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# International Journal of Nonlinear Sciences and Numerical Simulation

Editor-in-Chief: Birnir, Björn

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Volume 19, Issue 6

# Hybrid Method for Solution of Fractional Order Linear Differential Equation with Variable Coefficients

Amit Ujlayan
/ Ajay Dixit
Published Online: 2018-07-24 | DOI: https://doi.org/10.1515/ijnsns-2017-0167

## Abstract

In this paper, we proposed a new analytical hybrid methods for the solution of conformable fractional differential equations (CFDE), which are based on the recently proposed conformable fractional derivative (CFD) in R. Khalil, M. Al Horani, A. Yusuf and M. Sababhed, A New definition of fractional derivative, J. Comput. Appl. 264 (2014). Moreover, we use the method of variation of parameters and reduction of order based on CFD, for the CFDE. Furthermore, to show the efficiency of the proposed analytical hybrid method, some examples are also presented.

MSC 2010: 34A08; 26A33

## 1 Introduction

While the modeling of the real world problems it have been accepted by many mathematicians that the concept of fractional order derivatives the generalization of classical derivative gives better prediction.

Many definitions of fractional derivative have been introduced by many mathematicians like Riemann-Liouville, Caputo, Laplace, Grnewald-Letnicov, etc. and also definitions have been modified time to time as per requirement, but one could not get a definition which satisfies even the fundamental properties like product rule, quotient rule, chain rule, Rolles theorem, commutative properties of derivative, etc. Even computation of these defined derivatives is too complicated to evaluate. Besides this, there is not a suitable analytic method to solve linear fractional second-order or higher-order differential equations; in fact, one has to use either numerical method or Mittag–Leffler function (series solution) or fractional Laplace transform for fractional initial value problem. It has motivated the authors to try for an analytical method to solve fractional differential equations.

The concept of the conformable fractional derivative has taken place and given by R. Khalil and colleagues which has later generalized by Katugampola. It satisfies most of the properties of derivative which are not fulfilled by previously defined fractional derivatives. Although it has some assumptions of being a differentiable function and restrictions like x > 0, still it is easy to apply and convert a fractional derivative to a classical order derivative. A number of applications [1, 2, 3] as well as various properties [4, 5, 11] have made an interest to researchers to investigate more in this field. We have presented a method of variation of parameters and method of reduction of order to solve a fractional order differential equation using Katugampola’s fractional derivative fractional derivative and its properties We refer [8, 9, 10, 11].

To study the brief history of fractional calculus, we refer to the reader [1, 6, 7] and for more details about conformable.

Conformable fractional derivative (CFD) by R. Khalil [8]: For a given $g:\left[0,\mathrm{\infty }\right)⟶\mathbb{R},$ conformable fractional derivative of order α is defined as ${D}^{\alpha }\left(g\right)\left(t\right)=\underset{ϵ⟶0}{lim}\frac{g\left(t+ϵ{t}^{1-\alpha }\right)-g\left(t\right)}{ϵ}; \mathrm{\forall } t>0, 0<\alpha ⩽1.$(1)

If g is α-differentiable in some (0,a),a > 0, for t > 0,α∈(0,1) and $\underset{t\to {0}^{+}}{lim}{g}^{\left(\alpha \right)}\left(t\right) \text{exist}1$ then we define ${g}^{\alpha }\left(0\right)=\underset{t\to {0}^{+}}{lim}{g}^{\left(\alpha \right)}\left(t\right)$

Fractional derivative (FD) by Udita Katugampola [9] : For a given $g:\left[0,\mathrm{\infty }\right)⟶\mathbb{R},$ Katugampola fractional derivative of order α is defined as ${D}^{\alpha }\left(g\right)\left(t\right)=\underset{ϵ⟶0}{lim}\frac{g\left(t{e}^{ϵ{t}^{-\alpha }}\right)-g\left(t\right)}{ϵ}; \mathrm{\forall } t>0, 0<\alpha ⩽1.$(2)

Fractional derivative at origin defines as above.

Some basic properties of Katugampola fractional derivative [9]

• ${D}^{\alpha }\left[af+bg\right]=a{D}^{\alpha }\left[f\right]+b{D}^{\alpha }\left[g\right]\left(Linearity\right)$

• ${D}^{\alpha }\left[fg\right)\right]=f{D}^{\alpha }\left[g\right]+g{D}^{\alpha }\left[f\right]\left(Product rule\right)$

• ${D}^{\alpha }\left[f\circ g\right]={D}^{\alpha }\left[f\left(g\right)\right]{D}^{\alpha }\left[g\right]\left(Chain rule\right)$

• ${D}^{\alpha }\left[f\left(t\right)\right]={t}^{1-\alpha }{f}^{\prime }\left(t\right)$

where f is differentiable function and α∈(0,1]. For more details of the properties, we refer to the reader [8, 9]

The presentation in rest of the paper is organized as follows. We introduce the method of reduction of order for conformable fractional differential equation in Section 2 and variation of parameters method with the Katugampola’s fractional order derivative in Section 3. As an applications, we give various examples of the conformable fractional differential equations Section 4. Finally, we conclude the results and give some discussions about it in the Section 5.

## 2 Method of reduction of order for conformable fractional differential equation

Consider the following linear fractional differential equation: $\left({D}^{\alpha }{D}^{\alpha }+\alpha P{D}^{\alpha }+{\alpha }^{2}Q\right)y=R,$

where ${D}^{\alpha }y=\frac{{d}^{\alpha }y}{d{x}^{\alpha }}={x}^{1-\alpha }\frac{dy}{dx};0<\alpha ⩽1$

and P,Q,R are continuous functions of x in [0,1].

Now consider the following cases

• Case I:

If $y={e}^{m{x}^{\alpha }}$ satisfies the homogeneous part, i.e.

${D}^{\alpha }{D}^{\alpha }+\alpha P{D}^{\alpha }y+{\alpha }^{2}Qy=0$ We get ${m}^{2}+Pm+Q=0⟹{e}^{m{x}^{\alpha }}$ is the one part of complementary function.

• Case II:

If $y={x}^{m\alpha }$ satisfies the homogeneous part, then

$m\left(m-1\right)+Pmx+Q{x}^{2}=0⟹{x}^{m\alpha }$ is the one part of the complementary function. Let y = uv be the complete solution of in which u is the solution of homogeneous part, i.e

${D}^{\alpha }{D}^{\alpha }u+\alpha P{D}^{\alpha }u+{\alpha }^{2}Qu=0,$

then we have $\frac{{d}^{\alpha }y}{d{x}^{\alpha }}=v\frac{{d}^{\alpha }u}{d{x}^{\alpha }}+\frac{{d}^{\alpha }v}{d{x}^{\alpha }}u.$

Putting these values of derivatives in our equation, we get $\begin{array}{rl}& \left(v\frac{{d}^{\alpha }}{d{x}^{\alpha }}\frac{{d}^{\alpha }u}{d{x}^{\alpha }}+\frac{{d}^{\alpha }}{d{x}^{\alpha }}\frac{{d}^{\alpha }v}{d{x}^{\alpha }}u+2\frac{{d}^{\alpha }u}{d{x}^{\alpha }}\frac{{d}^{\alpha }v}{d{x}^{\alpha }}\right)+\alpha P\left(v\frac{{d}^{\alpha }u}{d{x}^{\alpha }}+\frac{{d}^{\alpha }v}{d{x}^{\alpha }}u\right)\\ & \phantom{\rule{1em}{0ex}}+{\alpha }^{2}Quv=R\end{array}$ $⟹u\left({D}^{\alpha }{D}^{\alpha }v+\alpha P{D}^{\alpha }v\right)+2u{D}^{\alpha }u{D}^{\alpha }v=R,$ $⟹{D}^{\alpha }{D}^{\alpha }v+\alpha P{D}^{\alpha }v+2{D}^{\alpha }u{D}^{\alpha }v=R/u$ $⟹{D}^{\alpha }{D}^{\alpha }v+\left(\alpha P+\frac{2}{u}{D}^{\alpha }u\right){D}^{\alpha }v=R/u,$ ${D}^{\alpha }v=q⟹{D}^{\alpha }q+\left(\alpha P+\frac{2}{u}{D}^{\alpha }u\right)q=R/u,$

and integrating factor (I.F) $I.F.={e}^{\underset{\alpha }{\int }\left(\alpha P+\frac{2}{u}{u}^{\alpha }\right)d{x}^{\alpha }}={I}_{\alpha },$ $q.{I}_{\alpha }={\int }_{\alpha }\frac{R}{u}{I}_{\alpha }d{x}^{\alpha }+{c}_{1}⟹\frac{{d}^{\alpha }v}{d{x}^{\alpha }}=\frac{1}{{I}_{\alpha }}{\int }_{\alpha }\frac{R}{u}{I}_{\alpha }d{x}^{\alpha }+\frac{1}{{I}_{\alpha }}{c}_{1},$ $⟹v={\int }_{\alpha }\left(\frac{1}{{I}_{\alpha }}{\int }_{\alpha }\frac{R}{u}{I}_{\alpha }d{x}^{\alpha }+\frac{1}{{I}_{\alpha }}{c}_{1}\right)d{x}^{\alpha }+{c}_{2},$

and so $y=uv=\left({\int }_{\alpha }\left(\frac{1}{{I}_{\alpha }}{\int }_{\alpha }\frac{R}{u}{I}_{\alpha }d{x}^{\alpha }+\frac{1}{{I}_{\alpha }}{c}_{1}\right)+{c}_{2}\right)u.$

where ${I}_{\alpha }={e}^{{\int }_{\alpha }\left(\alpha P+\frac{2}{u}{u}^{\alpha }\right)d{x}^{\alpha }}={e}^{{\int }_{\alpha }\alpha Pd{x}^{\alpha }}{e}^{\int {x}^{\alpha -1}\frac{2{x}^{1-\alpha }}{u}\frac{du}{dx}dx}$ $={e}^{\underset{\alpha }{\int }\alpha Pd{x}^{\alpha }}{e}^{2logu}={u}^{2}{e}^{{\int }_{\alpha }\alpha Pd{x}^{\alpha }}$

using the definition of fractional integral $\underset{\alpha }{\int }g\left(x\right)d{x}^{\alpha }=\int {x}^{\alpha -1}g\left(x\right)dx.$

Hence, from above, it is clear that if one part of complementary function u is known, then the other part will be $u{\int }_{\alpha }{u}^{-2}{e}^{-{\int }_{\alpha }\alpha Pd{x}^{\alpha }}d{x}^{\alpha }$

## 3 Method of variation of parameter for conformable fractional differential equation

Consider the following second-order linear fractional differential equation $\left({D}^{\alpha }{D}^{\alpha }+\alpha P{D}^{\alpha }+{\alpha }^{2}Q\right)y=R\left(x\right),$(3)

where ${D}^{\alpha }y={y}^{\alpha }=\frac{{d}^{\alpha }y}{d{x}^{\alpha }}={x}^{1-\alpha }\frac{dy}{dx}; 0<\alpha ⩽1$

and P,Q,R are continuous functions of x in in[0,1] Let its complementary function of (3). $y=au+bv,$ where u,v are two independent solution of homogeneous part, then we have $\left({D}^{\alpha }{D}^{\alpha }+\alpha P{D}^{\alpha }+{\alpha }^{2}Q\right)u=0, \left({D}^{\alpha }{D}^{\alpha }+\alpha P{D}^{\alpha }+{\alpha }^{2}Q\right)v=0.$

Let $y=Au+Bv$ , where A,B are two undetermined functions, be the complete solution of our equation. Now to determine A,B, we chose an independent condition ${A}^{\alpha }u+v{B}^{\alpha }=0,$

since $y=Au+Bv⟹{y}^{\alpha }=A{u}^{\alpha }+B{v}^{\alpha }+\left(u{A}^{\alpha }+v{B}^{\alpha }\right)=A{u}^{\alpha }+B{v}^{\alpha }.$

Also, ${D}^{\alpha }{y}^{\alpha }=A{D}^{\alpha }{u}^{\alpha }+{B}^{\alpha }{v}^{\alpha }+{u}^{\alpha }{A}^{\alpha }+{D}^{\alpha }{v}^{\alpha }B,$

putting these values in (3) $\begin{array}{rl}A{D}^{\alpha }{u}^{\alpha }& +{B}^{\alpha }{v}^{\alpha }+{u}^{\alpha }{A}^{\alpha }+{D}^{\alpha }{v}^{\alpha }B+\alpha P\left(A{u}^{\alpha }+B{v}^{\alpha }\right)\\ & +{\alpha }^{2}Q\left(au+bv\right)=R\\ A\left({D}^{\alpha }{u}^{\alpha }& +\alpha P{u}^{\alpha }+{\alpha }^{2}Qu\right)+B\left({D}^{\alpha }{v}^{\alpha }+\alpha P{v}^{\alpha }+{\alpha }^{2}Qv\right)\\ & +A{u}^{\alpha }+B{v}^{\alpha }=R\\ & ⟹{A}^{\alpha }{u}^{\alpha }+{B}^{\alpha }{v}^{\alpha }=R.\end{array}$

Solving with the independent condition, we get ${A}^{\alpha }=\frac{{d}^{\alpha }A}{d{x}^{\alpha }}=\frac{-vR}{u{v}^{\alpha }-v{u}^{\alpha }}=\frac{-vR}{{W}_{\alpha }}$

And ${B}^{\alpha }=\frac{{d}^{\alpha }A}{d{x}^{\alpha }}=\frac{uR}{u{v}^{\alpha }-v{u}^{\alpha }}=\frac{uR}{{W}_{\alpha }},$

where, ${W}_{\alpha }=\left|\begin{array}{rr}u& v\\ {u}^{\alpha }& {v}^{\alpha }\end{array}\right|=u{v}^{\alpha }-v{u}^{\alpha }$

called Wronskian for conformable fractional differential equations. Integrating we get, $A=X+{c}_{1},B=Y+{c}_{2}$

where $X={\int }_{\alpha }\frac{-vR}{{W}_{\alpha }}d{x}^{\alpha }$

and $Y={\int }_{\alpha }\frac{uR}{{W}_{\alpha }}d{x}^{\alpha }$

and ${\int }_{\alpha }f\left(x\right)d{x}^{\alpha }=\int f\left(x\right){x}^{\alpha -1}dx=$ anti-derivative of conformable fractional derivative, hence the complete solution $y={c}_{1}u+{c}_{2}v+Xu+Yv$

## 4 Applications

#### Example 4.1:

Consider the following FDE $4x{D}^{1/2}{D}^{1/2}z-2\left(x+2\sqrt{x}\right){D}^{1/2}z+\left(\sqrt{x}+2\right)z=4x\sqrt{x}{e}^{\sqrt{x}},$

we have $P=-\left(1+2/\sqrt{x}\right), Q=\frac{1}{\sqrt{x}}+\frac{2}{x}, R=\sqrt{x}{e}^{\sqrt{x}}$ And $P+Q{x}^{1/2}=0$ obviously, so complete solution will be given by z = uv, where $u=\sqrt{x}$ is known function.

Now to determine v we have the following differential equation ${D}^{1/2}{D}^{1/2}v+\left(\frac{1}{2}P+\frac{2}{u}{D}^{1/2}u\right){D}^{1/2}v=R/u,$ $⟹{D}^{1/2}{D}^{1/2}v+\left(-\frac{1}{2}-\frac{1}{\sqrt{x}}+\frac{2}{\sqrt{x}}{D}^{1/2}\sqrt{x}\right){D}^{1/2}v={e}^{\sqrt{x}},$ $⟹{D}^{1/2}{D}^{1/2}v+\left(-\frac{1}{2}-\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{x}}\right){D}^{1/2}v={e}^{\sqrt{x}},$

taking $q={D}^{1/2}v,$ we get $\frac{{d}^{1/2}q}{d{x}^{1/2}}-\frac{1}{2}q={e}^{\sqrt{x}}⟹I.F={e}^{\underset{1/2}{\int }-\frac{1}{2}d{x}^{1/2}}={e}^{\int -\frac{1}{2\sqrt{x}}dx}={e}^{-\sqrt{x}},$

So $\begin{array}{rl}q{e}^{-\sqrt{x}}& =\underset{1/2}{\int }{e}^{-\sqrt{x}}{e}^{\sqrt{x}}d{x}^{1/2}+a⟹q=\frac{{d}^{1/2}v}{d{x}^{1/2}}\\ & ={e}^{\sqrt{x}}\int \frac{1}{\sqrt{x}}dx+a{e}^{\sqrt{x}},\end{array}$

And therefore, $v=\int {e}^{\sqrt{x}}\left(2\sqrt{x}+a\right)\frac{1}{\sqrt{x}}dx+b=\int 2{e}^{\sqrt{x}}dx+2a\sqrt{x}+b$

Hence, $z=\left(\int 2{e}^{\sqrt{x}}dx+2a\sqrt{x}+b\right){e}^{\sqrt{x}}$

where a,b are constants.

#### Example 4.2:

Consider the following FDE ${D}^{1/3}{y}^{1/3}-\frac{1}{3}cot{x}^{1/3}{y}^{1/3}-\frac{1}{9}\left(1-cot{x}^{1/3}\right)y={e}^{{x}^{1/3}}sin{x}^{1/3}.$

Here we have $P=-cot{x}^{1/3},Q=cot{x}^{1/3}-1,R={e}^{{x}^{1/3}}sin{x}^{1/3}$ $1+P+Q=0⟹{e}^{{x}^{1/3}}$

The one part of C.F is ${e}^{{x}^{1/3}},$ Then now to determine v we have $\begin{array}{r}u={e}^{{x}^{1/3}}⟹\frac{{d}^{1/3}}{d{x}^{1/3}}\frac{{d}^{1/3}v}{d{x}^{1/3}}+\left(\frac{1}{3}P+\frac{2}{u}\frac{{d}^{1/3}u}{d{x}^{1/3}}\right)\frac{{d}^{1/3}v}{d{x}^{1/3}}=sin{x}^{1/3},\end{array}$ $\begin{array}{r}⟹\frac{{d}^{1/3}}{d{x}^{1/3}}\frac{{d}^{1/3}v}{d{x}^{1/3}}+\left(-\frac{1}{3}cot{x}^{1/3}+\frac{2}{{e}^{{x}^{1/3}}}\frac{{d}^{1/3}u}{d{x}^{1/3}}\right)\frac{{d}^{1/3}v}{d{x}^{1/3}}=sin{x}^{1/3},\end{array}$ $\begin{array}{r}⟹\frac{{d}^{1/3}}{d{x}^{1/3}}\frac{{d}^{1/3}v}{d{x}^{1/3}}+\left(-\frac{1}{3}cot{x}^{1/3}+\frac{2}{{e}^{{x}^{1/3}}}{x}^{2/3}\frac{{d}^{}u}{d{x}^{}}\right)\frac{{d}^{1/3}v}{d{x}^{1/3}}=sin{x}^{1/3},\end{array}$ $\begin{array}{r}⟹\frac{{d}^{1/3}q}{d{x}^{1/3}}+\left(-\frac{1}{3}cot{x}^{1/3}+\frac{2}{3}\right)q=sin{x}^{1/3}, q=\frac{{d}^{1/3}v}{d{x}^{1/3}},\end{array}$ $\begin{array}{r}I.F.={e}^{{\int }_{1/3}\frac{\left(2-cot{x}^{1/3}\right)}{3}d{x}^{1/3}}={e}^{\int \frac{\left(2-cot{x}^{1/3}\right)}{3{x}^{2/3}}dx}=\frac{{e}^{2{x}^{1/3}}}{sin{x}^{1/3}},\end{array}$ $\begin{array}{rl}⟹q\frac{{e}^{2{x}^{1/3}}}{sin{x}^{1/3}}& =3\frac{{e}^{2{x}^{1/3}}}{2}+c⟹q=\frac{{d}^{1/3}v}{d{x}^{1/3}}=3\frac{sin{x}^{1/3}}{2}\\ & +c{e}^{-2{x}^{1/3}}sin{x}^{1/3}\end{array}$ $\begin{array}{r}⟹v=\int \left(3\frac{sin{x}^{1/3}}{2}+c{e}^{-2{x}^{1/3}}sin{x}^{1/3}\right){x}^{-2/3}dx+{c}^{\prime }\end{array}$ $\begin{array}{r}=\frac{-9cos{x}^{1/3}}{2}-\frac{3}{5}c{e}^{-2{x}^{1/3}}\left(2sin{x}^{1/3}+cos{x}^{1/3}\right)+{c}^{\prime }\end{array}$

Hence, $\begin{array}{r}y=\left(\frac{-9cos{x}^{1/3}}{2}-\frac{3}{5}c{e}^{-2{x}^{1/3}}\left(2sin{x}^{1/3}+cos{x}^{1/3}\right)+{c}^{\prime }\right){e}^{{x}^{1/3}}\end{array}$

#### Example 4.3:

Consider the following fractional differential equation ${D}^{\alpha }{D}^{\alpha }y-3\alpha {D}^{\alpha }y+2{\alpha }^{2}y=\frac{{e}^{{x}^{\alpha }}}{1+{e}^{{x}^{\alpha }}}, 0<\alpha ⩽1$

Let us consider $y={e}^{m{x}^{\alpha }}$ as a solution, then auxiliary equation is ${m}^{2}-3m+2=0⟹m=1,2.$

So $C.F={c}_{1}{e}^{{x}^{\alpha }}+{c}_{2}{e}^{2{x}^{\alpha }}.$

Take $u={e}^{{x}^{\alpha }} v={e}^{2{x}^{\alpha }}$

Then Wronskian, ${W}_{\alpha }=\left|\begin{array}{rr}u& v\\ {u}^{\alpha }& {v}^{\alpha }\end{array}\right|=\left|\begin{array}{rr}{e}^{{x}^{\alpha }}& {e}^{2{x}^{\alpha }}\\ \alpha {e}^{{x}^{\alpha }}& 2\alpha {e}^{2{x}^{\alpha }}\end{array}\right|=\alpha {e}^{3{x}^{\alpha }}$ $P.I=Xu+Yv$

where $\begin{array}{rl}X& ={\int }_{\alpha }\frac{-vR}{{W}_{\alpha }}d{x}^{\alpha }={\int }_{\alpha }\frac{-{e}^{2{x}^{\alpha }}{e}^{{x}^{\alpha }}}{\alpha {e}^{3{x}^{\alpha }}\left(1+{e}^{{x}^{\alpha }}\right)}d{x}^{\alpha }\\ & =\int \frac{-{x}^{\alpha -1}}{\alpha \left(1+{e}^{{x}^{\alpha }}\right)}dx;\end{array}$ $\begin{array}{rl}{x}^{\alpha }& =t⟹{x}^{\alpha -1}dx=\frac{dx}{\alpha }⟹\int \frac{-{e}^{-t}}{{\alpha }^{2}\left(1+{e}^{-t}\right)}dt\\ & =\frac{1}{{\alpha }^{2}}log\left(1+{e}^{-{x}^{\alpha }}\right)\end{array}$

and $Y={\int }_{\alpha }\frac{uR}{{W}_{\alpha }}d{x}^{\alpha }={\int }_{\alpha }\frac{{e}^{{x}^{\alpha }}{e}^{{x}^{\alpha }}}{\alpha {e}^{3{x}^{\alpha }}\left(1+{e}^{{x}^{\alpha }}\right)}d{x}^{\alpha }=\int \frac{{x}^{\alpha -1}}{\alpha {e}^{{x}^{\alpha }}\left(1+{e}^{{x}^{\alpha }}\right)}dx$ $=\int \frac{1}{{\alpha }^{2}{e}^{t}\left(1+{e}^{t}\right)}dt=\frac{1}{{\alpha }^{2}}log\left(\frac{{e}^{{x}^{\alpha }}}{1+{e}^{{x}^{\alpha }}}\right)$

Hence, the complete solution $\begin{array}{rl}y=\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}& {c}_{1}{e}^{{x}^{\alpha }}+{c}_{2}{e}^{2{x}^{\alpha }}+\left(\frac{1}{{\alpha }^{2}}log\left(1+{e}^{-{x}^{\alpha }}\right)\right){e}^{{x}^{\alpha }}\\ & +\left(\frac{1}{{\alpha }^{2}}log\left(\frac{{e}^{{x}^{\alpha }}}{1+{e}^{{x}^{\alpha }}}\right)\right){e}^{2{x}^{\alpha }}\end{array}$

#### Example 4.4:

Solve the following FDE $\left(1-{x}^{\beta }\right){D}^{\beta }{y}^{\beta }+\beta {x}^{\beta }{y}^{\beta }-{\beta }^{2}y=2\left({x}^{\beta }-1{\right)}^{2}{e}^{-{x}^{\beta }};0<\beta ⩽1.$

We have ${D}^{\beta }{y}^{\beta }+\beta \frac{{x}^{\beta }}{1-{x}^{\beta }}{y}^{\beta }-{\beta }^{2}\frac{1}{1-{x}^{\beta }}y=2\frac{\left(1-{x}^{\beta }\right)}{{e}^{{x}^{\beta }}}$

so that $P=\frac{{x}^{\beta }}{1-{x}^{\beta }}, Q=\frac{1}{1-{x}^{\beta }}, R=2\frac{1-{x}^{\beta }}{{e}^{{x}^{\beta }}}.$

Now, $1+P+Q=0⟹{e}^{{x}^{\beta }}$

is the one part of the complementary function and $P+Q{x}^{\beta }=0⟹{x}^{\beta }$ will be second independent solution and so complementary function is ${c}_{1}{e}^{{x}^{\beta }}+{c}_{2}{x}^{\beta }$

Set $u={e}^{{x}^{\beta }},v={x}^{\beta },$ so that Wronskian ${W}_{\beta }=\left|\begin{array}{rr}u& v\\ {u}^{\beta }& {v}^{\beta }\end{array}\right|=\left|\begin{array}{rr}{e}^{{x}^{\beta }}& {x}^{\beta }\\ \beta {e}^{{x}^{\beta }}& \beta \end{array}\right|=\beta \left(1-{x}^{\beta }\right){e}^{{x}^{\beta }}$

and for particular integral, ${y}_{p}=Xu+Yv$ $\begin{array}{rl}X& ={\int }_{\beta }\frac{-vR}{{W}_{\beta }}d{x}^{\beta }=\int \frac{-2\left(1-{x}^{\beta }\right){x}^{\beta -1}{x}^{\beta }}{\beta {e}^{2{x}^{\beta }}\left(1-{x}^{\beta }\right)}dx\\ & =-\int 2{x}^{\beta }{x}^{\beta -1}{e}^{-2{x}^{\beta }}dx\end{array}$ ${x}^{\beta }=t⟹{x}^{\beta -1}dx=\frac{dt}{\beta }=-\int 2t{e}^{-2t}dt=\frac{1}{\beta {e}^{2{x}^{\beta }}}\left(\frac{1}{2}+{x}^{\beta }\right)$

and $\begin{array}{rl}Y& ={\int }_{\beta }\frac{uR}{{W}_{\beta }}d{x}^{\beta }=\int \frac{2\left(1-{x}^{\beta }\right){x}^{\beta -1}{e}^{{x}^{\beta }}}{\beta {e}^{2{x}^{\beta }}\left(1-{x}^{\beta }\right)}dx=\int 2{x}^{\beta -1}{e}^{-{x}^{\beta }}dx\\ & =\frac{-2{e}^{-{x}^{\beta }}}{\beta }\end{array}$

Therefore, the complete solution $y=\left({c}_{1}+X\right){e}^{{x}^{\beta }}+\left({c}_{2}+Y\right){x}^{\beta }$

#### Example 4.5:

Consider the following FDE ${D}^{1/2}{D}^{1/2}y+y=tan2\sqrt{x}.$

Let us assume that $y={e}^{m{x}^{1/2}}$as a solution of homogeneous part, then the corresponding

A.E. is ${m}^{2}+4=0⇒m=±2i$ $C.F=asin2\sqrt{x}+bcos2\sqrt{x},$

choose $u=sin2\sqrt{x},v=cos2\sqrt{x}$ ${W}_{1/2}=\left|\begin{array}{rr}u& v\\ {u}^{1/2}& {v}^{1/2}\end{array}\right|=\left|\begin{array}{rr}cos2\sqrt{x}& sin2\sqrt{x}\\ -sin2\sqrt{x}& cos2\sqrt{x}\end{array}\right|=1$ $\begin{array}{rl}X& =\underset{\alpha }{\int }\frac{-vR}{{W}_{\alpha }}d{x}^{\alpha }=\underset{1/2}{\int }\frac{-tan2\sqrt{x}cos2\sqrt{x}}{1}d{x}^{1/2}\\ & =\int -{x}^{-1/2}sin2\sqrt{x}dx=cos2\sqrt{x},\end{array}$

and $Y=\underset{\alpha }{\int }\frac{uR}{{W}_{\alpha }}d{x}^{\alpha }=\underset{1/2}{\int }\frac{tan2\sqrt{x}sin2\sqrt{x}}{1}d{x}^{1/2}$ $=\int \frac{{x}^{-1/2}{sin}^{2}2\sqrt{x}}{cos2\sqrt{x}}dx$ $=\left\{sin2\sqrt{x}-log\left(sec2\sqrt{x}+tan2\sqrt{x}\right)\right\},$

hence, the complete solution is $y=asin2\sqrt{x}+bcos2\sqrt{x}-log\left(sec2\sqrt{x}+tan2\sqrt{x}\right)cos2\sqrt{x}$

## 5 Conclusion

The important properties of the recently developed Udita N.Katugampola fractional derivative were used. The major advantage of the Katugampola’s fractional derivative is that it is limit based rather than defined via a fractional integral as Riemann–Liouville and Caputo derivatives. Moreover, a key property of the Katugampola derivative along with reduction of order and variation of parameters is used for solution of FDE. Using this hybrid approach, we solve various types of CFDE with variable coefficients. One remarkable and interesting fact is that the solution will coincide with the solution of classical differential equation at α = 1.

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Accepted: 2018-05-20

Published Online: 2018-07-24

Published in Print: 2018-09-25

Citation Information: International Journal of Nonlinear Sciences and Numerical Simulation, Volume 19, Issue 6, Pages 621–626, ISSN (Online) 2191-0294, ISSN (Print) 1565-1339,

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© 2018 Walter de Gruyter GmbH, Berlin/Boston.