# Solving Sequential Linear M-Fractional Differential Equations With Constants
Coefficients

V.Padmapriya1 1 Research Scholar,
VIT University,Chennai Campus,
India<EMAIL_ADDRESS>and  M.Kaliyappan 2 2 Division of
Mathematics, School of Advanced Sciences,
VIT University,Chennai Campus,
India<EMAIL_ADDRESS>
###### Abstract.

Fractional calculus is a powerful and effective tool for modelling nonlinear
systems. The M-derivative is the generalisation of alternative fractional
derivative introduced by Katugampola[6]. This M-derivative obey the properties
of integer calculus. In this paper, we present the method for solving
M-fractional sequential linear differential equations with constant
coefficients for $\alpha\geq 0$ and $\beta>0$. Existence and Uniqueness of the
solutions for the nth order sequential linear M-fractional differential
equations are discussed in detail. We have present illustration for
homogeneous and non homogeneous case.
Mathematics Subject Classification: 26A33, 34AXX.

###### Key words and phrases:

Sequential Linear Fractional Differential Equations, M-Fractional Derivative,
Existence and Uniqueness Theorem, Fractional Method of Variation of parameters

##  1\. Introduction

While L’Hospital has proposed the idea of fractional derivative in the
$17^{th}$ century, several researchers concerted fractional derivative in the
recent centuries. Riemann-Liouville, Caputo and other fractional derivatives
are defined on the basis of fractional integral form [8, 12, 13].
Recently, Khalil et al.[7] and Katugampola [6] proposed fractional derivatives
in the limit form as in usual derivative such as conformable fractional
derivative and alternative fractional derivative. Based on these derivative,
Sousa and Oliveira [15] introduced M-fractional derivatives which satisfies
properties of integer-order calculus.
Theory and applications of the sequential linear fractional differential
equations involving Hadamard, Riemann-Liouville, Caputo and Conformable
derivatives have been investigated in [1, 2, 3, 4, 9, 10, 11].
Lately, Gokdogan et al [5] have proved existence and uniqueness theorems for
solving sequential linear conformable fractional differential equations. Unal
et al [14] provide method to solve sequential linear conformable fractional
differential equations with constants coefficients. In this work, We present
Existence and Uniqueness theorems and solutions of sequential linear
M-fractional differential equations.
The arrangement of this paper is as following: In section 2 we present the
concept of M-fractional derivative. In section 3 we provide existence and
uniqueness theorems for sequential Linear M-fractional differential equations.
In section 4 we propose the solutions of sequential Linear M-fractional
differential equations. In section 5 we present solutions of Non-homogeneous
case. Finally, Conclusion is present in section 6.

##  2\. M-Fractional Calculus

In this section, we give some necessary definitions and theorems of M –
derivative which are explained in [15].

######  Definition 2.1.

Let $f:[0,\infty)\rightarrow\mathbb{R}$ be a function and $t>0$. Then for
$0<\alpha<1$, the M-fractional derivative of $f$ of order $\alpha$ is defined
as

$D^{\alpha,\beta}_{M}f(t)=\lim_{\epsilon\to
0}\frac{f(t\mathbb{E}_{\beta}(\epsilon t^{-\alpha})-f(t)}{\epsilon}$

Where $\mathbb{E}_{\beta}(.),\beta>0$ is Mittag-Leffler function with one
parameter.

If $f$ is M-differentiable in some interval $(0,a),a>0$ and

$\lim_{t\to 0^{+}}D^{\alpha,\beta}_{M}f(t)$

 exists, then we have

$D^{\alpha,\beta}_{M}f(0)=\lim_{t\to 0^{+}}D^{\alpha,\beta}_{M}f(t)$

######  Theorem 2.1.

Let $0<\alpha\leq 1,\beta>0,a,b\in\mathbb{R}$ and $f,g$ be M - differentiable
at a point $t>0$. Then

  1. (1)

$D^{\alpha,\beta}_{M}(af+bg)(t)=aD^{\alpha,\beta}_{M}f(t)+bD^{\alpha,\beta}_{M}g(t)$
for all $a,b\in\mathbb{R}$.

  2. (2)

$D^{\alpha,\beta}_{M}(f\cdot
g)(t)=f(t)D^{\alpha,\beta}_{M}g(t)+g(t)D^{\alpha,\beta}_{M}f(t)$

  3. (3)

$D^{\alpha,\beta}_{M}(\frac{f}{g})(t)=\frac{g(t)D^{\alpha,\beta}_{M}f(t)-f(t)D^{\alpha,\beta}_{M}g(t))}{[g(t)]^{2}}$

  4. (4)

$D^{\alpha,\beta}_{M}(c)=0$, where $f(t)=c$ is a constant.

  5. (5)

$D^{\alpha,\beta}_{M}t^{a}=\frac{a}{\Gamma(\beta+1)}t^{a-\alpha}$

  6. (6)

Moreover, $f$ is differentiable, then
$D^{\alpha,\beta}_{M}f(t)=\frac{t^{1-\alpha}}{\Gamma(\beta+1)}\frac{df(t)}{dt}$

Additionally,M-derivatives of certain functions as follows:

  1. (1)

$D^{\alpha,\beta}_{M}(sin(\frac{1}{\alpha}t^{\alpha}))=\frac{cos(\frac{1}{\alpha}t^{\alpha})}{\Gamma(\beta+1)}$

  2. (2)

$D^{\alpha,\beta}_{M}(cos(\frac{1}{\alpha}t^{\alpha}))=\frac{-sin(\frac{1}{\alpha}t^{\alpha})}{\Gamma(\beta+1)}$

  3. (3)

$D^{\alpha,\beta}_{M}(e^{\frac{t^{\alpha}}{\alpha}})=\frac{e^{\frac{t^{\alpha}}{\alpha}}}{\Gamma(\beta+1)}$

######  Theorem 2.2.

Let $a>0$ and $t\geq a$. Then, the M-integral of order $α$ of a function $f$
is defined by

${}_{M}I^{\alpha,\beta}_{a}f(t)=\Gamma(\beta+1)\int_{a}^{t}\frac{f(x)}{x^{1-\alpha}}dx$

######  Theorem 2.3.

Let $a\geq 0$ and $0<\alpha<1$. Also let $f$ be a continuous function such
that there exists ${}_{M}I^{\alpha,\beta}_{a}f$. Then

$D^{\alpha,\beta}_{M}(_{M}I^{\alpha,\beta}_{a}f(t))=f(t)$

######  Theorem 2.4.

Let $f,g:[a,b]\rightarrow\mathbb{R}$ be two functions such that $f,g$ are
differentiable and $0<\alpha<1$. Then

$\int_{a}^{b}f(x)D^{\alpha,\beta}_{M}g(x)d_{\alpha}x=f(x)g(x)\Bigr{|}_{a}^{b}-\int_{a}^{b}g(x)D^{\alpha,\beta}_{M}f(x)d_{\alpha}x$

where $d_{\alpha}x=\frac{\Gamma(\beta+1)}{x^{1-\alpha}}dx$

##  3\. Existence and Uniqueness Theorem

Let linear sequential M-fractional differential equation of order $n\alpha$

${}^{n}D^{\alpha,\beta}_{M}y+p_{n-1}(t)^{n-1}D^{\alpha,\beta}_{M}y+...+p_{2}(t)^{2}D^{\alpha,\beta}_{M}y+p_{1}(t)D^{\alpha,\beta}_{M}y+p_{0}(t)y=0$
(1)

where
${}^{n}D^{\alpha,\beta}_{M}y=D^{\alpha,\beta}_{M}D^{\alpha,\beta}_{M}...D^{\alpha,\beta}_{M}y,$
($n$ times)
Similarly, non-homogeneous fractional differential equation with M-derivative
is

${}^{n}D^{\alpha,\beta}_{M}y+p_{n-1}(t)^{n-1}D^{\alpha,\beta}_{M}y+...+p_{2}(t)^{2}D^{\alpha,\beta}_{M}y+p_{1}(t)D^{\alpha,\beta}_{M}y+p_{0}(t)y=f(t)$
(2)

We define an $n^{th}$-order differential operator for eqn. (1) as following

$L_{\alpha,\beta}[y]=^{n}D^{\alpha,\beta}_{M}y+p_{n-1}(t)^{n-1}D^{\alpha,\beta}_{M}y+...+p_{2}(t)^{2}D^{\alpha,\beta}_{M}y+p_{1}(t)D^{\alpha,\beta}_{M}y+p_{0}(t)y=0$
(3)

######  Theorem 3.1.

Let $\Gamma(\beta+1)t^{\alpha-1}p(t),\Gamma(\beta+1)t^{\alpha-1}f(t)\in
C(a,b)$ and let $y$ be M-differentiable for $0<\alpha\leq 1$ and $\beta>0$.
Then the initial value problem

$D^{\alpha,\beta}_{M}y+p(t)y=f(t)$ (4) $y(t_{0})=y_{0}$ (5)

has exactly one solution on the interval $(a,b)$ where $t_{0}\in(a,b)$

###### Proof.

Using property (6) in Theorem 2.1, we have

$D^{\alpha,\beta}_{M}y+p(t)y=f(t)$
$\frac{t^{1-\alpha}}{\Gamma(\beta+1)}y^{\prime}+p(t)y=f(t)$
$y^{\prime}+\Gamma(\beta+1)t^{\alpha-1}p(t)y=\Gamma(\beta+1)t^{\alpha-1}f(t)$

The proof is clear from classical linear fundamental theorem existence and
uniqueness. ∎

######  Theorem 3.2.

If
$\Gamma(\beta+1)t^{\alpha-1}p_{n-1}(t),...,\Gamma(\beta+1)t^{\alpha-1}p_{1}(t),\Gamma(\beta+1)t^{\alpha-1}p_{0}(t),\Gamma(\beta+1)t^{\alpha-1}f(t)\in
C(a,b)$ and $y$ be $n$ times M-differentiable function, then a solution $y(t)$
of the initial value problem

${}^{n}D^{\alpha,\beta}_{M}y+p_{n-1}(t)^{n-1}D^{\alpha,\beta}_{M}y+...+p_{2}(t)^{2}D^{\alpha,\beta}_{M}y+p_{1}(t)D^{\alpha,\beta}_{M}y+p_{0}(t)y=f(t)$
(6)
$y(t_{0})=y_{0},D^{\alpha,\beta}_{M}y(t_{0})=y_{1},...,^{n-1}D^{\alpha,\beta}_{M}y(t_{0})=y_{n-1},a<t_{0}<b$
(7)

###### Proof.

The existence of a local solution is obtained by transform our problem into
the first order system of differential equations. So, we introduce new
variables

$x_{1}=y,x_{2}=D^{\alpha,\beta}_{M}y,x_{3}=^{2}D^{\alpha,\beta}_{M}y,,...,x_{n}=^{n-1}D^{\alpha,\beta}_{M}y$

In this, we have

$D^{\alpha,\beta}_{M}x_{1}=x_{2}$ $D^{\alpha,\beta}_{M}x_{2}=x_{3}$ $\vdots$
$D^{\alpha,\beta}_{M}x_{n-1}=x_{n}$
$D^{\alpha,\beta}_{M}x_{n}=-p_{n-1}x_{n}-...-p_{2}x_{3}-p_{1}x_{2}-p_{0}x_{1}+f(t)$

The above equations can be written as the following

$D^{\alpha,\beta}_{M}\underbrace{\begin{bmatrix}x_{1}\\\ x_{2}\\\ \vdots\\\
x_{n-1}\\\
x_{n}\end{bmatrix}}_{X(t)}+\underbrace{\begin{bmatrix}0&-1&0&0&\cdots&0\\\
0&0&-1&0&\cdots&0\\\ \vdots&\vdots&\vdots&\vdots&\cdots&\vdots\\\
0&0&0&0&\cdots&-1\\\
p_{0}&p_{1}&p_{2}&p_{3}&\cdots&p_{n-1}\end{bmatrix}}_{P(t)}\begin{bmatrix}x_{1}\\\
x_{2}\\\ \vdots\\\ x_{n-1}\\\
x_{n}\end{bmatrix}=\underbrace{\begin{bmatrix}0\\\ 0\\\ \vdots\\\ 0\\\
f(t)\end{bmatrix}}_{F(t)}$ $D^{\alpha,\beta}_{M}X(t)+P(t)X(t)=F(t)$
$X^{\prime}(t)+\Gamma(\beta+1)t^{\alpha-1}P(t)X(t)=\Gamma(\beta+1)t^{\alpha-1}F(t)$

The existence and uniqueness of solution (6)-(7) follows from classical
theorems on existence and uniqueness for system equation. ∎

######  Theorem 3.3.

If $y_{1}$ and $y_{2}$ are $n$ times M-differentiable functions and
$c_{1},c_{2}$ are arbitrary numbers, then $L_{\alpha,\beta}$ is linear.

$i.e,\
L_{\alpha,\beta}[c_{1}y_{1}+c_{2}y_{2}]=c_{1}L_{\alpha,\beta}[y_{1}]+c_{2}L_{\alpha,\beta}[y_{2}]$

###### Proof.

We can easily derived the proof of this theorem by applying same procedure in
Theorem-4.3 [5] to M\- derivative. ∎

######  Theorem 3.4.

If $y_{1},y_{2}...y_{n}$ are the solutions of equation $L_{\alpha,\beta}[y]=0$
and $c_{1},c_{2}...c_{n}$ are arbitrary constants, then the linear combination
$y(t)=c_{1}y_{1}+c_{2}y_{2}+...+c_{n}y_{n}$ is also solution of
$L_{\alpha,\beta}[y]=0$.

###### Proof.

We can easily derived the proof of this theorem by applying same procedure in
Theorem-4.4[5] to M\- derivative. ∎

######  Definition 3.1.

For $n$ functions $y_{1},y_{2}...y_{n}$, we define the M-Wronskain of these
function to be the determinant

$W_{\alpha,\beta}(t)=\begin{vmatrix}y_{1}&y_{2}&\cdots&y_{n}\\\
D^{\alpha,\beta}_{M}y_{1}&D^{\alpha,\beta}_{M}y_{2}&\cdots&D^{\alpha,\beta}_{M}y_{n}\\\
\vdots&\vdots&\cdots&\vdots\\\
{}^{n-1}D^{\alpha,\beta}_{M}y_{1}&{}^{n-1}D^{\alpha,\beta}_{M}y_{2}&\cdots&{}^{n-1}D^{\alpha,\beta}_{M}y_{n}\end{vmatrix}$

######  Theorem 3.5.

Let $y_{1},y_{2}...y_{n}$ be $n$ solutions of $L_{\alpha,\beta}[y]=0$. If
there is a $t_{0}\in(a,b)$ such that $W_{\alpha,\beta}(t_{0})\neq 0$, then
${y_{1},y_{2}...y_{n}}$ is a fundamental set of solutions.

###### Proof.

We need to show that if $y(t)$ is a solution of $L_{\alpha,\beta}[y]=0$, then
we can write $y(t)$ as a linear combination of $y_{1},y_{2}...y_{n}$.

$i.e,y=c_{1}y_{1}+c_{2}y_{2}+...+c_{n}y_{n}$

so the problem reduces to finding the constants $c_{1},c_{2},...c_{n}$.These
constants are found by solving the following linear system of $n$ equations

$c_{1}y_{1}(t_{0})+c_{2}y_{2}(t_{0})+...+c_{n}y_{n}(t_{0})=y(t_{0})$
$c_{1}D^{\alpha,\beta}_{M}y_{1}(t_{0})+c_{2}D^{\alpha,\beta}_{M}y_{2}(t_{0})+...+c_{n}D^{\alpha,\beta}_{M}y_{n}(t_{0})=D^{\alpha,\beta}_{M}y(t_{0})$
$\vdots\\\ $
$c_{1}^{n-1}D^{\alpha,\beta}_{M}y_{1}(t_{0})+c_{2}^{n-1}D^{\alpha,\beta}_{M}y_{2}(t_{0})+...+c_{n}^{n-1}D^{\alpha,\beta}_{M}y_{n}(t_{0})=^{n-1}D^{\alpha,\beta}_{M}y(t_{0})$

Using Cramer’s rule, we can find

$c_{i}=\frac{W_{\alpha,\beta}^{i}(t_{0})}{W_{\alpha,\beta}(t_{0})},1\leq i\leq
n$

Since $W_{\alpha,\beta}(t_{0})\neq 0$, it follows that $c_{1},c_{2},...c_{n}$
exist. ∎

######  Theorem 3.6.

Let $y_{1},y_{2}...y_{n}$ be $n$ solutions of $L_{\alpha,\beta}[y]=0$. Then

  1. (1)

$W_{\alpha,\beta}(t)$ satisfies the differential
equation$D^{\alpha,\beta}_{M}W_{\alpha,\beta}+p_{(}n-1)W_{\alpha,\beta}=0$

  2. (2)

If $t_{0}$ is any point in $(a,b)$, then

$W_{\alpha,\beta}(t)=W_{\alpha,\beta}(t_{0})e^{-\Gamma(\beta+1)\int_{t_{0}}^{t}x^{\alpha-1}p_{n-1}(x)dx}$

Further, if $W_{\alpha,\beta}(t_{0})\neq 0$ then $W_{\alpha,\beta}(t)\neq 0$
for all $t\in(a,b)$

###### Proof.

(1)Let us introduce new variables

$x_{1}=y,x_{2}=D^{\alpha,\beta}_{M}y,x_{3}=^{2}D^{\alpha,\beta}_{M}y,...,x_{n}=^{n-1}D^{\alpha,\beta}_{M}y$

From this, we have

$D^{\alpha,\beta}_{M}x_{1}=x_{2}$ $D^{\alpha,\beta}_{M}x_{2}=x_{3}$ $\vdots$
$D^{\alpha,\beta}_{M}x_{n-1}=x_{n}$
$D^{\alpha,\beta}_{M}x_{n}=-p_{n-1}x_{n}-...-p_{2}x_{3}-p_{1}x_{2}-p_{0}x_{1}$
$D^{\alpha,\beta}_{M}\underbrace{\begin{bmatrix}x_{1}\\\ x_{2}\\\ \vdots\\\
x_{n-1}\\\
x_{n}\end{bmatrix}}_{X(t)}=\underbrace{\begin{bmatrix}0&-1&0&0&\cdots&0\\\
0&0&-1&0&\cdots&0\\\ \vdots&\vdots&\vdots&\vdots&\cdots&\vdots\\\
0&0&0&0&\cdots&-1\\\
-p_{0}&-p_{1}&-p_{2}&-p_{3}&\cdots&-p_{n-1}\end{bmatrix}}_{P(t)}\begin{bmatrix}x_{1}\\\
x_{2}\\\ \vdots\\\ x_{n-1}\\\
x_{n}\end{bmatrix}=\underbrace{\begin{bmatrix}0\\\ 0\\\ \vdots\\\ 0\\\
f(t)\end{bmatrix}}_{F(t)}$ $D^{\alpha,\beta}_{M}X(t)=P(t)X(t)$

We have

$D^{\alpha,\beta}_{M}W_{\alpha,\beta}(t)=(a_{11}+a_{22}+...+a_{nn})W_{\alpha,\beta}(t)$

In our case

$a_{11}+a_{22}+...+a_{nn}=-p_{n-1}(t)$

So,

$D^{\alpha,\beta}_{M}W_{\alpha,\beta}(t)+p_{n-1}(t)W_{\alpha,\beta}(t)=0$

(2)The above differential equation can be solved by the method of integrating
factor, we have

$W_{\alpha,\beta}(t)=W_{\alpha,\beta}(t_{0})e^{-\Gamma(\beta+1)\int_{t_{0}}^{t}x^{\alpha-1}p_{n-1}(x)dx}$

Thus the proof of theorem is completed. ∎

######  Theorem 3.7.

If $\\{y_{1},y_{2}...y_{n}\\}$ is a fundamental set of solutions of
$L_{\alpha,\beta}[y]=0$ where
$\Gamma(\beta+1)t^{\alpha-1}p_{n-1}(t)...\Gamma(\beta+1)t^{\alpha-1}p_{1}(t),\Gamma(\beta+1)t^{\alpha-1}p_{0}(t)\in
C(a,b)$,then $W_{\alpha,\beta}(t)\neq 0$ for all $t\in(a,b)$.

###### Proof.

By applying procedure in Theorem-4.8 [5] to M-derivative, we can easily prove
this theorem. ∎

######  Theorem 3.8.

Let
$\Gamma(\beta+1)t^{\alpha-1}p_{n-1}(t),...,\Gamma(\beta+1)t^{\alpha-1}p_{1}(t),\Gamma(\beta+1)t^{\alpha-1}p_{0}(t)\in
C(a,b)$. The solution set $\\{y_{1},y_{2},...,y_{n}\\}$ is a fundamental set
of solutions to the equation $L_{\alpha,\beta}[y]=0$ if and only if the
functions $y_{1},y_{2},...,y_{n}$ are linearly independent.

###### Proof.

By applying procedure in Theorem-4.9 [5] to M-derivative, we can easily prove
this theorem. ∎

######  Theorem 3.9.

Let $y_{1},y_{2},...,y_{n}$ be a fundamental set of solutions of the equation
(1) and $y_{p}$ be any particular solution of the non homogeneous equation
(2). Then the general solution of the equation is
$y=c_{1}y_{1}+c_{2}y_{2}+...+c_{n}y_{n}+y_{p}$

###### Proof.

Let $L_{\alpha,\beta}$ be the differential operator and $y(t)$ and $y_{p}(t)$
be the solutions of the non homogeneous equation $L_{\alpha,\beta}[y]=f(t)$.
If we take $u(t)=y(t)-y_{p}(t)$, then by linearity of $L_{\alpha,\beta}$ we
have,

$L_{\alpha,\beta}[u]=L_{\alpha,\beta}[y(t)-y_{p}(t)]=L_{\alpha,\beta}[y(t)]-L_{\alpha,\beta}[y_{p}(t)]=f(t)-f(t)=0$

Then $u(t)$ is a solution of the homogenous equation $L_{\alpha,\beta}[y]=0$.
Then by Theorem 3.4

$u(t)=c_{1}y_{1}(t)+c_{2}y_{2}(t)+...+c_{n}y_{n}(t)$

i.e,

$y(t)-y_{p}(t)=c_{1}y_{1}(t)+c_{2}y_{2}(t)+...+c_{n}y_{n}(t)$

Then

$y(t)=c_{1}y_{1}(t)+c_{2}y_{2}(t)+...+c_{n}y_{n}(t)+y_{p}(t)$

∎

##  4\. Solution of Homogeneous Case

Consider the $n$ times M-differentiable function $y$ for $\alpha\in(0,1]$ and
$\beta>0$. The homogeneous sequential linear fractional differential equation
with M-derivative is

${}^{n}D^{\alpha,\beta}_{M}y+p_{n-1}(t)^{n-1}D^{\alpha,\beta}_{M}y+...+p_{2}(t)^{2}D^{\alpha,\beta}_{M}y+p_{1}(t)D^{\alpha,\beta}_{M}y+p_{0}(t)y=0$
(8)

where
${}^{n}D^{\alpha,\beta}_{M}y=D^{\alpha,\beta}_{M}D^{\alpha,\beta}_{M}...D^{\alpha,\beta}_{M}y$
$n$ times, and the coefficients $p_{0},p_{1},...,p_{n-1}$ are real constants.
We define an $n^{th}$-order differential operator for eqn. (1) as following

$L_{\alpha,\beta}[y]=^{n}D^{\alpha,\beta}_{M}y+p_{n-1}^{n-1}D^{\alpha,\beta}_{M}y+...+p_{2}^{2}D^{\alpha,\beta}_{M}y+p_{1}D^{\alpha,\beta}_{M}y+p_{0}y=0$
(9)

If $y_{1}(t),y_{2}(t),...,y_{n}(t)$ are linearly independent solutions of
Eqn.(1), then general solution is

$y=c_{1}y_{1}(t)+c_{2}y_{2}(t)+...+c_{n}y_{n}(t)$

where $c_{1},c_{2}...c_{n}$ are arbitrary constants.

######  Lemma 4.1.

Suppose that $L_{\alpha,\beta}[.]$ is a linear operator with constant
coefficients and $\alpha\in(0,1]$ and $\beta>0$, then for $t>0$

$L_{\alpha,\beta}[e^{\frac{r\Gamma(\beta+1)}{\alpha}t^{\alpha}}]=P_{n}(r)[e^{\frac{r\Gamma(\beta+1)}{\alpha}t^{\alpha}}]$

Where $P_{n}(r)=r^{n}+P_{n-1}r^{n-1}+...+P_{0}$ and $r$ is a real or complex
constant

###### Proof.

M-derivatives of $y=e^{\frac{r\Gamma(\beta+1)}{\alpha}t^{\alpha}}$ are

$D^{\alpha,\beta}_{M}y=re^{\frac{r\Gamma(\beta+1)}{\alpha}t^{\alpha}},^{2}D^{\alpha,\beta}_{M}y=r^{2}e^{\frac{r\Gamma(\beta+1)}{\alpha}t^{\alpha}},...,^{n}D^{\alpha,\beta}_{M}y=r^{n}e^{\frac{r\Gamma(\beta+1)}{\alpha}t^{\alpha}}$
(10)

We substitute $y=e^{\frac{r\Gamma(\beta+1)}{\alpha}t^{\alpha}}$ and Eqn.(10)
in $L_{\alpha,\beta}[y]$

$L_{\alpha,\beta}[e^{\frac{r\Gamma(\beta+1)}{\alpha}t^{\alpha}}]=(^{n}D^{\alpha,\beta}_{M}+p_{n-1}^{n-1}D^{\alpha,\beta}_{M}+...+p_{2}^{2}D^{\alpha,\beta}_{M}+p_{1}D^{\alpha,\beta}_{M}+p_{0})e^{\frac{r\Gamma(\beta+1)}{\alpha}t^{\alpha}}$
$=(r^{n}+P_{n-1}r^{n-1}+...+P_{0})e^{\frac{r\Gamma(\beta+1)}{\alpha}t^{\alpha}}$
$L_{\alpha,\beta}[e^{\frac{r\Gamma(\beta+1)}{\alpha}t^{\alpha}}]=p_{n}(r)e^{\frac{r\Gamma(\beta+1)}{\alpha}t^{\alpha}}$

Hence, the proof is completed. ∎

The solution to the equation (8) is
$y=e^{\frac{r\Gamma(\beta+1)}{\alpha}t^{\alpha}}$.
It follows from Eqn.(9) and Lemma 3.1 that

$L_{\alpha,\beta}[e^{\frac{r\Gamma(\beta+1)}{\alpha}t^{\alpha}}]=p_{n}(r)e^{\frac{r\Gamma(\beta+1)}{\alpha}t^{\alpha}}=0$

Where $P_{n}(r)=r^{n}+P_{n-1}r^{n-1}+...+P_{0}$ is called as the
characteristic polynomial. For all $r$, we have
$e^{\frac{r\Gamma(\beta+1)}{\alpha}t^{\alpha}}\neq 0$. Hence $P_{n}(r)=0$.
Here

$r^{n}+P_{n-1}r^{n-1}+...+P_{0}=0$ (11)

is called as the characteristic equation.

######  Lemma 4.2.

Let $r$ be a root of the characteristic equation (11), then

$\frac{\partial}{\partial
r}\\{L_{\alpha,\beta}[e^{\frac{r\Gamma(\beta+1)}{\alpha}t^{\alpha}}]\\}=L_{\alpha,\beta}[\frac{\partial}{\partial
r}e^{\frac{r\Gamma(\beta+1)}{\alpha}t^{\alpha}}]$

and $\frac{\partial^{l}}{\partial
r^{l}}e^{\frac{r\Gamma(\beta+1)}{\alpha}t^{\alpha}}=(\frac{\Gamma(\beta+1)}{\alpha}t^{\alpha})^{l}e^{\frac{r\Gamma(\beta+1)}{\alpha}t^{\alpha}}$
where $l$ is integer.

###### Proof.

From Theorem 3.3 it follows that $L_{\alpha,\beta}[.]$ is linear and also
$\frac{\partial}{\partial r}$ is linear by property of classical derivative.
Hence

$\frac{\partial}{\partial
r}\\{L_{\alpha,\beta}[e^{\frac{r\Gamma(\beta+1)}{\alpha}t^{\alpha}}]\\}=L_{\alpha,\beta}[\frac{\partial}{\partial
r}e^{\frac{r\Gamma(\beta+1)}{\alpha}t^{\alpha}}]$

Additionally, from classical derivative, it follows that

$\frac{\partial^{l}}{\partial
r^{l}}e^{\frac{r\Gamma(\beta+1)}{\alpha}t^{\alpha}}=(\frac{\Gamma(\beta+1)}{\alpha}t^{\alpha})^{l}e^{\frac{r\Gamma(\beta+1)}{\alpha}t^{\alpha}}$

∎

######  Lemma 4.3.

If $r_{1}$ is a root of multiplicity of $\mu_{1}$of the characteristic
equation (11), then the functions $y_{1,l}(t)$ , where $l=0,1,...,\mu_{l-1}$
such that

$y_{1,l}=(\frac{\Gamma(\beta+1)}{\alpha}t^{\alpha})^{l}e^{\frac{r_{1}\Gamma(\beta+1)}{\alpha}t^{\alpha}}$

are solutions of Eq.(8).

###### Proof.

Consider
$L_{\alpha,\beta}[e^{\frac{r\Gamma(\beta+1)}{\alpha}t^{\alpha}}]=p_{n}(r)e^{\frac{r\Gamma(\beta+1)}{\alpha}t^{\alpha}}$.
From Lemma 4.2 and applying classical Leibniz rule it follows that

$\Bigg{\\{}L_{\alpha,\beta}\bigg{[}\frac{\partial^{l}}{\partial
r^{l}}e^{\frac{r\Gamma(\beta+1)}{\alpha}t^{\alpha}}\bigg{]}\Bigg{\\}}_{r=r_{1}}=\Bigg{\\{}\frac{\partial^{l}}{\partial
r^{l}}\Bigg{[}L_{\alpha,\beta}\bigg{[}e^{\frac{r\Gamma(\beta+1)}{\alpha}t^{\alpha}}\bigg{]}\Bigg{]}\Bigg{\\}}_{r=r_{1}}=\Bigg{\\{}\frac{\partial^{l}}{\partial
r^{l}}\bigg{[}p_{n}(r)e^{\frac{r\Gamma(\beta+1)}{\alpha}t^{\alpha}}\bigg{]}\Bigg{\\}}_{r=r_{1}}$
$\Bigg{\\{}L_{\alpha,\beta}\bigg{[}\frac{\partial^{l}}{\partial
r^{l}}e^{\frac{r\Gamma(\beta+1)}{\alpha}t^{\alpha}}\bigg{]}\Bigg{\\}}_{r=r_{1}}=\sum_{j=0}^{l}\begin{pmatrix}l\\\
j\end{pmatrix}\bigg{[}\frac{\partial^{l-j}}{\partial
r^{l-j}}e^{\frac{r\Gamma(\beta+1)}{\alpha}t^{\alpha}}\bigg{]}_{r=r_{1}}\frac{\partial^{j}}{\partial
r^{j}}\big{[}P_{n}(r)\big{]}_{r=r_{1}}$

Since $\frac{\partial^{j}}{\partial r^{j}}\big{[}P_{n}(r)\big{]}_{r=r_{1}}=0$
for $j=0,1,...,\mu_{1}-1$

$\Bigg{\\{}L_{\alpha,\beta}\bigg{[}\frac{\partial^{l}}{\partial
r^{l}}e^{\frac{r\Gamma(\beta+1)}{\alpha}t^{\alpha}}\bigg{]}\Bigg{\\}}_{r=r_{1}}=0$

From Lemma 4.2

$\Bigg{\\{}L_{\alpha,\beta}\bigg{[}\Big{(}\frac{\Gamma(\beta+1)}{\alpha}t^{\alpha}\Big{)}^{l}e^{\frac{r\Gamma(\beta+1)}{\alpha}t^{\alpha}}\bigg{]}\Bigg{\\}}_{r=r_{1}}=0$
$L_{\alpha,\beta}\Big{[}y_{1,l}(t)\Big{]}=0$

Hence $y_{1,l}(t)$ are solutions of Eq.(8). ∎

######  Corollary 4.1.

Let $r_{j},j=1,2,...,k$ are distinct roots of multiplicity
$\mu_{j},j=1,2,...,k$ of the characteristic Eq.(5). Then the following
functions

$\bigcup_{j=1}^{k}\Bigg{\\{}\Big{(}\frac{\Gamma(\beta+1)}{\alpha}t^{\alpha}\Big{)}^{l}e^{\frac{r_{j}\Gamma(\beta+1)}{\alpha}t^{\alpha}}\Bigg{\\}}_{l=0}^{\mu_{j}-1}$

###### Proof.

Corollary 4.1 follows from Lemma 4.3 and Theorem 3.5. ∎

######  Lemma 4.4.

If $r_{1}$ and $\bar{r_{1}}$ $(r_{1}=a+ib,b\neq 0)$ are complex roots of
multiplicity $\sigma_{1}$ of the characteristic equation (11), then for
$l=0,1,...,\sigma_{1}-1$, the functions

$y_{1,l}(t)=\Big{(}\frac{\Gamma(\beta+1)}{\alpha}t^{\alpha}\Big{)}^{l}e^{\frac{a\Gamma(\beta+1)}{\alpha}t^{\alpha}}\Big{[}cos\Big{(}\frac{b\Gamma(\beta+1)}{\alpha}t^{\alpha}\Big{)}+isin\Big{(}\frac{b\Gamma(\beta+1)}{\alpha}t^{\alpha}\Big{)}\Big{]}$

and

$y_{2,l}(t)=\Big{(}\frac{\Gamma(\beta+1)}{\alpha}t^{\alpha}\Big{)}^{l}e^{\frac{a\Gamma(\beta+1)}{\alpha}t^{\alpha}}\Big{[}cos\Big{(}\frac{b\Gamma(\beta+1)}{\alpha}t^{\alpha}\Big{)}-isin\Big{(}\frac{b\Gamma(\beta+1)}{\alpha}t^{\alpha}\Big{)}\Big{]}$

are linearly independent solutions of Eq.(8).

###### Proof.

Since $r_{1}=a+ib$ is a root of multiplicity $\sigma_{1}$ of the
characteristic equation (11), From Lemma 4.3 and using Euler’s identity it
follows that, the functions

$y_{1,l}(t)=\Big{(}\frac{\Gamma(\beta+1)}{\alpha}t^{\alpha}\Big{)}^{l}e^{\frac{(a+ib)\Gamma(\beta+1)}{\alpha}t^{\alpha}}$

i.e

$y_{1,l}(t)=\Big{(}\frac{\Gamma(\beta+1)}{\alpha}t^{\alpha}\Big{)}^{l}e^{\frac{a\Gamma(\beta+1)}{\alpha}t^{\alpha}}\Big{[}cos\Big{(}\frac{b\Gamma(\beta+1)}{\alpha}t^{\alpha}\Big{)}+isin\Big{(}\frac{b\Gamma(\beta+1)}{\alpha}t^{\alpha}\Big{)}\Big{]}$

are solutions of the Eq.(8). Similarly, for $\bar{r_{1}}=a-ib$, the functions

$y_{2,l}(t)=\Big{(}\frac{\Gamma(\beta+1)}{\alpha}t^{\alpha}\Big{)}^{l}e^{\frac{(a-ib)\Gamma(\beta+1)}{\alpha}t^{\alpha}}$

i.e

$y_{2,l}(t)=\Big{(}\frac{\Gamma(\beta+1)}{\alpha}t^{\alpha}\Big{)}^{l}e^{\frac{a\Gamma(\beta+1)}{\alpha}t^{\alpha}}\Big{[}cos\Big{(}\frac{b\Gamma(\beta+1)}{\alpha}t^{\alpha}\Big{)}-isin\Big{(}\frac{b\Gamma(\beta+1)}{\alpha}t^{\alpha}\Big{)}\Big{]}$

are solutions of the Eq.(8). Hence proof is completed. ∎

######  Corollary 4.2.

If $\big{\\{}r_{j},\bar{r_{j}}\big{\\}}_{j=1}^{m},r_{j}=a_{j}+ib_{j},b_{j}\neq
0$ distinct $2m$ roots of multiplicity
$\big{\\{}\sigma_{j}\big{\\}}_{j=1}^{m}$ of the characteristic equation
(11),then, the functions

$\bigcup_{j=1}^{m}\Bigg{\\{}\Big{(}\frac{\Gamma(\beta+1)}{\alpha}t^{\alpha}\Big{)}^{l}e^{\frac{a_{j}\Gamma(\beta+1)}{\alpha}t^{\alpha}}\Big{[}cos\Big{(}\frac{b_{j}\Gamma(\beta+1)}{\alpha}t^{\alpha}\Big{)}+isin\Big{(}\frac{b_{j}\Gamma(\beta+1)}{\alpha}t^{\alpha}\Big{)}\Big{]}\Bigg{\\}}_{l=0}^{\sigma_{j}-1}$

and

$\bigcup_{j=1}^{m}\Bigg{\\{}\Big{(}\frac{\Gamma(\beta+1)}{\alpha}t^{\alpha}\Big{)}^{l}e^{\frac{a_{j}\Gamma(\beta+1)}{\alpha}t^{\alpha}}\Big{[}cos\Big{(}\frac{b_{j}\Gamma(\beta+1)}{\alpha}t^{\alpha}\Big{)}-isin\Big{(}\frac{b_{j}\Gamma(\beta+1)}{\alpha}t^{\alpha}\Big{)}\Big{]}\Bigg{\\}}_{l=0}^{\sigma_{j}-1}$

###### Proof.

To prove corollary 4.2, it is sufficient to apply the Lemma 4.4 and Theorem
3.5. ∎

######  Theorem 4.1.

If $\big{\\{}r_{j}\big{\\}}_{j=1}^{k}$ are distinct $k$ roots of multiplicity
$\big{\\{}\mu_{j}\big{\\}}_{j=1}^{k}$ and
$\big{\\{}\lambda_{j},\bar{\lambda_{j}}\big{\\}}_{j=1}^{m},\lambda_{j}=a_{j}+ib_{j},b_{j}\neq
0$ are distinct $2m$ roots of multiplicity
$\big{\\{}\sigma_{j}\big{\\}}_{j=1}^{m}$ of the characteristic equation (11)
such that $\sum_{j=1}^{k}\mu_{j}+2\sum_{j=1}^{m}\sigma_{j}=n$, then the
functions

$\bigcup_{j=1}^{k}\Bigg{\\{}\Big{(}\frac{\Gamma(\beta+1)}{\alpha}t^{\alpha}\Big{)}^{l}e^{\frac{r_{j}\Gamma(\beta+1)}{\alpha}t^{\alpha}}\Bigg{\\}}_{l=0}^{\mu_{j}-1}$
$\bigcup_{j=1}^{m}\Bigg{\\{}\Big{(}\frac{\Gamma(\beta+1)}{\alpha}t^{\alpha}\Big{)}^{l}e^{\frac{a_{j}\Gamma(\beta+1)}{\alpha}t^{\alpha}}\Big{[}cos\Big{(}\frac{b_{j}\Gamma(\beta+1)}{\alpha}t^{\alpha}\Big{)}+isin\Big{(}\frac{b_{j}\Gamma(\beta+1)}{\alpha}t^{\alpha}\Big{)}\Big{]}\Bigg{\\}}_{l=0}^{\sigma_{j}-1}$

and

$\bigcup_{j=1}^{m}\Bigg{\\{}\Big{(}\frac{\Gamma(\beta+1)}{\alpha}t^{\alpha}\Big{)}^{l}e^{\frac{a_{j}\Gamma(\beta+1)}{\alpha}t^{\alpha}}\Big{[}cos\Big{(}\frac{b_{j}\Gamma(\beta+1)}{\alpha}t^{\alpha}\Big{)}-isin\Big{(}\frac{b_{j}\Gamma(\beta+1)}{\alpha}t^{\alpha}\Big{)}\Big{]}\Bigg{\\}}_{l=0}^{\sigma_{j}-1}$

are the fundamental set of solutions of the equation (8).

###### Proof.

The proof of the Theorem 4.1 is follows from Corollary 4.1, Corollary 4.2 and
Theorem 3.5. ∎

######  Example 4.1.

${}^{2}D^{\alpha,\beta}_{M}y+4D^{\alpha,\beta}_{M}y+3y=0$ (12)

The characteristic equation of (12) is

$r^{2}+4r+3=0$

Therefore, the roots are $r=-3$ and $r=-1$
Hence, the general solution is

$y(t)=c_{1}e^{\frac{-3\Gamma(\beta+1)}{\alpha}t^{\alpha}}+c_{2}e^{\frac{-\Gamma(\beta+1)}{\alpha}t^{\alpha}}$

######  Example 4.2.

${}^{2}D^{\alpha,\beta}_{M}y-4D^{\alpha,\beta}_{M}y+4y=0$ (13)

The characteristic equation of (13) is

$r^{2}-4r+4=0$

The roots are $r_{1,2}=2$
Hence, the general solution is

$y(t)=\Bigg{(}c_{1}+c_{2}\frac{\Gamma(\beta+1)}{\alpha}t^{\alpha}\Bigg{)}e^{\frac{2\Gamma(\beta+1)}{\alpha}t^{\alpha}}$

######  Example 4.3.

${}^{2}D^{\alpha,\beta}_{M}y+4D^{\alpha,\beta}_{M}y+5y=0$ (14)

The characteristic equation of (14) is

$r^{2}+4r+5=0$

The roots are $r_{1}=-2+i$ and $r_{2}=-2-i$
Hence, the general solution is

$y(t)=e^{\frac{-2\Gamma(\beta+1)}{\alpha}t^{\alpha}}\Big{[}cos\Big{(}\frac{\Gamma(\beta+1)}{\alpha}t^{\alpha}\Big{)}+isin\Big{(}\frac{\Gamma(\beta+1)}{\alpha}t^{\alpha}\Big{)}\Big{]}$

##  5\. Solution of Non-Homogeneous Case

In this section, Method of variation of parameters is applied to derive the
particular solution of the equation.

${}^{n}D^{\alpha,\beta}_{M}y+p_{n-1}(t)^{n-1}D^{\alpha,\beta}_{M}y+...+p_{2}(t)^{2}D^{\alpha,\beta}_{M}y+p_{1}(t)D^{\alpha,\beta}_{M}y+p_{0}(t)y=f(t)$
(15)

where $y$ is $n$ times M-differentiable function for $\alpha\in(0,1]$ and
$\beta>0$.

######  Theorem 5.1.

If $u(t)$ is a solution of homogeneous case of the equation (15) such that

$u(t)=\sum_{i=1}^{n}c_{i}y_{i}(t)$ (16)

then particular solution of the equation (15) is

$v(t)=\sum_{i=1}^{n}c_{i}(t)y_{i}(t)$

Where $c_{1}(t),c_{2}(t),...,c_{n}(t)$ provide following system of equations

$\sum_{i=1}^{n}D^{\alpha,\beta}_{M}c_{i}(t)y_{i}(t)=0$
$\sum_{i=1}^{n}D^{\alpha,\beta}_{M}c_{i}(t)D^{\alpha,\beta}_{M}y_{i}(t)=0$
$\vdots$
$\sum_{i=1}^{n}D^{\alpha,\beta}_{M}c_{i}(t)^{n-2}D^{\alpha,\beta}_{M}y_{i}(t)=0$
$\sum_{i=1}^{n}D^{\alpha,\beta}_{M}c_{i}(t)^{n-1}D^{\alpha,\beta}_{M}y_{i}(t)=f(t)$

###### Proof.

The solution of the equation (15) is in the form

$v(t)=\sum_{i=1}^{n}c_{i}(t)y_{i}(t)$

The M-derivative of $v(t)$ for $\alpha\in(0,1]$ and $\beta>0$ will be

$D^{\alpha,\beta}_{M}v(t)=\sum_{i=1}^{n}c_{i}(t)D^{\alpha,\beta}_{M}y_{i}(t)+\sum_{i=1}^{n}D^{\alpha,\beta}_{M}c_{i}(t)y_{i}(t)$

Applying the first condition
$\sum_{i=1}^{n}D^{\alpha,\beta}_{M}c_{i}(t)y_{i}(t)=0$, we obtain

$D^{\alpha,\beta}_{M}v(t)=\sum_{i=1}^{n}c_{i}(t)D^{\alpha,\beta}_{M}y_{i}(t)$

If we calculate the M-derivative of $D^{\alpha,\beta}_{M}v(t)$ for
$\alpha\in(0,1]$ and $\beta>0$, then we get

${}^{2}D^{\alpha,\beta}_{M}v(t)=\sum_{i=1}^{n}c_{i}(t)^{2}D^{\alpha,\beta}_{M}y_{i}(t)+\sum_{i=1}^{n}D^{\alpha,\beta}_{M}c_{i}(t)D^{\alpha,\beta}_{M}y_{i}(t)$

Apply second condition
$\sum_{i=1}^{n}D^{\alpha,\beta}_{M}c_{i}(t)D^{\alpha,\beta}_{M}y_{i}(t)=0$, we
obtain

${}^{2}D^{\alpha,\beta}_{M}v(t)=\sum_{i=1}^{n}c_{i}(t)^{2}D^{\alpha,\beta}_{M}y_{i}(t)$

By continuing in this way, we get

${}^{n-1}D^{\alpha,\beta}_{M}v(t)=\sum_{i=1}^{n}c_{i}(t)^{n-1}D^{\alpha,\beta}_{M}y_{i}(t)+\sum_{i=1}^{n}D^{\alpha,\beta}_{M}c_{i}(t)^{n-2}D^{\alpha,\beta}_{M}y_{i}(t)$

We substitute
$v(t),^{2}D^{\alpha,\beta}_{M}v(t),...,^{n}D^{\alpha,\beta}_{M}v(t)$ in the
equation (15), we have

$\sum_{i=1}^{n}c_{i}(t)^{n}D^{\alpha,\beta}_{M}y_{i}(t)+\sum_{i=1}^{n}D^{\alpha,\beta}_{M}c_{i}(t)^{n-1}D^{\alpha,\beta}_{M}y_{i}(t)+p_{n-1}\sum_{i=1}^{n}c_{i}(t)^{n-1}D^{\alpha,\beta}_{M}y_{i}(t)+...+\\\
p_{2}\sum_{i=1}^{n}c_{i}(t)^{2}D^{\alpha,\beta}_{M}y_{i}(t)+p_{1}\sum_{i=1}^{n}c_{i}(t)D^{\alpha,\beta}_{M}y_{i}(t)+p_{0}\sum_{i=1}^{n}c_{i}(t)y_{i}(t)=f(t)$
$\sum_{i=1}^{n}D^{\alpha,\beta}_{M}c_{i}(t)^{n-1}D^{\alpha,\beta}_{M}y_{i}(t)+\sum_{i=1}^{n}c_{i}(t)\big{[}^{n}D^{\alpha,\beta}_{M}y_{i}(t)+p_{n-1}^{n-1}D^{\alpha,\beta}_{M}y_{i}(t)+...+\\\
p_{1}D^{\alpha,\beta}_{M}y_{i}(t)+p_{0}y_{i}(t)\big{]}=f(t)$

Since $y_{1}(t),y_{2}(t),...,y_{n}(t)$ are solutions of homogeneous case of
equation (8), then

$\sum_{i=1}^{n}c_{i}(t)\big{[}^{n}D^{\alpha,\beta}_{M}y_{i}(t)+p_{n-1}^{n-1}D^{\alpha,\beta}_{M}y_{i}(t)+...+\\\
p_{1}D^{\alpha,\beta}_{M}y_{i}(t)+p_{0}y_{i}(t)\big{]}=0$

We obtain $n^{th}$ condition as

$\sum_{i=1}^{n}D^{\alpha,\beta}_{M}c_{i}(t)^{n-1}D^{\alpha,\beta}_{M}y_{i}(t)=f(t)$

Hence we obtain the following system

$\sum_{i=1}^{n}D^{\alpha,\beta}_{M}c_{i}(t)y_{i}(t)=0$
$\sum_{i=1}^{n}D^{\alpha,\beta}_{M}c_{i}(t)D^{\alpha,\beta}_{M}y_{i}(t)=0$
$\vdots$ (17)
$\sum_{i=1}^{n}D^{\alpha,\beta}_{M}c_{i}(t)^{n-2}D^{\alpha,\beta}_{M}y_{i}(t)=0$
$\sum_{i=1}^{n}D^{\alpha,\beta}_{M}c_{i}(t)^{n-1}D^{\alpha,\beta}_{M}y_{i}(t)=f(t)$

Solving the above system (17) provides $D^{\alpha,\beta}_{M}c_{i}(t)$,
$i=1,2,...,n$. Therefore we can write the particular solution of equation (15)
as $v(t)=\sum_{i=1}^{n}c_{i}(t)y_{i}(t)$ ∎

######  Example 5.1.

${}^{2}D^{\alpha,\beta}_{M}y+4D^{\alpha,\beta}_{M}y+3y=f(t)$

(a) Let $f(t)=e^{2t^{\alpha}}$. For
$v(t)=c_{1}(t)e^{\frac{-3\Gamma(\beta+1)}{\alpha}t^{\alpha}}+c_{2}(t)e^{\frac{-\Gamma(\beta+1)}{\alpha}t^{\alpha}}$,
the system of equations are built by the conditions as following

$D^{\alpha,\beta}_{M}c_{1}(t)e^{\frac{-3\Gamma(\beta+1)}{\alpha}t^{\alpha}}+D^{\alpha,\beta}_{M}c_{2}(t)e^{\frac{-\Gamma(\beta+1)}{\alpha}t^{\alpha}}=0$
$-3D^{\alpha,\beta}_{M}c_{1}(t)e^{\frac{-3\Gamma(\beta+1)}{\alpha}t^{\alpha}}-D^{\alpha,\beta}_{M}c_{2}(t)e^{\frac{-\Gamma(\beta+1)}{\alpha}t^{\alpha}}=e^{2t^{\alpha}}$

Solving the above system of equations and using M-integral we obtain
$c_{1}(t)=\frac{-\Gamma(\beta+1)}{4\alpha+6\Gamma(\beta+1)}e^{\frac{2\alpha+3\Gamma(\beta+1)}{\alpha}t^{\alpha}},c_{2}(t)=\frac{\Gamma(\beta+1)}{4\alpha+2\Gamma(\beta+1)}e^{\frac{2\alpha+\Gamma(\beta+1)}{\alpha}t^{\alpha}}$.
Then particular solution $v(t)$ is

$v(t)=\frac{\Gamma(\beta+1)^{2}}{4\alpha^{2}+8\alpha\Gamma(\beta+1)+3\Gamma(\beta+1)^{2}}e^{2t^{\alpha}}$

(b) Let $f(t)=2t^{2\alpha}+t^{\alpha}-3$. The system of equations for this
case is

$D^{\alpha,\beta}_{M}c_{1}(t)e^{\frac{-3\Gamma(\beta+1)}{\alpha}t^{\alpha}}+D^{\alpha,\beta}_{M}c_{2}(t)e^{\frac{-\Gamma(\beta+1)}{\alpha}t^{\alpha}}=0$
$-3D^{\alpha,\beta}_{M}c_{1}(t)e^{\frac{-3\Gamma(\beta+1)}{\alpha}t^{\alpha}}-D^{\alpha,\beta}_{M}c_{2}(t)e^{\frac{-\Gamma(\beta+1)}{\alpha}t^{\alpha}}=2t^{2\alpha}+t^{\alpha}-3$

Solve this system of equations, we have

$c_{1}(t)=\frac{-1}{3}t^{2\alpha}e^{\frac{3\Gamma(\beta+1)}{\alpha}t^{\alpha}}+\Big{(}\frac{4\alpha-3\Gamma(\beta+1)}{18\Gamma(\beta+1)}\Big{)}t^{\alpha}e^{\frac{3\Gamma(\beta+1)}{\alpha}t^{\alpha}}\\\
+\Bigg{(}\frac{-4\alpha^{2}+3\alpha\Gamma(\beta+1)+27\Gamma(\beta+1)^{2}}{54\Gamma(\beta+1)^{2}}\Bigg{)}e^{\frac{3\Gamma(\beta+1)}{\alpha}t^{\alpha}}$
$c_{2}(t)=t^{2\alpha}e^{\frac{\Gamma(\beta+1)}{\alpha}t^{\alpha}}+\Big{(}\frac{\Gamma(\beta+1)-4\alpha}{2\Gamma(\beta+1)}\Big{)}t^{\alpha}e^{\frac{\Gamma(\beta+1)}{\alpha}t^{\alpha}}\\\
+\Bigg{(}\frac{4\alpha^{2}-\alpha\Gamma(\beta+1)-3\Gamma(\beta+1)^{2}}{2\Gamma(\beta+1)^{2}}\Bigg{)}e^{\frac{\Gamma(\beta+1)}{\alpha}t^{\alpha}}$

Hence, particular solution $v(t)$ is obtained by

$v(t)=\frac{2}{3}t^{2\alpha}+\Big{(}\frac{3\Gamma(\beta+1)-16\alpha}{9\Gamma(\beta+1)}\Big{)}t^{\alpha}e^{\frac{\Gamma(\beta+1)}{\alpha}t^{\alpha}}\\\
+\Bigg{(}\frac{52\alpha^{2}-12\alpha\Gamma(\beta+1)-27\Gamma(\beta+1)^{2}}{27\Gamma(\beta+1)^{2}}\Bigg{)}e^{\frac{\Gamma(\beta+1)}{\alpha}t^{\alpha}}$

(c) Let $f(t)=sin2t^{\alpha}$. The system of equations for this case is

$D^{\alpha,\beta}_{M}c_{1}(t)e^{\frac{-3\Gamma(\beta+1)}{\alpha}t^{\alpha}}+D^{\alpha,\beta}_{M}c_{2}(t)e^{\frac{-\Gamma(\beta+1)}{\alpha}t^{\alpha}}=0$
$-3D^{\alpha,\beta}_{M}c_{1}(t)e^{\frac{-3\Gamma(\beta+1)}{\alpha}t^{\alpha}}-D^{\alpha,\beta}_{M}c_{2}(t)e^{\frac{-\Gamma(\beta+1)}{\alpha}t^{\alpha}}=sin2t^{\alpha}$

Solve this system of equations, we have

$c_{1}(t)=\frac{-3\Gamma(\beta+1)^{2}}{8\alpha^{2}+18\Gamma(\beta+1)^{2}}e^{\frac{3\Gamma(\beta+1)}{\alpha}t^{\alpha}}sin2t^{\alpha}+\frac{\alpha\Gamma(\beta+1)}{4\alpha^{2}+9\Gamma(\beta+1)^{2}}e^{\frac{3\Gamma(\beta+1)}{\alpha}t^{\alpha}}cos2t^{\alpha}$
$c_{2}(t)=\frac{\Gamma(\beta+1)^{2}}{8\alpha^{2}+2\Gamma(\beta+1)^{2}}e^{\frac{\Gamma(\beta+1)}{\alpha}t^{\alpha}}sin2t^{\alpha}\\\
-\frac{\alpha\Gamma(\beta+1)}{4\alpha^{2}+\Gamma(\beta+1)^{2}}e^{\frac{3\Gamma(\beta+1)}{\alpha}t^{\alpha}}cos2t^{\alpha}$

Hence, particular solution $v(t)$ is obtained by

$v(t)=\frac{-4\alpha^{2}\Gamma(\beta+1)^{2}+3\Gamma(\beta+1)^{4}}{16\alpha^{4}+40\alpha^{2}\Gamma(\beta+1)^{2}+9\Gamma(\beta+1)^{4}}sin2t^{\alpha}\\\
-\frac{8\alpha\Gamma(\beta+1)^{3}}{16\alpha^{4}+40\alpha^{2}\Gamma(\beta+1)^{2}+9\Gamma(\beta+1)^{4}}cos2t^{\alpha}$

(d) Let $f(t)=e^{2t^{\alpha}}t^{\alpha}$. The system of equations for this
case is

$D^{\alpha,\beta}_{M}c_{1}(t)e^{\frac{-3\Gamma(\beta+1)}{\alpha}t^{\alpha}}+D^{\alpha,\beta}_{M}c_{2}(t)e^{\frac{-\Gamma(\beta+1)}{\alpha}t^{\alpha}}=0$
$-3D^{\alpha,\beta}_{M}c_{1}(t)e^{\frac{-3\Gamma(\beta+1)}{\alpha}t^{\alpha}}-D^{\alpha,\beta}_{M}c_{2}(t)e^{\frac{-\Gamma(\beta+1)}{\alpha}t^{\alpha}}=e^{2t^{\alpha}}t^{\alpha}$

Solve this system of equations, we have

$c_{1}(t)=\frac{-\Gamma(\beta+1)}{4\alpha+6\Gamma(\beta+1)}t^{\alpha}e^{\frac{2\alpha+3\Gamma(\beta+1)}{\alpha}t^{\alpha}}+\frac{\alpha\Gamma(\beta+1)}{2\big{(}2\alpha+3\Gamma(\beta+1)\big{)}^{2}}e^{\frac{2\alpha+3\Gamma(\beta+1)}{\alpha}t^{\alpha}}$
$c_{2}(t)=\frac{\Gamma(\beta+1)}{4\alpha+2\Gamma(\beta+1)}t^{\alpha}e^{\frac{2\alpha+\Gamma(\beta+1)}{\alpha}t^{\alpha}}-\frac{\alpha\Gamma(\beta+1)}{2\big{(}2\alpha+\Gamma(\beta+1)\big{)}^{2}}e^{\frac{2\alpha+\Gamma(\beta+1)}{\alpha}t^{\alpha}}$

Hence, particular solution $v(t)$ is obtained by

$v(t)=\frac{\Gamma(\beta+1)^{2}}{4\alpha^{2}+8\alpha\Gamma(\beta+1)+3\Gamma(\beta+1)^{2}}t^{\alpha}e^{2t^{\alpha}}-\frac{4\alpha^{2}\Gamma(\beta+1)^{2}+4\alpha\Gamma(\beta+1)^{3}}{\big{(}4\alpha^{2}+8\alpha\Gamma(\beta+1)+3\Gamma(\beta+1)^{2}\big{)}^{2}}e^{2t^{\alpha}}$

(e) Let $f(t)=e^{-4t^{\alpha}}$. Take $\alpha\neq\frac{3}{4}$ and
$\alpha\neq\frac{1}{4}$, the system of equations for this case is

$D^{\alpha,\beta}_{M}c_{1}(t)e^{\frac{-3\Gamma(\beta+1)}{\alpha}t^{\alpha}}+D^{\alpha,\beta}_{M}c_{2}(t)e^{\frac{-\Gamma(\beta+1)}{\alpha}t^{\alpha}}=0$
$-3D^{\alpha,\beta}_{M}c_{1}(t)e^{\frac{-3\Gamma(\beta+1)}{\alpha}t^{\alpha}}-D^{\alpha,\beta}_{M}c_{2}(t)e^{\frac{-\Gamma(\beta+1)}{\alpha}t^{\alpha}}=e^{-4t^{\alpha}}$

Solve this system of equations, we have

$c_{1}(t)=\frac{\Gamma(\beta+1)}{8\alpha-6\Gamma(\beta+1)}e^{\frac{3\Gamma(\beta+1)-4\alpha}{\alpha}t^{\alpha}}$
$c_{2}(t)=\frac{\Gamma(\beta+1)}{2\Gamma(\beta+1)-8\alpha}e^{\frac{\Gamma(\beta+1)-4\alpha}{\alpha}}t^{\alpha}$

Hence, we obtain particular solution $v(t)$ as following:

$v(t)=\frac{\Gamma(\beta+1)^{2}}{16\alpha^{2}-16\alpha\Gamma(\beta+1)+3\Gamma(\beta+1)^{2}}e^{-4t^{\alpha}}$

Take $\alpha=\frac{3}{4}$ and
$v(t)=c_{1}(t)e^{-4\Gamma(\beta+1)t^{\frac{3}{4}}}+c_{2}(t)e^{\frac{-4\Gamma(\beta+1)}{3}t^{\frac{3}{4}}}$
The system of equations is

$D^{\frac{3}{4},\beta}_{M}c_{1}(t)e^{-4\Gamma(\beta+1)t^{\frac{3}{4}}}+D^{\frac{3}{4},\beta}_{M}c_{2}(t)e^{\frac{-4\Gamma(\beta+1)}{3}t^{\frac{3}{4}}}=0$
$-3D^{\frac{3}{4},\beta}_{M}c_{1}(t)e^{-4\Gamma(\beta+1)t^{\frac{3}{4}}}-D^{\frac{3}{4},\beta}_{M}c_{2}(t)e^{\frac{-4\Gamma(\beta+1)}{3}t^{\frac{3}{4}}}=e^{-4t^{\frac{3}{4}}}$

We solve the above equation,
$c_{1}(t)=-\frac{\Gamma(\beta+1)}{-6+6\Gamma(\beta+1)}e^{\big{(}-4+4\Gamma(\beta+1)\big{)}t^{\frac{3}{4}}}$
and
$c_{2}(t)=\frac{\Gamma(\beta+1)}{-6+2\Gamma(\beta+1)}e^{\big{(}-12+4\Gamma(\beta+1)\big{)}t^{\frac{3}{4}}}$
is obtained.
The particular solution is
$v(t)=\frac{\Gamma(\beta+1)^{2}}{9-12\Gamma(\beta+1)+3\Gamma(\beta+1)^{2}}e^{-4t^{\frac{3}{4}}}$

Take $\alpha=\frac{1}{4}$ and
$v(t)=c_{1}(t)e^{-12\Gamma(\beta+1)t^{\frac{1}{4}}}+c_{2}(t)e^{-4\Gamma(\beta+1)t^{\frac{1}{4}}}$
The system of equations is

$D^{\frac{1}{4},\beta}_{M}c_{1}(t)e^{-12\Gamma(\beta+1)t^{\frac{1}{4}}}+D^{\frac{1}{4},\beta}_{M}c_{2}(t)e^{-4\Gamma(\beta+1)t^{\frac{1}{4}}}=0$
$-3D^{\frac{1}{4},\beta}_{M}c_{1}(t)e^{-12\Gamma(\beta+1)t^{\frac{1}{4}}}-D^{\frac{1}{4},\beta}_{M}c_{2}(t)e^{-4\Gamma(\beta+1)t^{\frac{1}{4}}}=e^{-4t^{\frac{1}{4}}}$

We solve the above
equation,$c_{1}(t)=-\frac{\Gamma(\beta+1)}{-2+6\Gamma(\beta+1)}e^{\big{(}-4+12\Gamma(\beta+1)\big{)}t^{\frac{1}{4}}}$
and
$c_{2}(t)=\frac{\Gamma(\beta+1)}{-2+2\Gamma(\beta+1)}e^{\big{(}-4+4\Gamma(\beta+1)\big{)}t^{\frac{1}{4}}}$
is obtained.
The particular solution is
$v(t)=\frac{4\Gamma(\beta+1)^{2}}{4-16\Gamma(\beta+1)+12\Gamma(\beta+1)^{2}}e^{-4t^{\frac{1}{4}}}$

##  6\. Conclusion

In this paper, Existences and Uniqueness theorems for sequential linear
M-fractional differential equations are presented. We give solution of
M-fractional differential equations with constants for homogeneous case using
fractional exponential function and for non homogeneous case, we applied
method of variation of parameters.

## References

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