EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 17, No. 1, 2024, 372-384
ISSN 1307-5543 – ejpam.com
Published by New York Business Global

Solving Fractional Riccati Differential equation with
Caputo- Fabrizio fractional derivative

Eman Abuteen

Department of Basic Scientific Sciences, Faculty of Engineering Technology,
Al-Balqa Applied University, Jordan

Abstract. This article offers an analytical solution for the fractional Riccati differential equation
in three distinct cases. These cases are determined by the discriminant and the analytical solution
based on the properties of the Caputo-Fabrizio fractional derivative and integral. Several examples
were tested using this analytical solution. It is noteworthy that various methods have yielded
related results as indicated in the literature.

2020 Mathematics Subject Classifications: 34A08, 26A33, 34A34, 65L05

Key Words and Phrases: Caputo-Fabrizio fractional operator, Riccati differential equation,
Fractional differential equation

1. Introduction

Fractional calculus serves as a smooth extension of classical calculus, exploring inte-
grals and derivatives of non-integer order [19],[34],[2]. This extension opens the door to a
multitude of applications and real-world phenomena. In various disciplines, including en-
gineering, physics, chemistry, biology, economics, control theory, and other different fields,
Fractional calculus has evolved into a pivotal tool. It achieves this by transforming com-
plex problems in these domains into mathematical models using fractional orders. Despite
these advancements, challenges persist in solving several models that employ fractional
differential operators. Recent progress in the theory and applications of fractional calcu-
lus has been observed, introducing analytical methods for resolving fractional differential
equations, such as Adomain decomposition method [9], variational iteration method [44],
the continuous and discrete symmetry methods [17],[41],[15],[16].
Numerous fractional operators find usage in the literature, some are more widely adopted,
such as Riemann-Liouville and Caputo operators . The integral kernel of commonly uti-
lized fractional operators is characterized by singularity. To tackle the singularity challenge
and attain efficient and dependable modeling results in recent times, Caputo and Fabrizio

DOI: https://doi.org/10.29020/nybg.ejpam.v17i1.5013

Email address: dr.eman.abuteen@bau.edu.jo (E. Abuteen)

https://www.ejpam.com 372 © 2024 EJPAM All rights reserved.



E. Abuteen / Eur. J. Pure Appl. Math, 17 (1) (2024), 372-384 373

have introduced an effective fractional order Caputo-Fabrizio derivative featuring a non-
singular kernel. A comprehensive presentation of the essential traits of the Caputo-Fabrizio
derivative is provided in [31],[14],[29],[13],[12]. Later, many authors turned to the Caputo-
Fabrizio derivative to model diverse engineering problems [14],[29],[13],[3],[10],[13],[39].
The Riccati differential equation is employed across diverse disciplines like physics, en-
gineering, biology, control theory, signal processing, and finance [25],[6],[20]. The frac-
tional Riccati equation holds significance in numerous physics and engineering contexts
[36],[33],[43],[40],[8],[11],[23],[35],[45]. Many investigators have examined the numerical so-
lution of this problem [24],[22],[21],[30],[42],[5],[7].More convenientreferences for this equa-
tion can be found in [18],[37],[1],[30],[38],[27].
This article is organized as follows: Section 2 reviews some concepts and properties of
the Caputo-Fabrizio fractional derivative along with its corresponding integral. An an-
alytical solution of any nonhomogeneous fractional differential equation is presented. In
detail, solutions of the fractional Riccati differential equation for different cases by using
the solution of the Caputo-Fabrizio fractional nonhomogeneous differential equation are
presented in the third section. Section 4 is dedicated to conducting various numerical
problems.

2. Preliminaries

In this section, we introduce some basic definitions and theorems related to Caputo-
Fabrizio fractional derivative and integral.

Definition 1. [14]
The Caputo-Fabrizio fractional derivative for a smooth function f : [a,∞) → R is defined
by

Dαf(t) =
1

1− α

∫ t

a
e

( −α

1− α
(t−s)

)
f ′(s)ds, (1)

were a, α ∈ R and α ∈ (0, 1).

Definition 2. [14]
The Caputo-Fabrizio fractional integral for a smooth function f : [a,∞) → R is defined by

Iαf(t) = (1− α)(f(t)− f(a)) + α

∫ t

a
f(s)ds, (2)

were a, α ∈ R and α ∈ (0, 1).

Theorem 1. [28]
Let a, α ∈ R with α ∈ (0, 1).Then we have

(DαIαf(t)) = f(t)− e

( −α

1− α
(t−a)

)
f(a), (3)



E. Abuteen / Eur. J. Pure Appl. Math, 17 (1) (2024), 372-384 374

Theorem 2. [28]
Let a, α ∈ R with α ∈ (0, 1). For a smooth function, f : [a,∞) → R the equality

(Dαf(t))′ = 1

1− α
f ′(t)− α

1− α
Dαf(t) (4)

Corollary 1. Let a, α ∈ R with α ∈ (0, 1).For a smooth function f : [a,∞) → R we have∫ t

a
Dαf(s)ds =

1

α
(f(t)− f(a))− 1− α

α
Dαf(t) (5)

Proof.
Integrate both sides of the equation (4) in theorem 2, we get

Dαf(t) =
1

1− α
(f(t)− f(a))− α

1− α

∫ t

a
Dαf(s)ds

Therefore, we have ∫ t

a
Dαf(s)ds =

1

α
(f(t)− f(a))− 1− α

α
Dαf(t).

Theorem 3. [32]
Let a, α ∈ R with α ∈ (0, 1).Then we have

Iα(Dαf(t)) = f(t)− f(a), (6)

Proof.
Using the definition of CF integral, we have

Iα(Dαf(t)) = (1− α)Dαf(t) + α

∫ t

a
Dαf(s)ds

Applying equation (5), we get

Iα(Dαf(t)) = (1− α)Dαf(t) + α

(
1

α
(f(t)− f(a))− 1− α

α
Dαf(t)

)
= f(t)− f(a).

Now we apply the definitions and properties of Caputo-Fabrizio fractional derivative
and integral on the nonhomogeneous fractional differential equation

Dαf(t) = g(t) (7)

Derive both sides and use theorem2, we get

−α

1− α
Dαf(t) +

1

1− α
f ′(t) = g′(t) (8)

Integrate equation (8) and use the previous properties of Caputo-Fabrizio fractional deriva-
tive.
The solution of the nonhomogeneous fractional differential equation (7) is

f(t) = (1− α)(g(t)− g(a)) + α

∫ t

a
g(s)ds+ f(a) (9)



E. Abuteen / Eur. J. Pure Appl. Math, 17 (1) (2024), 372-384 375

3. Solution of Fractional order Riccati differential equation

The typical format for the general form fractional Riccati differential equations is as
follows:

Dαy(t) = Ay2(t) +By(t) + C, 0 < α ≤ 1. (10)

With the initial condition, y(0) = y0.
Using equation (9), the solution of fractional Riccati differential equation (10) is

y(t)−y(0) = (1−α)
(
(Ay2 +By + C)− (A(y(0))2 +By(0) + C)

)
+α

∫ t

0
(Ay2(s)+By(s)+C)ds

(11)
By deriving both sides, we get

y′(t) = (1− α) (2Ay(t)y′(t) +By′(t)) + α(Ay2(t) +By(t) + C) (12)

which is equivalent to

y′(t)
Ay2(t) +By(t) + C

− (1− α)
(2Ay(t)y′(t) +By′(t))
Ay2(t) +By(t) + C

= α (13)

To find a general solution for Fractional order Riccati differential equation, we analyze the
equation Ay2(t) +By(t) + C, under 3 distinct cases

Case 1: Assume that the discriminant △ = B2 − 4AC > 0
The fractional Riccati differential equations can be reformulated as

Dαy(t) = Ay2 +By + C = (a1y + b1)(a2y + b2) (14)

Using equation (13), we have

y′(t)
(a1y + b1)(a2y + b2)

− (1− α)
(2Ay(t)y′(t) +By′(t))
Ay2(t) +By(t) + C

= α

Using partial fractions, we get

y′(t)
(

A1

a1y + b1
− A2

a2y + b2

)
− (1− α)

(2Ay(t)y′(t) +By′(t))
Ay2(t) +By(t) + C

= α (15)

Where A1 =
a1

b2a1 − b1a2
and A2 =

a2
b1a2 − b2a1

The solution of the fractional Riccati differential equation will be

ln|a1y + b1|
A1

a1
−(1−α)

+ ln|a2y + b2|
A2

a2
−(1−α)

= αt+ c (16)

So, we have the solution is

|a1y + b1|
1

b2a1 − b1a2
−(1−α)

|a2y + b2|
1

b1a2 − b2a1
−(1−α)

= ceαt (17)



E. Abuteen / Eur. J. Pure Appl. Math, 17 (1) (2024), 372-384 376

Case 2: If the discriminant △ = B2 − 4AC = 0
The fractional Riccati differential equation (10)takes the form

Dαy(t) = (ay(t) + b)2, 0 < α ≤ 1. (18)

With the initial condition, y(0) = y0.
Using equation (13) we have

y′(t)
(ay(t) + b)2

− (1− α)
(2Ay(t)y′(t) +By′(t))
Ay2(t) +By(t) + C

= α (19)

Integrate equation (19), the solution will have the following form

|ay + b|2(α−1) = ce

1

ay + b eαt, y ̸= −b

a
(20)

Case 3: If the discriminant △ = B2 − 4AC < 0
The fractional Riccati differential equation (10)

Dαy(t) = Ay2 +By + C, 0 < α ≤ 1.

The equation Ay2 +By + C is irreducible, so by integrating equation (13), we get

1

A
tan−1 2Ay +B√

4AC −B2
= (1− α)ln|Ay2 +By + C|+ αt+ c (21)

4. Numerical examples

We employ the general solution of the fractional Riccati differential equation in all
three cases to analyze the following examples and contrast them with alternative methods.

Example 1. Consider the following fractional logistic differential equation

Dαy(t) = y − y2, y(0) =
1

2
(22)

This differential equation is classified under case 1 and by applying equation (17), we
find that the solution to be

|y|α|1− y|α−2 =

(
1

4

)2α−2

eαt (23)

If α = 1, then the solution is

y =
1

e−t + 1
(24)



E. Abuteen / Eur. J. Pure Appl. Math, 17 (1) (2024), 372-384 377

This solution agrees with several solutions for fractional logistic differential equations
see [24],[5],[7],[32],[4].

Comparisons for different values of α are shown in Figure 1.

Figure 1: Solutions of the fractional differential equation (22) for different values of α.

Example 2. Consider the following fractional differential equation

Dαy(t) = −y2 + 2y + 1, y(0) = 0 (25)

Based on positive discriminant, this equation is associated with case 1. By applying equa-
tion (17), the solution is characterized by

|y − (1−
√
2)|

1

2
√
2
−1+α

|(1 +
√
2)− y|

α−1−
1

2
√
2 =

(√
2− 1√
2 + 1

) 1

2
√
2
eαt (26)

Substituting α = 1 in equation (26), we obtain the solution as

y =
e2

√
2t − 1

(
√
2− 1)e2

√
2t + (

√
2 + 1)

(27)



E. Abuteen / Eur. J. Pure Appl. Math, 17 (1) (2024), 372-384 378

This solution is equivalent to

y = 1 +
√
2tanh

(
√
2t+

1

2
ln

(√
2− 1√
2 + 1

))
(28)

Which is compatible with the one found in [36],[33],[43],[40], [35], [26].

Figure 2: Solutions of the fractional differential equation (25) for different values of α.

Example 3. Consider the following fractional differential equation

Dαy(t) = y2 + 4y + 4, y(0) = 0 (29)

This example is classified as case 2. Therefore, the solution for this case follows the pattern
of equation (20), and thereafter, the solution is

|y + 2|2(α−1) = 22(α−1)e
αt+

1

y + 2
−
1

2 (30)

Where the exact solution for α = 1 is

y =
4t

1− 2t
(31)



E. Abuteen / Eur. J. Pure Appl. Math, 17 (1) (2024), 372-384 379

Figure 3: Solutions of the fractional differential equation (29) for different values of α.

Example 4. Consider the following fractional differential equation

Dαy(t) = y2 + 1, y(0) = 0 (32)

We notice that this example is categorized under case 3. Therefore, according to the
equation (21), the solution comes out to be

tan−1y = (1− α)ln(y2 + 1) + αt (33)

The exact solution for α = 1 is
y = tan(t)
We can see that the exact solution agrees with our solution.
For comparisons with this solution and figures see [1] Figure 4 shows a comparison of
different values of α.



REFERENCES 380

Figure 4: Solutions of the fractional differential equation (32) for different values of α for
0 ≤ t ≤ 1

5. Conclusion

In this paper, we categorize the fractional Riccati differential equation under three cases
and apply Caputo-Fabrizio fractional derivative and integral properties on these cases to
find analytical solutions of these equations. We have applied the analytical solution to
various examples and provided figures that demonstrate a strong agreement with the
solutions presented in the literature. We plan in a future work to apply Caputo-Fabrizio
fractional derivative and integral on other different fractional differential equations.

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