥) ≠ 0 , several methods are available in the literature [1-11]. The equation in this case is also known as Bernoulli equation.

The classical method for solving Riccati equation makes the transformation 𝑦𝑦�=𝑦𝑦0+ 1⁄𝑦𝑦, where 𝑦𝑦0 is a known

solution of the equation. However, it is not possible to see a proper solution to the equation at every time. Therefore, this method has a limited usage. Among other methods, Rao transformation and Allen-Stein transformation can be

mentioned. The basic idea here is to bring the main equation into a separable form. [1,7,8] However, the differentiability condition is a serious problem with these methods.

Harko, Lobo and Mak have investigated the Riccati Equation and developed a restricted analytical solution [2]. Since the method requires specific conditions among the coefficients in the equation, it cannot be considered a general method. Sugai transformed Riccati’s equation into a second order differential equation by suggesting a new transformation. Since the transformed equation is more complicated and unsolvable in most cases, it is also not general and applicable [11]. Rao and Ukidave reduced Riccati’s equation into a separable form under restricted condition [9]. It is of no importance in respect of engineering application.Siller also investigated a separability condition of the equation [10].

Integrability condition for the Riccati equation has been studied by Mak and Harko and a new method to generate analytical solutions of the Riccati equation was presented [5]. Mortici gives a new method of the variation of constants which leads directly to an equation with separable variables. This method also imposes several restrictions on the solution. Therefore, it cannot also be considered a general method [6].

When all methods are investigated, we observe that no method explains the question whether or not the analytical solution is obtained in explicit, implicit or power series form. In addition to finding the analytical solution to the problem for arbitrary values of 𝑃𝑃(𝑥𝑥), 𝑄𝑄(𝑥𝑥) and 𝑅𝑅(𝑥𝑥) , the present method answers these important questions.

1.1. Special Type Riccati Equation

In order to solve Eq. (1), consider a new transformation in the form

𝑦𝑦�=𝑓𝑓(𝑥𝑥)𝑒𝑒∫ 𝑔𝑔(𝑑𝑑)𝑑𝑑(𝑑𝑑)𝑑𝑑𝑑𝑑 ₍₂₎

where (𝑥𝑥) , 𝑔𝑔(𝑥𝑥) are functions to be determined in a convenient manner. We differentiate both sides of Eq. (2) twice :

𝑦𝑦�′= (𝑓𝑓′₊_{𝑓𝑓𝑔𝑔𝑦𝑦}₎_𝑒𝑒∫ 𝑔𝑔(𝑑𝑑)𝑑𝑑(𝑑𝑑)𝑑𝑑𝑑𝑑 ₍₃₎

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𝑦𝑦�′′= (𝑓𝑓𝑔𝑔𝑦𝑦′_{+ 2}_𝑓𝑓′_{𝑔𝑔𝑦𝑦}₊_{𝑓𝑓𝑔𝑔}′_𝑦𝑦₊_{𝑓𝑓𝑔𝑔}2_𝑦𝑦2₊_{𝑓𝑓′′}₎_𝑒𝑒∫ 𝑔𝑔(𝑑𝑑)𝑑𝑑(𝑑𝑑)𝑑𝑑𝑑𝑑

(4) Eq. (4) can be written as

𝑦𝑦′₊_�2𝑓𝑓′ 𝑓𝑓 +

𝑔𝑔′

𝑔𝑔� 𝑦𝑦+𝑔𝑔𝑦𝑦2+ 𝑓𝑓′′ 𝑓𝑓𝑔𝑔=

1

𝑓𝑓𝑔𝑔𝑦𝑦�′′𝑒𝑒− ∫ 𝑔𝑔(𝑑𝑑)𝑑𝑑(𝑑𝑑)𝑑𝑑𝑑𝑑 (5)

Recall that the left hand side of Eq. (5) has the form of Riccati equation. Comparing Eq. (1) and Eq. (5), we have

�2𝑓𝑓′_𝑓𝑓 +𝑔𝑔′_𝑔𝑔�=𝑃𝑃(𝑥𝑥) (6a)

𝑔𝑔(𝑥𝑥) =𝑄𝑄(𝑥𝑥) (6b)

𝑓𝑓′′

𝑓𝑓𝑔𝑔=−𝑅𝑅(𝑥𝑥) (6c)

If we wish to solve an equation of the form

𝑔𝑔′

𝑔𝑔� 𝑦𝑦+𝑔𝑔𝑦𝑦2+ 𝑓𝑓′′

𝑓𝑓𝑔𝑔= 0 (7)

then, by Eq.(5), in the first place, we can have

𝑦𝑦�′′_{= 0}_→ _𝑦𝑦�₌_{𝑎𝑎𝑥𝑥}₊_𝑏𝑏₍₈₎

The functions 𝑓𝑓 and 𝑔𝑔 are to be obtained such that Eqs. (6) are satisfied. Now, using the inverse transformation from Eq. (2), we obtain

𝑦𝑦=_𝑔𝑔1_{𝑑𝑑𝑑𝑑}𝑑𝑑 �ln𝑑𝑑�_𝑓𝑓� (9) We recall that, for some simple types of Riccati equations, the solution is obtained without having any difficulty in solving the second order equation (Eq. (8)). In other words, the solution is found without solving a complex second order differential equation with variable coefficients. The following four examples illustrate the method.

Example 1: We first try to solve the equation

𝑦𝑦′_{+ 2}_𝑦𝑦₊_𝑦𝑦2_{+ 1 = 0}₍₁₀₎

whose solution can also be found by integration. Comparing Eq.(10) with Eq.(7) gives

�2𝑓𝑓′_𝑓𝑓 +𝑔𝑔′_𝑔𝑔�= 2 (11a)

𝑔𝑔= 1 (11b)

𝑓𝑓′′

𝑓𝑓𝑔𝑔= 1 (11c)

Inserting 𝑔𝑔= 1 into Eq. (11a) and solving the equation yields

𝑓𝑓=𝑐𝑐𝑒𝑒𝑑𝑑₍₁₂₎

Eq. (11c) is automatically satisfied. Now, using Eq. (9), we have

𝑦𝑦=1_𝑔𝑔_{𝑑𝑑𝑑𝑑}𝑑𝑑 �ln𝑑𝑑�_𝑓𝑓�=1₁_{𝑑𝑑𝑑𝑑}𝑑𝑑 �ln𝑎𝑎𝑑𝑑+𝑏𝑏_{𝑐𝑐𝑐𝑐}𝑥𝑥� (13)

or

𝑦𝑦=_{𝑑𝑑+𝑐𝑐̅}1 −1 , 𝑐𝑐̅=𝑏𝑏/𝑎𝑎 (14) It can be verified that Eq. (14) satisfies Eq. (10).

Example 2: We now solve the equation

𝑦𝑦′_{+ 5}_{𝑥𝑥𝑦𝑦}₊_𝑦𝑦2₊25 4𝑥𝑥2+

5

2= 0 (15)

Comparison of Eq. (15) with Eq. (7) yields

�2𝑓𝑓′_𝑓𝑓 +𝑔𝑔′_𝑔𝑔�= 5𝑥𝑥 (16a)

𝑔𝑔= 1 (16b)

𝑓𝑓′′ 𝑓𝑓𝑔𝑔=

25 4 𝑥𝑥2+

5

2 (16c)

The function 𝑓𝑓 can be obtained via Eq. (16a) and Eq. (16b)

𝑓𝑓=𝑐𝑐𝑒𝑒54𝑑𝑑2 (17)

Using Eq. (9), we find

𝑦𝑦=1_𝑔𝑔_{𝑑𝑑𝑑𝑑}𝑑𝑑 �ln_𝑓𝑓𝑑𝑑��=1₁_{𝑑𝑑𝑑𝑑}𝑑𝑑 �ln𝑎𝑎𝑑𝑑+𝑏𝑏

𝑐𝑐𝑐𝑐54𝑥𝑥2� (18)

or

𝑦𝑦=_{(𝑑𝑑+𝑐𝑐̅)}1 −5₂𝑥𝑥, 𝑐𝑐̅=𝑏𝑏/𝑎𝑎 (19) Again Eq. (19) satisfies Eq. (15).

Example 3: The equation that follows is required to be solved.

𝑦𝑦′_{+ 2}_{𝑥𝑥𝑦𝑦}₊_𝑥𝑥2_𝑦𝑦2₋ 1 𝑑𝑑2+

2

𝑑𝑑4+ 1 = 0 (20)

Comparison of Eq. (20) with Eq. (7) yields

�2𝑓𝑓′_𝑓𝑓 +𝑔𝑔′_𝑔𝑔�= 2𝑥𝑥 (21a)

𝑔𝑔=𝑥𝑥2_(21b) 𝑓𝑓′′

𝑓𝑓𝑔𝑔=− 1 𝑑𝑑2+

2

𝑑𝑑4+ 1 (21c)

The function 𝑓𝑓 can readily be obtained

𝑓𝑓=𝑐𝑐.1_𝑑𝑑.𝑒𝑒𝑥𝑥22 (22)

Using Eq. (9), we obtain

𝑦𝑦=1_𝑔𝑔_{𝑑𝑑𝑑𝑑}𝑑𝑑 �ln𝑑𝑑�_𝑓𝑓�=_𝑑𝑑12_{𝑑𝑑𝑑𝑑}𝑑𝑑 �ln 𝑎𝑎𝑑𝑑+𝑏𝑏 𝑐𝑐.𝑑𝑑−1_.𝑐𝑐𝑥𝑥22

� (23) or

𝑦𝑦=_𝑑𝑑2_{(𝑑𝑑+𝑐𝑐̅)}1 −1_𝑑𝑑+_𝑑𝑑13 𝑐𝑐̅=𝑏𝑏/𝑎𝑎 (24)

In the examples above, the examples satisfying the condition 𝑅𝑅�(𝑥𝑥) =𝑓𝑓′′ 𝑓𝑓𝑔𝑔⁄ automatically have been chosen. Otherwise, the unknown function 𝑓𝑓(𝑥𝑥) must be found such that the equations

𝑃𝑃�(𝑥𝑥) =�2𝑓𝑓′_𝑓𝑓 +𝑔𝑔′_𝑔𝑔� (25a)

𝑄𝑄�(𝑥𝑥) =𝑔𝑔 (25b)

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following example in this section. Example 4: We try to solve the equation

𝑦𝑦′₊_{𝛼𝛼𝑥𝑥}2_𝑦𝑦₊_{𝛽𝛽𝑥𝑥𝑦𝑦}2₊_𝑅𝑅�₍_𝑥𝑥_{) = 0}₍₂₆₎

Using Eq. (25a) and Eq. (25b), we can write

�2𝑓𝑓′_𝑓𝑓 +𝑔𝑔′_𝑔𝑔�=𝛼𝛼𝑥𝑥2_(27a)

𝑔𝑔=𝛽𝛽𝑥𝑥 (27b) The function 𝑓𝑓 can be obtained as

𝑓𝑓=𝑐𝑐.𝑥𝑥−12.𝑒𝑒𝛼𝛼𝑥𝑥36 (28)

We then have

𝑅𝑅�(𝑥𝑥) =𝑓𝑓′′_{𝑓𝑓𝑔𝑔}=_4𝛽𝛽𝛼𝛼2𝑥𝑥3₊ 3 4𝛽𝛽𝑥𝑥−3+

𝛼𝛼

2𝛽𝛽 (29)

Thus the present method gives a solution as long as the condition given in Eq. (29) is satisfied. The new form of Eq. (24) reads

𝑦𝑦′₊_{𝛼𝛼𝑥𝑥}2_𝑦𝑦₊_{𝛽𝛽𝑥𝑥𝑦𝑦}2₊𝛼𝛼2 4𝛽𝛽𝑥𝑥3+

3 4𝛽𝛽𝑥𝑥−3+

𝛼𝛼

2𝛽𝛽= 0 (30)

The solution can be obtained to be

𝑦𝑦=_𝑔𝑔1_{𝑑𝑑𝑑𝑑}𝑑𝑑 �ln_𝑓𝑓𝑑𝑑��=_{𝛽𝛽𝑑𝑑}1 _{𝑑𝑑𝑑𝑑}𝑑𝑑 �ln 𝑎𝑎𝑑𝑑+𝑏𝑏

𝑐𝑐.𝑑𝑑−12.𝑐𝑐𝛼𝛼𝑥𝑥36 �

(31) or

𝑦𝑦=_{𝛽𝛽𝑑𝑑(𝑑𝑑+𝑐𝑐̅)}1 −𝛼𝛼𝑑𝑑_2𝛽𝛽+_{2𝛽𝛽𝑑𝑑}12 𝑐𝑐̅=𝑏𝑏/𝑎𝑎 (32)

1.2. The Riccati Equation of General Type

The transformation in Eq. (2) prevents the free choice of the term 𝑄𝑄(𝑥𝑥) in Eq. (1). So, the method can be valid only for special type Riccati equations. Now, we wish to remove this restriction to have the analytical solution of general type Riccati equation involving arbitrary 𝑃𝑃(𝑥𝑥), 𝑄𝑄(𝑥𝑥) and 𝑅𝑅(𝑥𝑥). Again, we consider the same transformation

𝑦𝑦�=𝑓𝑓(𝑥𝑥)𝑒𝑒∫ 𝑔𝑔(𝑑𝑑)𝑑𝑑(𝑑𝑑)𝑑𝑑𝑑𝑑₍₃₃₎

Differentiating Eq. (33) twice yields

𝑦𝑦�′= (𝑓𝑓′₊_{𝑓𝑓𝑔𝑔𝑦𝑦}₎_𝑒𝑒∫ 𝑔𝑔(𝑑𝑑)𝑑𝑑(𝑑𝑑)𝑑𝑑𝑑𝑑₍₃₄₎

and

𝑦𝑦�′′= (𝑓𝑓𝑔𝑔𝑦𝑦′_{+ 2}_𝑓𝑓′_{𝑔𝑔𝑦𝑦}₊_{𝑓𝑓𝑔𝑔}′_𝑦𝑦₊_{𝑓𝑓𝑔𝑔}2_𝑦𝑦2₊_{𝑓𝑓′′}₎_𝑒𝑒∫ 𝑔𝑔(𝑑𝑑)𝑑𝑑(𝑑𝑑)𝑑𝑑𝑑𝑑

(35) Eq. (35) can be written as

𝑔𝑔′

𝑔𝑔� 𝑦𝑦+𝑔𝑔𝑦𝑦2+ 𝑓𝑓′′ 𝑓𝑓𝑔𝑔=

1

𝑓𝑓𝑔𝑔𝑦𝑦�′′𝑒𝑒− ∫ 𝑔𝑔(𝑑𝑑)𝑑𝑑(𝑑𝑑)𝑑𝑑𝑑𝑑 (36)

In order to remove the restriction in the section 1.1, we now assume that we seek the solution of the equation

𝑔𝑔′

𝑔𝑔� 𝑦𝑦+𝑔𝑔𝑦𝑦2+ 𝑓𝑓′′

𝑓𝑓𝑔𝑔=𝑆𝑆(𝑥𝑥) (37)

where 𝑆𝑆(𝑥𝑥) is a function to be determined. Thus, Eq. (36) gives

𝑦𝑦�′′₌_𝑓𝑓_._𝑔𝑔_._𝑆𝑆₍_𝑥𝑥_)._𝑒𝑒∫ 𝑔𝑔(𝑑𝑑)𝑑𝑑(𝑑𝑑)𝑑𝑑𝑑𝑑₍₃₈₎

However, by Eq. (33), we can also write

𝑦𝑦�′′₌_{𝑔𝑔𝑆𝑆}₍_𝑥𝑥₎₍_𝑦𝑦�₎₍₃₉₎

Replacing 𝑦𝑦� by 𝑢𝑢(𝑥𝑥), we obtain

𝑢𝑢′′_{− 𝑔𝑔𝑆𝑆𝑢𝑢}_{= 0}₍₄₀₎

Eq. (40) is a linear second order ordinary differential equation whose general solution is generally obtained in power series. However, if 𝑔𝑔𝑆𝑆 is a constant or something that leads to the analytical solution of Eq.(40), then the explicit form of 𝑢𝑢 or 𝑦𝑦 can always be obtained.

Now, an attention should be paid to Eq. (40). According to this result, we are led to conclude that one of the main advantages of the transformation given by Eq. (33) is that it converts Riccati equation directly into the linear form of Eq. (40). Eq. (40) has the simplest form of a second order differential equation. This form also provides us the knowledge of whether we can obtain the explicit solution of Eq. (37) or not. After Eq. (40) is solved via classical methods, the inverse transformation yields the solution as

𝑦𝑦=_{𝑔𝑔(𝑑𝑑)}1 _{𝑑𝑑𝑑𝑑}𝑑𝑑 �ln𝑢𝑢(𝑑𝑑)_{𝑓𝑓(𝑑𝑑)}� (41) Since Eq. (41) doesn’t involve any expression to be integrated, there exists no difficulty in obtaining the explicit form of 𝑦𝑦(𝑥𝑥) .

Example 5: As a first example in this section, we solve the equation

𝑦𝑦′_{+ 2}_{𝑥𝑥𝑦𝑦 − 𝑦𝑦}2₋_{(1 +}_𝑥𝑥2_{) = 0}₍₄₂₎

Comparing Eq. (42) and Eq. (37), we have

𝑔𝑔=−1 (43a)

�2𝑓𝑓′_𝑓𝑓 +𝑔𝑔′_𝑔𝑔�= 2𝑥𝑥 (43b)

𝑓𝑓′′

𝑓𝑓𝑔𝑔=−(1 +𝑥𝑥2) , 𝑆𝑆(𝑥𝑥) = 0 (43c)

Solving Eq. (43a) and Eq. (43b) gives 𝑓𝑓=𝑐𝑐𝑒𝑒𝑑𝑑2_⁄₂

(𝑐𝑐=constant) and we can see immediately that 𝑓𝑓′′ 𝑓𝑓𝑔𝑔⁄

directly equals to 𝑅𝑅(𝑥𝑥). Hence, 𝑆𝑆(𝑥𝑥) can be taken as zero. This means that 𝑦𝑦�′′ _{= 0}_{. The solution of}_𝑦𝑦�_{is given by}

𝑦𝑦�=𝑎𝑎𝑥𝑥+𝑏𝑏 . Substituting this result into the inverse transformation, after some operations, we obtain

𝑦𝑦=𝑥𝑥 −_{𝑑𝑑+𝑐𝑐̅}1 (𝑐𝑐̅=𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑎𝑎𝑐𝑐𝑐𝑐) (44) It can be checked that Eq. (44) satisfies Eq. (42).

Example 6: We seek the solution of the equation

𝑦𝑦′_{+ 8}_{𝑥𝑥𝑦𝑦}_{+ 4}_𝑦𝑦2_{+ 4}_𝑥𝑥2₋_{3 = 0}₍₄₅₎

𝑔𝑔= 4 (46a)

�2𝑓𝑓′_𝑓𝑓 +𝑔𝑔′_𝑔𝑔�= 8𝑥𝑥 (46b) Solving Eq. (46a) and Eq. (46b) gives 𝑓𝑓=𝑐𝑐𝑒𝑒2𝑑𝑑2

(𝑐𝑐=constant). The other condition yields

𝑓𝑓′′

(4)

The transformed equation in this case becomes

𝑢𝑢′′₋₁₆_𝑢𝑢_{= 0}₍₄₈₎

The characteristic equation of Eq. (48) is

𝑚𝑚2₋_{16 = 0}₍₄₉₎

whose roots are 𝑚𝑚1= 4 and 𝑚𝑚2=−4 . Thus, the solution of Eq. (48) has the form

𝑢𝑢=𝑐𝑐1𝑒𝑒4𝑑𝑑+𝑐𝑐2𝑒𝑒−4𝑑𝑑 (50)

Using Eq. (41), we obtain 𝑦𝑦(𝑥𝑥) as

𝑦𝑦=_{𝑔𝑔(𝑑𝑑)}1 _{𝑑𝑑𝑑𝑑}𝑑𝑑 �ln𝑢𝑢(𝑑𝑑)_{𝑓𝑓(𝑑𝑑)}�=₄1_{𝑑𝑑𝑑𝑑}𝑑𝑑�ln𝑐𝑐1𝑐𝑐4𝑥𝑥+𝑐𝑐2𝑐𝑐−4𝑥𝑥

𝑐𝑐𝑐𝑐2𝑥𝑥2 � (51)

𝑦𝑦

=

𝑐𝑐1𝑒𝑒4𝑥𝑥−𝑐𝑐2𝑒𝑒−4𝑥𝑥

𝑐𝑐1𝑒𝑒4𝑥𝑥+𝑐𝑐2𝑒𝑒−4𝑥𝑥

− 𝑥𝑥

(52) It can be verified that

𝑦𝑦=𝑐𝑐_𝑐𝑐8𝑥𝑥8𝑥𝑥−𝑐𝑐̅_+𝑐𝑐̅− 𝑥𝑥 , (𝑐𝑐̅=𝑐𝑐2⁄𝑐𝑐1) (53)

Note that when we take 𝑐𝑐1𝑒𝑒4𝑑𝑑 as the solution of 𝑢𝑢, then we obtain the proper solution 𝑦𝑦𝑝𝑝= 1− 𝑥𝑥. In addition, if we take

𝑐𝑐2𝑒𝑒−4𝑑𝑑 only, then we find the other proper solution of Eq. (44) as 𝑦𝑦𝑝𝑝=−1− 𝑥𝑥. Indeed, we can verify that these are proper

solutions of Eq. (45). Thus, the method presented also gives the proper solutions of Riccati Equation. Then, when desired, one can use the famous transformation 𝑦𝑦=𝑆𝑆̅(𝑥𝑥) + 1⁄𝑥𝑥 to find the general solution. Here, 𝑆𝑆̅(𝑥𝑥) is a proper solution.

Example 7: We now try to solve the equation

𝑦𝑦′₊_�₄₋1 𝑑𝑑� 𝑦𝑦+

1

𝑑𝑑𝑦𝑦2− 𝑥𝑥2+ 4𝑥𝑥= 0 (54)

We recall that a proper solution of Eq. (54) is not readily seen. Therefore, the well-known classical method cannot be used. It is obvious that the following equations are valid:

𝑔𝑔=_𝑑𝑑1 (55a)

�2𝑓𝑓′_𝑓𝑓 +𝑔𝑔′_𝑔𝑔�= 4−_𝑑𝑑1 (55b) Solving Eq. (55a) and Eq. (55b) gives 𝑓𝑓=𝑐𝑐𝑒𝑒2𝑑𝑑₍_𝑐𝑐_{=constant). Now, noting that}

𝑓𝑓′′

𝑓𝑓𝑔𝑔= 4𝑥𝑥 (56)

from Eq.(37), we obtain 𝑆𝑆(𝑥𝑥) as

𝑓𝑓′′

𝑓𝑓𝑔𝑔− 𝑆𝑆(𝑥𝑥) =−𝑥𝑥2+ 4𝑥𝑥 ⟹ 4𝑥𝑥 − 𝑆𝑆(𝑥𝑥) =−𝑥𝑥2+ 4𝑥𝑥 ⟹ 𝑆𝑆(𝑥𝑥) =𝑥𝑥2 (57)

Hence, Eq. (40) reads

𝑢𝑢′′_{− 𝑥𝑥𝑢𝑢}_{= 0}₍₅₈₎

Eq. (58) is Airy equation and its solution is given by

𝑦𝑦�=𝑢𝑢=𝑐𝑐1𝐴𝐴𝐴𝐴(𝑥𝑥) +𝑐𝑐2𝐵𝐵𝐴𝐴(𝑥𝑥) (59)

Here, the functions 𝐴𝐴𝐴𝐴𝐴𝐴𝑦𝑦𝐴𝐴𝐴𝐴(𝑥𝑥) and 𝐴𝐴𝐴𝐴𝐴𝐴𝑦𝑦𝐵𝐵𝐴𝐴(𝑥𝑥) are given as

𝐴𝐴𝐴𝐴(𝑥𝑥) = 1 +∑∞ _{2.3.5.6…(3𝑛𝑛−3).(3𝑛𝑛−1).3𝑛𝑛}𝑑𝑑3𝑛𝑛

𝑛𝑛=1 , 𝐵𝐵𝐴𝐴(𝑥𝑥) =𝑥𝑥+∑ 𝑑𝑑 3𝑛𝑛+1 3.4.6.7…(3𝑛𝑛−2).3𝑛𝑛.(3𝑛𝑛+1) ∞

𝑛𝑛=1 (60)

Their derivatives are given by

(𝐴𝐴𝐴𝐴(𝑥𝑥))′=∑∞ _{2.3.5.6…(3𝑛𝑛−3).(3𝑛𝑛−1)}𝑑𝑑3𝑛𝑛−1

𝑛𝑛=1 , (𝐵𝐵𝐴𝐴(𝑥𝑥))′= 1 +∑ 𝑑𝑑 3𝑛𝑛 3.4.5.6.7…(3𝑛𝑛−2).3𝑛𝑛 ∞

𝑛𝑛=1 (61)

Hence, using the inverse transformation, we obtain

(5)

Example 8: We try to solve the equation which has been studied by Mortici [6]

𝑦𝑦′₋𝛽𝛽

𝑑𝑑𝑦𝑦 − 𝛼𝛼𝑦𝑦2− 𝛾𝛾

𝑑𝑑2= 0 (63)

𝑔𝑔=−𝛼𝛼 (64a)

�2𝑓𝑓′_𝑓𝑓 +𝑔𝑔′_𝑔𝑔�=−𝛽𝛽_𝑑𝑑 (64b) Solving Eq. (64a) and Eq. (64b) gives 𝑓𝑓=𝑐𝑐𝑥𝑥−𝛽𝛽 2⁄ _.

(𝑐𝑐=constant). Noting that

𝑓𝑓′′ 𝑓𝑓𝑔𝑔=

−(𝛽𝛽2+2𝛽𝛽) 4𝛼𝛼 ∙

1

𝑑𝑑2 (64c)

from Eq. (37), we obtain 𝑆𝑆(𝑥𝑥) as

𝑓𝑓′′

𝑓𝑓𝑔𝑔− 𝑆𝑆(𝑥𝑥) =− 𝛾𝛾 2⟹

−(𝛽𝛽2_+2𝛽𝛽) 4𝛼𝛼 ∙

1

𝑑𝑑2 − 𝑆𝑆(𝑥𝑥) =− 𝛾𝛾

2 (65)

𝑆𝑆(𝑥𝑥) = (𝛾𝛾 −�𝛽𝛽2_4𝛼𝛼+2𝛽𝛽�)∙_𝑑𝑑12 (66)

We have the transformed equation below via Eq. (40):

𝑢𝑢′′₋₍_−𝛼𝛼₎₍_{𝛾𝛾 −}�𝛽𝛽2+2𝛽𝛽� 4𝛼𝛼 )

1

𝑑𝑑2 𝑢𝑢= 0 (67)

𝑢𝑢′′_{+ (}4𝛾𝛾𝛼𝛼−�𝛽𝛽2+2𝛽𝛽�

4 )

1

𝑑𝑑2 𝑢𝑢= 0 (68)

After this point, the following cases can be considered. Case 1: If we assume

4 𝛾𝛾𝛼𝛼

= (

𝛽𝛽

2

_{+ 2}

_𝛽𝛽

₎

_{, t}

_he

transformed equation in this case becomes

𝑢𝑢′′_{= 0}₍₆₉₎

The form of 𝑢𝑢 must be 𝑢𝑢=𝑎𝑎𝑥𝑥+𝑏𝑏. Hence, Eq. (41) gives

𝑦𝑦=_𝑔𝑔1_{𝑑𝑑𝑑𝑑}𝑑𝑑�ln𝑢𝑢_𝑓𝑓�=−_𝛼𝛼1_{𝑑𝑑𝑑𝑑}𝑑𝑑 �ln_{𝑐𝑐𝑑𝑑}𝑎𝑎𝑑𝑑+𝑏𝑏−𝛽𝛽 2⁄ � (70)

or

𝑦𝑦=_{𝛼𝛼(𝑑𝑑+𝑐𝑐̅)}−1 −_{2𝛼𝛼𝑑𝑑}𝛽𝛽 𝑐𝑐̅=𝑏𝑏/𝑎𝑎 (71) Case 2: If we assume

4 𝛾𝛾𝛼𝛼 ≠

(

𝛽𝛽

2

_{+ 2}

_𝛽𝛽

₎

_{as a}

general case, then we can put

𝑘𝑘=4𝛾𝛾𝛼𝛼−�𝛽𝛽₄2+2𝛽𝛽�=

𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐. the transformed equation in this case becomes

𝑢𝑢′′₊ 𝑘𝑘

𝑑𝑑2 𝑢𝑢= 0 (72)

This is an Euler-Cauchy Equation which has the general form.

𝑢𝑢′′₊𝐴𝐴 𝑑𝑑𝑢𝑢′+

𝐵𝐵

𝑑𝑑2 𝑢𝑢= 0 (73)

This equation can be transformed into the linear form as

𝑌𝑌′′+ (𝐴𝐴 −1)𝑌𝑌′+𝐵𝐵𝑌𝑌= 0 , (𝐴𝐴= 0 ,𝐵𝐵=𝑘𝑘) (74) The characteristic equation of Eq. (74) is

𝑚𝑚2_{− 𝑚𝑚}₊_𝑘𝑘_{= 0}₍₇₅₎

whose roots are 𝑚𝑚1 and 𝑚𝑚2. 𝑚𝑚1 and 𝑚𝑚2 can be real,

complex or repeated. The form of solution depends on the types of the roots.

Case a: If (1−4𝑘𝑘) > 0, 𝑚𝑚1≠ 𝑚𝑚2 and 𝑚𝑚1,𝑚𝑚2∈ ℝ.

𝑚𝑚1,2=1±√1−4𝑘𝑘₂

The solution for 𝑢𝑢(𝑥𝑥) in this case is

𝑢𝑢=𝑐𝑐1𝑥𝑥𝑚𝑚1+𝑐𝑐2𝑥𝑥𝑚𝑚2 (76)

Then, the solution for 𝑦𝑦(𝑥𝑥) is given by

𝑦𝑦=1_𝑔𝑔_{𝑑𝑑𝑑𝑑}𝑑𝑑�ln𝑢𝑢_𝑓𝑓�=−_𝛼𝛼1_{𝑑𝑑𝑑𝑑}𝑑𝑑 �ln𝑐𝑐1𝑑𝑑𝑚𝑚1+𝑐𝑐2𝑑𝑑𝑚𝑚2

𝑐𝑐𝑑𝑑−𝛽𝛽 2⁄ � (77)

or

𝑦𝑦=−_{𝛼𝛼𝑑𝑑}1 �𝑐𝑐1𝑚𝑚1𝑑𝑑𝑚𝑚1+𝑐𝑐2𝑚𝑚2𝑑𝑑𝑚𝑚2 𝑐𝑐1𝑑𝑑𝑚𝑚1+𝑐𝑐2𝑑𝑑𝑚𝑚2 +

𝛽𝛽

2� (78)

or

𝑦𝑦=−_{𝛼𝛼𝑑𝑑}1 �𝑚𝑚1𝑑𝑑_𝑑𝑑𝑚𝑚1_𝑚𝑚1_{+𝑐𝑐̅𝑑𝑑}+𝑐𝑐̅𝑚𝑚2𝑑𝑑_𝑚𝑚2𝑚𝑚2+𝛽𝛽₂� 𝑐𝑐̅=𝑐𝑐2⁄𝑐𝑐1 (79)

Case b: If (1−4𝑘𝑘) = 0 𝑚𝑚1=𝑚𝑚2=𝑚𝑚 and 𝑚𝑚 ∈ ℝ. The

repeated root is 𝑚𝑚=1₂

The solution for 𝑢𝑢(𝑥𝑥) is

𝑢𝑢=𝑐𝑐1𝑥𝑥𝑚𝑚+𝑐𝑐2𝑥𝑥𝑚𝑚ln𝑥𝑥 (80)

The solution in this case is obtained as

𝑦𝑦=_𝑔𝑔1_{𝑑𝑑𝑑𝑑}𝑑𝑑 �ln𝑢𝑢_𝑓𝑓�=−_𝛼𝛼1_{𝑑𝑑𝑑𝑑}𝑑𝑑 �ln𝑐𝑐1𝑑𝑑𝑚𝑚+𝑐𝑐2𝑑𝑑𝑚𝑚ln 𝑑𝑑

𝑐𝑐𝑑𝑑−𝛽𝛽 2⁄ � (81)

or

𝑦𝑦=−_𝛼𝛼1�𝑐𝑐1𝑚𝑚𝑑𝑑𝑚𝑚−1+𝑐𝑐2𝑑𝑑𝑚𝑚−1(1+𝑚𝑚 ln 𝑑𝑑) 𝑐𝑐1𝑑𝑑𝑚𝑚+𝑐𝑐2𝑑𝑑𝑚𝑚ln 𝑑𝑑 +

𝛽𝛽

2𝑑𝑑� (82)

or

𝑦𝑦=−_{2𝛼𝛼𝑑𝑑}1 �1 +_{𝑐𝑐̅+𝑙𝑙𝑛𝑛𝑑𝑑}2 +𝛽𝛽�𝑚𝑚= 1 2⁄ 𝑎𝑎𝑐𝑐𝑎𝑎𝑐𝑐̅=𝑐𝑐1⁄𝑐𝑐2 (83)

or

𝑦𝑦=−𝛽𝛽+1_{2𝛼𝛼𝑑𝑑}−_{𝑑𝑑(𝛼𝛼𝑐𝑐̅+𝛼𝛼 ln 𝑑𝑑)}1 (84) Please note that Eq. (84) is equal to Mortici’s solution [6] in this case.

Case c: 𝑚𝑚1,𝑚𝑚2∉ ℝ , 𝑚𝑚1=𝑎𝑎+𝑏𝑏𝐴𝐴 𝑎𝑎𝑐𝑐𝑎𝑎 𝑚𝑚2=𝑎𝑎 − 𝑏𝑏𝐴𝐴 ,

where 𝐴𝐴=√−1. The solution for 𝑢𝑢(𝑥𝑥) is

𝑢𝑢=𝑒𝑒−𝑎𝑎𝑑𝑑₍_𝑐𝑐

1cos𝑏𝑏𝑥𝑥+𝑐𝑐2sin𝑏𝑏𝑥𝑥) (85)

Hence, the solution for 𝑦𝑦(𝑥𝑥) is finally obtained as

𝑦𝑦=_𝑔𝑔1_{𝑑𝑑𝑑𝑑}𝑑𝑑 �ln𝑢𝑢_𝑓𝑓�=−1_𝛼𝛼_{𝑑𝑑𝑑𝑑}𝑑𝑑 �ln𝑐𝑐−𝑎𝑎𝑥𝑥(𝑐𝑐1cos 𝑏𝑏𝑑𝑑+𝑐𝑐2sin 𝑏𝑏𝑑𝑑)

𝑐𝑐𝑑𝑑−𝛽𝛽 2⁄ � (86)

or

𝑦𝑦=−1_𝛼𝛼�(𝑏𝑏𝑐𝑐̅−𝑎𝑎)_{cos 𝑏𝑏𝑑𝑑+𝑐𝑐̅ sin 𝑏𝑏𝑑𝑑}cos 𝑏𝑏𝑑𝑑−(𝑎𝑎𝑐𝑐̅+𝑏𝑏)sin 𝑏𝑏𝑑𝑑+_2𝑑𝑑𝛽𝛽� ,𝑐𝑐̅=𝑐𝑐2⁄𝑐𝑐1 (87)

2. Results and Discussion

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Looking at the form of 𝑔𝑔𝑆𝑆, we can immediately see whether the solution is expressed in explicit or implicit form. Since we don’t need any proper solution already required for the analytical solution, the present method is further simple and can be preferred in teaching the solutions of Riccati differential equation. Since the method doesn’t put any condition on the functions involved in the equation, the method is very general.

According to the present method, proper solutions of Riccati equation can also be obtained. Thus, using the conventional method of transforming the original equation into a linear first order differential equation can be followed. This situation is exemplified in Example 6.

REFERENCES

[1] Allen, J.L. and Stein, E.M, On the Solution of Certain Riccati Equations, the American Math.Montly, U.S.A., pp.1113-1115, 1964.

[2] Harko, T., Lobo, F.S.N., and Mak, M.K., Analytical Solution of the Riccati Equation with Coefficients Satisfying Integral or Differential Conditions with Arbitrary Functions, Universal Journal of Applied Mathematics, Vol.2, U.S.A., pp.109-118, 2014.

[3] Ince, E.L., Ordinary Differential Equation, ISBN: 978-0-486-60349-0, Dover Publications, New York-U.S.A., pp. 1-204, 1956.

[4] Kreyszig, E., Advanced Engineering Mathematics, ISBN: 0-471-33328-X, John Wiley&Sons. Inc, New York-U.S.A., pp.1-146, 1999.

[5] Mak, M.K. and Harko, T., New Integrability Case for the Riccati Equation, Applied Mathematics and Computation, Vol.218, Netherlands, pp.10974-10981, 2012.

[6] Mortici, C., The method of the variation of constants for Riccati Equations, General Mathematics Vol. 16, No. 1, Romania, pp.111-116, 2008)

[7] Pala, Y., Modern Differential Equations and Its Applications, ISBN: 075-591-936-8, Nobel Publications, Bursa-Turkey, pp.1-188, 2006.

[8] Rao, P.R.P., The Riccati Differential Equation, The American Mathematical Montly, U.S.A., pp.995-996, 1962.

[9] Rao, P.R.P. and Ukidave, V.H., Separable forms of the Riccati Equation, The American Mathematical Montly, Vol.75, U.S.A., pp.38-39, 1968.

[10] Siller, H., On the Separability of the Riccati Differential Equation, Mathematics Magazine, Vol.43, No.4, U.S.A., pp.197-202, 1970.