# NEW INTEGRAL TRANSFORM: SHEHU TRANSFORM A GENERALIZATION OF SUMUDU AND
LAPLACE TRANSFORM FOR SOLVING DIFFERENTIAL EQUATIONS

SHEHU MAITAMA∗, WEIDONG ZHAO School of Mathematics, Shandong University,
Jinan, Shandong 250100, P.R. China  ∗Corresponding author:
<EMAIL_ADDRESS>
###### Abstract.

In this paper, we introduce a Laplace-type integral transform called the Shehu
transform which is a generalization of the Laplace and the Sumudu integral
transforms for solving differential equations in the time domain. The proposed
integral transform is successfully derived from the classical Fourier integral
transform and is applied to both ordinary and partial differential equations
to show its simplicity, efficiency, and the high accuracy.

###### Key words and phrases:

Shehu transform; Fourier integral transform; Laplace transform; natural
transform; Sumudu transform; ordinary and partial differential equations.

###### 2010 Mathematics Subject Classification:

44A10, 44A15, 44A20, 44A30, 44A35

##  1\. Introduction

Historically, the origin of the integral transforms can be traced back to the
work of P. S. Laplace in 1780s and Joseph Fourier in 1822\. In recent years,
differential and integral equations have been solved using many integral
transforms ([1]-[11]). The Laplace transform, and Fourier integral transforms
are the most commonly used in the literature. The Fourier integral transform
[12] was named after the French mathematician Joseph Fourier. Mathematically,
Fourier integral transform is defined as:

$\digamma[f(t)]=f(\omega)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\exp\left(-i\omega
t\right)f(t)dt.$ (1.1)

The Fourier transform have many applications in physics and engineering
processes [13]. The Laplace integral transform is similar with the Fourier
transform and is defined as:

$\pounds[f(t)]=F(s)=\int_{-\infty}^{\infty}\exp\left(-st\right)f(t)dt.$ (1.2)

The Laplace transform is highly efficient for solving some class of ordinary
and partial differential equations [14]. By replacing the variable $i\omega$
with the variable $s$ in Equ.(1.1), the well-known Fourier transform will
become a Laplace transform and the vice-versa. The only difference between the
Laplace transform, and the Fourier transform is that the Laplace transform can
be defined for both stable and unstable system while the Fourier transform can
only be defined on a stable system. In mathematical literature, the discrete-
time equivalent of the Laplace transform called z-transform [15] converts a
discrete-time signal into a complex frequency-domain representation. The basic
idea of the z-transform was known to Laplace and later it was re-introduced by
the Jewish-Polish mathematician Witold Hurewicz to treat a sampled-data
control systems used with radar in 1947 ([16]-[17]). In mathematics and signal
processing, the bilateral or two-sided z-transform of a discrete-time signal
$x[n]$ is the normal power series $X(z)$ which is defined as:

$X(z)=Z\\{x[n]\\}=\sum_{n=-\infty}^{\infty}x[n]z^{-n},$ (1.3)

where $n$ is an integer and $z$ is in general a complex number [18].

The multiplicative version of the two-sided Laplace transform called the
Mellin integral transform is defined as [19]:

$M[f(s);s]=f^{*}(s)=\int_{0}^{\infty}x^{s-1}f(x)dx.$ (1.4)

The Mellin integral transform is similar with the Laplace transform and
Fourier transform and is widely applied in computer science and number theory
due to its invariant property [20, 21]. In railway engineering, the Laplace-
Carson transform [22] which is a Laplace-type integral transform named after
Pierre Simon Laplace and John Renshaw Carson is defined as:

$\hat{f}_{C}(p)=p\int_{0}^{\infty}\exp\left(-pt\right)f(t)dt,\,\,t\geq 0.$
(1.5)

The Laplace-Carson integral transform have many applications in physics and
engineering and can easily be converted into a Mellin deconvolution problem,
see ([23],[24]). In mathematics, the Hankel’s integral transform [25] which is
similar to the Fourier transform was first introduced by the German
mathematician Hermann Hankel and was widely used in physical science and
engineering [26]. The Hankel’s transform is defined as:

$F_{v}(s)=H_{v}[f(r)]=\int_{0}^{\infty}rf(r)J_{v}(sr)dr,\,\,\,\,\,r\geq 0,$
(1.6)

where $J_{v}$ is the Bessel function of the first kind of order $v$ with
$v\geq-\frac{1}{2}$.

In 1993, Watugala introduced a Laplace-like integral transform called the
Sumudu integral transform [27]. In recent years, Sumudu transform has been
applied to many real-life problems because of its scale and unit preserving
properties ([28]-[31]). The mathematical definition of the Sumudu transform is
given by:

$\textbf{S}[f(t)](u)=G(u)=\frac{1}{u}\int_{0}^{\infty}\exp\left(\frac{-t}{u}\right)f(t)dt,$
(1.7)

provided the integral exists for some $u$. Based on the basic idea of the
Laplace and the Sumudu integral transform, the Elzaki transform was proposed
in 2011. The Elzaki transform is closely related with the Laplace transform,
Sumudu transform, and the natural transform. Elzaki transform is defined as
[32]:

$E[f(t)]=T(u)=u\int_{0}^{\infty}\exp\left(\frac{-t}{u}\right)f(t)dt,$ (1.8)

provided the integral exists for some $u$.

The natural transform [33] which is similar to Laplace and Sumudu integral
transform was introduced in 2008. In recent years, natural transform was
successfully applied to many applications (see [34, 35]). The natural
transform is defined by the following integral:

${\mathbb{N}^{+}}[f(t)](s,u)=R(s,u)=\frac{1}{u}\int_{0}^{\infty}\exp\left(\frac{-st}{u}\right)f(t)dt,\,\,s\,>0,\,\,u>0,$
(1.9)

provided the integral exists for some variables $u$ and $s$. Recently, a new
integral transform called the ${\mathbb{M}}$-transform which is also similar
to natural transform is introduced by Srivastava et al. in 2015.
Mathematically speaking, ${\mathbb{M}}$-transform is closely connected with
the well-known Laplace transform and the Sumudu integral transform.
${\mathbb{M}}$-transform was successfully applied to first order initial-
boundary value problem (see Srivastava et al. [36]). The
${\mathbb{M}}$-transform is defined as:

${\mathbb{M}_{\rho,m}}[f(t)](u,v)=\int_{0}^{\infty}\frac{\exp\left(-ut\right)f(vt)}{\left(t^{m}+v^{m}\right)^{\rho}}dt,$
(1.10)

$\left(\rho\in{\mathbb{C}};\,\Re(\rho)\geq
0,\,m\in{\mathbb{Z}_{+}}={1,2,3,\cdots}\right),$ where both $u\in{\mathbb{C}}$
and $v\in{\mathbb{R}_{+}}$ are the ${\mathbb{M}}$-transform variables.

In 2013, Atangana and Kilicman introduced a novel integral transform called
the Abdon-Kilicman integral transform [37] for solving some differential
equations with some kind of singularities. The novel integral transform is
defined as:

$M_{n}(s)=M_{n}[f(x)](s)=\int_{0}^{\infty}x^{n}\exp\left(-xs\right)f(x)dx.$
(1.11)

The Atangana-Kilicman integral becomes Laplace transform when $n=0$. Recently,
a Laplace-type integral transform called the Yang transform ([38]-[40]) for
solving steady heat transfer problems was introduced in 2016. The integral
transform is defined as:

$Y[\phi(\tau)]=\phi(\omega)=\int_{0}^{\infty}\exp\left(\frac{-\tau}{\omega}\right)\phi(\tau)d\tau,$
(1.12)

provided the integral exists for some $\omega$.

Due to the rapid development in the physical science and engineering models,
there are many other integral transforms in the literature. However, most of
the existing integral transforms have some limitations and cannot be used
directly to solved nonlinear problems or many complex mathematical models. As
a result, many authors became highly interested to come up with the
alternative approach for solving many real-life problems. In 2016, Atangana
and Alkaltani introduced a new double integral equation and their properties
based on the Laplace transform and decomposition method. The double integral
transform was successfully applied to second order partial differential
equation with singularity called the two-dimensional Mboctara equation [41].
Recently, Eltayeb applied double Laplace decomposition method to nonlinear
partial differential equations [42]. In 2017, Belgacem el at. extended the
applications of the natural and the Sumudu transforms to fractional diffusion
and Stokes fluid flow realms [43].

Motivated by the above-mentioned researches, in this paper we proposed a
Laplace-type integral transform called Shehu transform for solving both
ordinary and partial differential equations. The Laplace-type integral
transform converges to Laplace transform when $u=1$, and to Yang integral
transform when $s=1$. The proposed integral transform is successfully applied
to both ordinary and partial differential equations. All the results obtained
in the applications section can easily be verified using the Laplace or
Fourier integral transforms. Throughout this paper, the Shehu transform is
denoted by an operator ${\mathbb{S}[.]}$.

##  2\. Main result

######  Definition 1.

The Shehu transform of the function $v(t)$ of exponential order is defined
over the set of functions,
$A=\left\\{v(t):\exists\,\,N,\,\eta_{1},\,\eta_{2}>0,\,\,\left|v(t)\right|<N\exp\left(\frac{\left|t\right|}{\eta_{i}}\right),\,\,\textnormal{if}\,\,t\in(-1)^{i}\times\left[0,\infty\right)\right\\}$,
by the following integral

$\displaystyle{\mathbb{S}}\left[v(t)\right]$ $\displaystyle=$ $\displaystyle
V(s,u)=\int_{0}^{\infty}\exp\left(\frac{-st}{u}\right)v(t)dt$ (2.1)
$\displaystyle=$
$\displaystyle\lim_{\alpha\rightarrow\infty}\int_{0}^{\alpha}\exp\left(\frac{-st}{u}\right)v(t)dt;\,\,\,\,s\,>0,\,\,u>0.$

It converges if the limit of the integral exists, and diverges if not.

The inverse Shehu transform is given by

${\mathbb{S}^{-1}}\left[V(s,u)\right]=v(t),\,\,\,for\,\,\,t\geq 0.$ (2.2)

Equivalently

$v(t)={\mathbb{S}^{-1}}\left[V(s,u)\right]=\frac{1}{2\pi
i}\int_{\alpha-i\infty}^{\alpha+i\infty}\frac{1}{u}\exp\left(\frac{st}{u}\right)V(s,u)\,ds,$
(2.3)

where $s$ and $u$ are the Shehu transform variables, and $\alpha$ is a real
constant and the integral in Equ.(2.3) is taken along $s=\alpha$ in the
complex plane $s=x+iy$.

######  Theorem 1.

The sufficient condition for the existence of Shehu transform. If the function
$v(t)$ is piecewise continues in every finite interval $0\leq t\leq\beta$ and
of exponential order $\alpha$ for $t>\beta$. Then its Shehu transform $V(s,u)$
exists.

Proof. For any positive number $\beta$, we algebraically deduce

$\int_{0}^{\infty}\exp\left(-\frac{st}{u}\right)v(t)dt=\int_{0}^{\beta}\exp\left(-\frac{st}{u}\right)v(t)dt+\int_{\beta}^{\infty}\exp\left(-\frac{st}{u}\right)v(t)dt.$
(2.4)

Since the function $v(t)$ is piecewise continues in every finite interval
$0\leq t\leq\beta$, then the first integral on the right hand side exists.
Besides, the second integral on the right hand side exists, since the function
$v(t)$ is of exponential order $\alpha$ for $t>\beta$. To verify this claim,
we consider the following case

$\displaystyle\left|\int_{\beta}^{\infty}\exp\left(-\frac{st}{u}\right)v(t)dt\right|$
$\displaystyle\leq$
$\displaystyle\int_{\beta}^{\infty}\left|\exp\left(-\frac{st}{u}\right)v(t)\right|dt$
$\displaystyle\leq$
$\displaystyle\int_{0}^{\infty}\exp\left(-\frac{st}{u}\right)\left|v(t)\right|dt$
$\displaystyle\leq$
$\displaystyle\int_{\beta}^{\infty}\exp\left(-\frac{st}{u}\right)N\exp\left(\alpha
t\right)dt$ $\displaystyle=$ $\displaystyle
N\int_{\beta}^{\infty}\exp\left(-\frac{(s-\alpha u)t}{u}\right)dt$
$\displaystyle=$ $\displaystyle-\frac{uN}{(s-\alpha
u)}\lim_{\gamma\rightarrow\infty}\left[\exp\left(-\frac{(s-\alpha
u)t}{u}\right)\right]^{\gamma}_{0}$ $\displaystyle=$
$\displaystyle\frac{uN}{s-\alpha u}.$

The proof is complete. $\Box$

######  Property 1.

Linearity property of Shehu transform. Let the functions $\alpha v(t)$ and
$\beta w(t)$ be in set A, then $\left(\alpha v(t)+\beta w(t)\right)\in$A,
where $\alpha$ and $\beta$ are nonzero arbitrary constants, and

${\mathbb{S}}\left[\alpha v(t)+\beta
w(t)\right]=\alpha{\mathbb{S}}\left[v(t)\right]+\beta{\mathbb{S}}\left[w(t)\right].$
(2.5)

Proof. Using the Definition 1 of Shehu transform, we get

$\displaystyle{\mathbb{S}}\left[\alpha v(t)+\beta w(t)\right]$
$\displaystyle=$
$\displaystyle\int_{0}^{\infty}\exp\left(\frac{-st}{u}\right)(\alpha
v(t)+\beta w(t))dt$ $\displaystyle=$
$\displaystyle\int_{0}^{\infty}\exp\left(\frac{-st}{u}\right)(\alpha
v(t))\,dt+\int_{0}^{\infty}\exp\left(\frac{-st}{u}\right)(\beta w(t))dt$
$\displaystyle=$
$\displaystyle\alpha\int_{0}^{\infty}\exp\left(\frac{-st}{u}\right)v(t)dt+\beta\int_{0}^{\infty}\exp\left(\frac{-st}{u}\right)w(t)dt$
$\displaystyle=$ $\displaystyle\alpha
u\int_{0}^{\infty}\exp\left(-st\right)v(ut)dt+\beta
u\int_{0}^{\infty}\exp\left(-st\right)w(ut)dt$ $\displaystyle=$
$\displaystyle\alpha{\mathbb{S}}\left[v(t)\right]+\beta{\mathbb{S}}\left[w(t)\right].$

The proof is complete. $\Box$

In particular, using the Definition 1 and Property 1, we obtain

${\mathbb{S}}\left[3\cos(t)+5\sin(2t)\right]=3{\mathbb{S}}\left[\cos(t)\right]+5{\mathbb{S}}\left[\sin(2t)\right]$
$=\frac{3us}{s^{2}+u^{2}}+\frac{5u^{2}}{s^{2}+(2u)^{2}},$

see entries of table 1.

######  Property 2.

Change of scale property of Shehu transform. Let the function $v(\beta t)$ be
in set A, where $\beta$ is an arbitrary constant. Then

${\mathbb{S}}\left[v(\beta
t)\right]=\frac{u}{\beta}V\left(\frac{s}{\beta},u\right).$ (2.6)

Proof. Using the Definition 1 of Shehu transform, we deduce

$\displaystyle{\mathbb{S}}\left[v(\beta t)\right]$ $\displaystyle=$
$\displaystyle\int_{0}^{\infty}\exp\left(\frac{-st}{u}\right)v(\beta t)dt$
(2.7)

Substituting $\eta=\beta t$ which implies $t=\frac{\eta}{\beta}$ and
$dt=\frac{d\eta}{\beta}$ in Equ.(2.7) yields

$\displaystyle{\mathbb{S}}\left[v(\beta t)\right]$ $\displaystyle=$
$\displaystyle\frac{1}{\beta}\int_{0}^{\infty}\exp\left(\frac{-s\eta}{u\beta}\right)v(\eta)d\eta$
$\displaystyle=$
$\displaystyle\frac{1}{\beta}\int_{0}^{\infty}\exp\left(\frac{-st}{u\beta}\right)v\left(t\right)dt$
$\displaystyle=$
$\displaystyle\frac{u}{\beta}\int_{0}^{\infty}\exp\left(\frac{-st}{\beta}\right)v(ut)dt$
$\displaystyle=$
$\displaystyle\frac{u}{\beta}V\left(\frac{s}{\beta},u\right).$

The proof is complete. $\Box$

######  Theorem 2.

Derivative of Shehu transform. If the function $v^{(n)}(t)$ is the $nth$
derivative of the function $v(t)\in A$ with respect to
${}^{\prime}t^{\prime}$, then its Shehu transform is defined by

${\mathbb{S}}\left[v^{(n)}(t)\right]=\frac{s^{n}}{u^{n}}V(s,u)-\sum_{k=0}^{n-1}\left(\frac{s}{u}\right)^{n-(k+1)}v^{(k)}(0).$
(2.8)

When n=1, 2, and 3 in Equ. (2.8) above, we obtain the following derivatives
with respect to $t$.

${\mathbb{S}}\left[v^{\prime}(t)\right]=\frac{s}{u}V(s,u)-v(0).$ (2.9)
${\mathbb{S}}\left[v^{\prime\prime}(t)\right]=\frac{s^{2}}{u^{2}}V(s,u)-\frac{s}{u}v(0)-v^{\prime}(0).$
(2.10)
${\mathbb{S}}\left[v^{\prime\prime\prime}(t)\right]=\frac{s^{3}}{u^{3}}V(s,u)-\frac{s^{2}}{u^{2}}v(0)-\frac{s}{u}v^{\prime}(0)-v^{\prime\prime}(0).$
(2.11)

Proof. Now suppose Equ. (2.8) is true for $n=k$. Then using Equ. (2.9) and the
induction hypothesis, we deduce

$\displaystyle{\mathbb{S}}\left[(v^{(k)}(t))^{\prime}\right]$ $\displaystyle=$
$\displaystyle\frac{s}{u}{\mathbb{S}}\left[v^{(k)}(t)\right]-v^{(k)}(0)$
$\displaystyle=$
$\displaystyle\frac{s}{u}\left[\frac{s^{k}}{u^{k}}{\mathbb{S}}\left[v(t)\right]-\sum_{i=0}^{k-1}\left(\frac{s}{u}\right)^{k-(i+1)}v^{(i)}(0)\right]-v^{(k)}(0)$
$\displaystyle=$
$\displaystyle\left(\frac{s}{u}\right)^{k+1}{\mathbb{S}}\left[v(t)\right]-\sum_{i=0}^{k}\left(\frac{s}{u}\right)^{k-i}v^{(i)}(0),$

which implies that Equ. (2.8) holds for $n=k+1$. By induction hypothesis the
proof is complete. $\Box$
The following important properties are obtain using the Leibniz’s rule

$\displaystyle{\mathbb{S}}\left[\frac{\partial v(x,t)}{\partial
x}\right]=\int_{0}^{\infty}\exp\left(\frac{-st}{u}\right)\frac{\partial
v(x,t)}{\partial x}\,dt=\frac{\partial}{\partial
x}\int_{0}^{\infty}\exp\left(\frac{-st}{u}\right)v(x,t)\,dt$
$\displaystyle=\frac{\partial}{\partial
x}\left[V(x,s,u)\right]\Rightarrow{\mathbb{S}}\left[\frac{\partial
v(x,t)}{\partial x}\right]=\frac{d}{dx}\left[V(x,s,u)\right],$
$\displaystyle{\mathbb{S}}\left[\frac{\partial^{2}v(x,t)}{\partial
x^{2}}\right]=\int_{0}^{\infty}\exp\left(\frac{-st}{u}\right)\frac{\partial^{2}v(x,t)}{\partial
x^{2}}\,dt=\frac{\partial^{2}}{\partial
x^{2}}\int_{0}^{\infty}\exp\left(\frac{-st}{u}\right)v(x,t)\,dt$
$\displaystyle=\frac{\partial^{2}}{\partial
x^{2}}\left[V(x,s,u)\right]\Rightarrow{\mathbb{S}}\left[\frac{\partial^{2}v(x,t)}{\partial
x^{2}}\right]=\frac{d^{2}}{dx^{2}}\left[V(x,s,u)\right],$

and

$\displaystyle{\mathbb{S}}\left[\frac{\partial^{n}v(x,t)}{\partial
x^{n}}\right]=\int_{0}^{\infty}\exp\left(\frac{-st}{u}\right)\frac{\partial^{n}v(x,t)}{\partial
x^{n}}\,dt=\frac{\partial^{n}}{\partial
x^{n}}\int_{0}^{\infty}\exp\left(\frac{-st}{u}\right)v(x,t)dt$
$\displaystyle=\frac{\partial^{n}}{\partial
x^{n}}\left[V(x,s,u)\right]\Rightarrow{\mathbb{S}}\left[\frac{\partial^{n}v(x,t)}{\partial
x^{n}}\right]=\frac{d^{n}}{dx^{n}}\left[V(x,s,u)\right].$

##  3\. Some useful results of Shehu transform

######  Property 3.

Let the function $v(t)=1$ be in set A. Then its Shehu transform is given by

${\mathbb{S}}\left[1\right]=\frac{u}{s}.$ (3.1)

Proof. Using Equ.(2.1), we deduce

${\mathbb{S}}\left[1\right]=\int_{0}^{\infty}\exp\left(\frac{-st}{u}\right)dt=-\frac{u}{s}\lim_{\gamma\rightarrow\infty}\left[\exp\left(\frac{-st}{u}\right)\right]^{\gamma}_{0}=\frac{u}{s}.$

This ends the proof. $\Box$

######  Property 4.

Let the function $v(t)=t$ be in set A. Then its Shehu transform is given by

${\mathbb{S}}\left[t\right]=\frac{u^{2}}{s^{2}}.$ (3.2)

Proof. Using the Definition 1 of the Shehu transform and integration by parts,
we get

$\displaystyle{\mathbb{S}}\left[t\right]$ $\displaystyle=$
$\displaystyle\int_{0}^{\infty}t\exp\left(\frac{-st}{u}\right)dt=\frac{u}{s}\lim_{\gamma\rightarrow\infty}\left[t\exp\left(\frac{-st}{u}\right)\right]^{\gamma}_{0}+\frac{u}{s}\int_{0}^{\infty}\exp\left(\frac{-st}{u}\right)dt$
$\displaystyle=-\frac{u^{2}}{s^{2}}\lim_{\gamma\rightarrow\infty}\left[\exp\left(\frac{-st}{u}\right)\right]^{\gamma}_{0}=\frac{u^{2}}{s^{2}}.$

Thus the proof ends. $\Box$

######  Property 5.

Let the function $v(t)=\frac{t^{n}}{n!}\,\,\,n=0,1,2..$ be in set A. Then its
Shehu transform is given by

${\mathbb{S}}\left[\frac{t^{n}}{n!}\right]=\left(\frac{u}{s}\right)^{n+1}.$
(3.3)

Proof. From the Definition 1 of the Shehu transform and integration by parts,
we deduce

$\displaystyle{\mathbb{S}}\left[t^{n}\right]=\int_{0}^{\infty}t^{n}\exp\left(\frac{-st}{u}\right)dt=\frac{u}{s}n\int_{0}^{\infty}t^{n-1}\exp\left(\frac{-st}{u}\right)dt$
$\displaystyle=\frac{u^{2}}{s^{2}}n(n-1)\int_{0}^{\infty}t^{n-2}\exp\left(\frac{-st}{u}\right)dt$
$\displaystyle=$
$\displaystyle\frac{u^{3}}{s^{3}}n(n-1)(n-2)\int_{0}^{\infty}t^{n-3}\exp\left(\frac{-st}{u}\right)dt$
$\displaystyle=\frac{u^{4}}{s^{4}}n(n-1)(n-2)(n-3)\int_{0}^{\infty}t^{n-4}\exp\left(\frac{-st}{u}\right)dt$
$\displaystyle=$
$\displaystyle\frac{u^{5}}{s^{5}}n(n-1)(n-2)(n-3)(n-4)\int_{0}^{\infty}t^{n-5}\exp\left(\frac{-st}{u}\right)dt=\cdots=n!\left(\frac{u}{s}\right)^{n+1}.$

The proof is completed. $\Box$

######  Property 6.

Let the function $v(t)=\frac{t^{n}}{\Gamma(n+1)}\,\,\,n=0,1,2,\cdots$ be in
set A. Then its Shehu transform is given by

${\mathbb{S}}\left[\frac{t^{n}}{\Gamma(n+1)}\right]=\left(\frac{u}{s}\right)^{n+1}.$
(3.4)

The proof of property 6 follows immediately from the previous property 5.
$\Box$

######  Property 7.

Let the function $v(t)=\exp(\alpha t)$ be in A. Then its Shehu transform is
given by

${\mathbb{S}}\left[\exp(\alpha t)\right]=\frac{u}{s-\alpha u}.$ (3.5)

Proof. Using Equ.(2.1), we get

$\displaystyle{\mathbb{S}}\left[\exp(\alpha t)\right]$ $\displaystyle=$
$\displaystyle\int_{0}^{\infty}\exp\left(-\frac{(s-\alpha u)t}{u}\right)dt$
$\displaystyle=-\frac{u}{s-\alpha
u}\lim_{\gamma\rightarrow\infty}\left[\exp\left(-\frac{(s-\alpha
u)t}{u}\right)\right]^{\gamma}_{0}=\frac{u}{s-\alpha u}.$

This ends the proof. $\Box$

######  Property 8.

Let the function $v(t)=t\exp(\alpha t)$ be in set A. Then its Shehu transform
is given by

${\mathbb{S}}\left[t\exp(\alpha t)\right]=\frac{u^{2}}{(s-\alpha u)^{2}}.$
(3.6)

Proof. Using the Definition 1 of the Shehu transform and integration by parts,
we get

$\displaystyle{\mathbb{S}}\left[t\exp(\alpha
t)\right]=\int_{0}^{\infty}t\exp\left(-\frac{(s-\alpha u)t}{u}\right)dt$
$\displaystyle=-\frac{u}{s-\alpha
u}\lim_{\gamma\rightarrow\infty}\left[t\exp\left(-\frac{(s-\alpha
u)t}{u}\right)\right]^{\gamma}_{0}$ $\displaystyle+\frac{u}{s-\alpha
u}\int_{0}^{\infty}\exp\left(-\frac{(s-\alpha u)t}{u}\right)dt$
$\displaystyle=-\frac{u^{2}}{(s-\alpha
u)^{2}}\lim_{\gamma\rightarrow\infty}\left[\exp\left(-\frac{(s-\alpha
u)t}{u}\right)\right]^{\gamma}_{0}=\frac{u^{2}}{(s-\alpha u)^{2}}.$

The proof is complete. $\Box$

######  Property 9.

Let the function $v(t)=\frac{t^{n}\exp(\alpha t)}{n!}\,\,\,n=0,1,2,...$ be in
set A. Then its Shehu transform is given by

${\mathbb{S}}\left[\frac{t^{n}\exp(\alpha
t)}{n!}\right]=\frac{u^{n+1}}{(s-\alpha u)^{n+1}}.$ (3.7)

Proof. Using the Definition 1 of the Shehu transform and integration by parts,
we deduce

$\displaystyle{\mathbb{S}}\left[t^{n}\exp(\alpha
t)\right]=\int_{0}^{\infty}t^{n}\exp\left(-\frac{(s-\alpha u)t}{u}\right)dt$
$\displaystyle=\frac{un}{(s-\alpha
u)}\int_{0}^{\infty}t^{n-1}\exp\left(-\frac{(s-\alpha u)t}{u}\right)dt$
$\displaystyle=$ $\displaystyle\frac{u^{2}n(n-1)}{(s-\alpha
u)^{2}}\int_{0}^{\infty}t^{n-2}\exp\left(-\frac{(s-\alpha
u)t}{u}\right)dt=\cdots=\frac{n!}{(s-\alpha u)^{n+1}}.$

Thus the proof is complete. $\Box$

######  Property 10.

Let the function $v(t)=\frac{t^{n}}{\Gamma(n+1)}\exp(\alpha
t)\,\,\,n=0,1,2,...$ be in set A. Then its Shehu transform is given by

${\mathbb{S}}\left[\frac{t^{n}\exp(\alpha
t)}{\Gamma(n+1)}\right]=\frac{u^{n+1}}{(s-\alpha u)^{n+1}}.$ (3.8)

The proof of Property 10 follows as a direct consequence of Property 9. $\Box$

######  Property 11.

Let the function $v(t)=sin(\alpha t)$ be in set A. Then its Shehu transform is
given by

${\mathbb{S}}\left[\sin(\alpha t)\right]=\frac{\alpha
u^{2}}{s^{2}+\alpha^{2}u^{2}}.$ (3.9)

Proof. Using the Definition 1 of the Shehu transform and integration by parts,
we get

$\displaystyle{\mathbb{S}}\left[\sin(\alpha
t)\right]=\int_{0}^{\infty}\exp\left(-\frac{st}{u}\right)\sin(\alpha t)dt$
$\displaystyle=-\frac{u}{s}\lim_{\gamma\rightarrow\infty}\left[\exp\left(-\frac{st}{u}\right)\sin(\alpha
t)\right]^{\gamma}_{0}+\frac{u\alpha}{s}\int_{0}^{\infty}\exp\left(-\frac{st}{u}\right)\cos(\alpha
t)dt$ $\displaystyle=$ $\displaystyle-\frac{\alpha
u^{2}}{s^{2}}\lim_{\gamma\rightarrow\infty}\left[\exp\left(-\frac{st}{u}\right)\cos(\alpha
t)\right]^{\gamma}_{0}-\frac{\alpha^{2}u^{2}}{s^{2}}\int_{0}^{\infty}\exp\left(-\frac{st}{u}\right)\sin(\alpha
t)dt$ $\displaystyle=$ $\displaystyle\frac{\alpha
u^{2}}{s^{2}}-\frac{\alpha^{2}u^{2}}{s^{2}}\int_{0}^{\infty}\exp\left(-\frac{st}{u}\right)\sin(\alpha
t)dt.$

Simplifying the required integrals complete the proof of Property 11\. $\Box$

######  Property 12.

Let the function $v(t)=\cos(\alpha t)$ be in set A. Then its Shehu transform
is given by

${\mathbb{S}}\left[\cos(\alpha t)\right]=\frac{us}{s^{2}+\alpha^{2}u^{2}}.$
(3.10)

Proof. Using the Definition 1 of the Shehu transform and integration by parts,
we deduce

$\displaystyle{\mathbb{S}}\left[\cos(\alpha
t)\right]=\int_{0}^{\infty}\exp\left(-\frac{st}{u}\right)\cos(\alpha t)dt$
$\displaystyle=-\frac{u}{s}\lim_{\gamma\rightarrow\infty}\left[\exp\left(-\frac{st}{u}\right)\cos(\alpha
t)\right]^{\gamma}_{0}-\frac{\alpha
u}{s}\int_{0}^{\infty}\exp\left(-\frac{st}{u}\right)\sin(\alpha t)dt$
$\displaystyle=$ $\displaystyle\frac{u}{s}-\frac{\alpha
u^{2}}{s^{2}}\lim_{\gamma\rightarrow\infty}\left[\exp\left(-\frac{st}{u}\right)\sin(\alpha
t)\right]^{\gamma}_{0}-\frac{\alpha^{2}u^{2}}{s^{2}}\int_{0}^{\infty}\exp\left(-\frac{st}{u}\right)\cos(\alpha
t)dt$ $\displaystyle=$
$\displaystyle\frac{u}{s}-\frac{\alpha^{2}u^{2}}{s^{2}}\int_{0}^{\infty}\exp\left(-\frac{st}{u}\right)\cos(\alpha
t)dt.$

Simplifying the required integrals complete the proof of Property 12\. $\Box$

######  Property 13.

Let the function $v(t)=\frac{\sinh(\alpha t)}{\alpha}$ be in set A. Then its
Shehu transform is given by

${\mathbb{S}}\left[\frac{\sinh(\alpha
t)}{\alpha}\right]=\frac{u^{2}}{s^{2}-\alpha^{2}u^{2}}.$ (3.11)

Proof. From the Definition 1 of the Shehu transform and integration by parts,
we get

$\displaystyle{\mathbb{S}}\left[\sinh(\alpha
t)\right]=\int_{0}^{\infty}\exp\left(-\frac{st}{u}\right)\sinh(\alpha t)dt$
$\displaystyle=-\frac{u}{s}\lim_{\gamma\rightarrow\infty}\left[\exp\left(-\frac{st}{u}\right)\sinh(\alpha
t)\right]^{\gamma}_{0}+\frac{u\alpha}{s}\int_{0}^{\infty}\exp\left(-\frac{st}{u}\right)\cosh(\alpha
t)dt$ $\displaystyle=$ $\displaystyle-\frac{\alpha
u^{2}}{s^{2}}\lim_{\gamma\rightarrow\infty}\left[\exp\left(-\frac{st}{u}\right)\cos(\alpha
t)\right]^{\gamma}_{0}+\frac{\alpha^{2}u^{2}}{s^{2}}\int_{0}^{\infty}\exp\left(-\frac{st}{u}\right)\sinh(\alpha
t)dt$ $\displaystyle=$ $\displaystyle\frac{\alpha
u^{2}}{s^{2}}+\frac{\alpha^{2}u^{2}}{s^{2}}\int_{0}^{\infty}\exp\left(-\frac{st}{u}\right)\sinh(\alpha
t)dt.$

Simplifying the required integrals complete the proof of Property 13\. $\Box$

######  Property 14.

Let the function $v(t)=\cosh(\alpha t)$ be in set A. Then its Shehu transform
is given by

${\mathbb{S}}\left[\cosh(\alpha t)\right]=\frac{us}{s^{2}-\alpha^{2}u^{2}}.$
(3.12)

Proof. Applying the Definition 1 of the Shehu transform and integration by
parts, we get

$\displaystyle{\mathbb{S}}\left[\cosh(\alpha
t)\right]=\int_{0}^{\infty}\exp\left(-\frac{st}{u}\right)\cosh(\alpha t)dt$
$\displaystyle=-\frac{u}{s}\lim_{\gamma\rightarrow\infty}\left[\exp\left(-\frac{st}{u}\right)\cos(\alpha
t)\right]^{\gamma}_{0}+\frac{\alpha
u}{s}\int_{0}^{\infty}\exp\left(-\frac{st}{u}\right)\sinh(\alpha t)dt$
$\displaystyle=$ $\displaystyle\frac{u}{s}-\frac{\alpha
u^{2}}{s}\lim_{\gamma\rightarrow\infty}\left[\exp\left(-\frac{st}{u}\right)\sinh(\alpha
t)\right]^{\gamma}_{0}+\frac{\alpha^{2}u^{2}}{s^{2}}\int_{0}^{\infty}\exp\left(-\frac{st}{u}\right)\cos(\alpha
t)dt$ $\displaystyle=$
$\displaystyle\frac{u}{s}+\frac{\alpha^{2}u^{2}}{s^{2}}\int_{0}^{\infty}\exp\left(-\frac{st}{u}\right)\cosh(\alpha
t)dt.$

Collecting the required integrals complete the proof of Property 14\. $\Box$

######  Property 15.

Let the function $\frac{\exp\left(\beta t\right)\sin(\alpha t)}{\alpha}$ be in
set A. Then its Shehu transform is given by

${\mathbb{S}}\left[\frac{\exp\left(\beta t\right)\sin(\alpha
t)}{\alpha}\right]=\frac{u^{2}}{(s-\beta u)^{2}+\alpha^{2}u^{2}}.$ (3.13)

Proof. Using the Definition 1 of the Shehu transform and integration by parts,
we deduce

$\displaystyle{\mathbb{S}}\left[\exp\left(\beta t\right)\sin(\alpha t)\right]$
$\displaystyle=$ $\displaystyle\int_{0}^{\infty}\exp\left(-\frac{(s-\beta
u)}{u}t\right)\sin(\alpha t)dt$ $\displaystyle=\frac{-u}{(s-\beta
u)}\lim_{\gamma\rightarrow\infty}\left[\exp\left(-\frac{(s-\beta
u)}{u}t\right)\sin(\alpha t)dt\right]^{\gamma}_{0}$
$\displaystyle+\frac{u\alpha}{s-\beta
u}\int_{0}^{\infty}\exp\left(-\frac{(s-\beta u)}{u}t\right)\cos(\alpha t)dt$
$\displaystyle=-\frac{u^{2}\alpha}{(s-\beta
u)^{2}}\lim_{\gamma\rightarrow\infty}\left[\exp\left(-\frac{(s-\beta
u)}{u}t\right)\cos(\alpha t)dt\right]^{\gamma}_{0}$
$\displaystyle-\frac{\alpha u^{2}\alpha^{2}}{(s-\beta
u)^{2}}\int_{0}^{\infty}\exp\left(-\frac{(s-\beta u)}{u}t\right)\sin(\alpha
t)dt$ $\displaystyle=$ $\displaystyle\frac{u^{2}\alpha}{(s-\beta
u)^{2}}-\frac{u^{2}\alpha^{2}}{(s-\beta
u)^{2}}\int_{0}^{\infty}\exp\left(-\frac{(s-\beta u)}{u}t\right)\sin(\alpha
t)dt.$

Simplifying the required integrals complete the proof of property 15\. This
ends the proof. $\Box$

######  Property 16.

Let the function $\exp\left(\beta t\right)\cos(\alpha t)$ be in set A. Then
its Shehu transform is given by

${\mathbb{S}}\left[\exp\left(\beta t\right)\cos(\alpha
t)\right]=\frac{u(s-\alpha u)}{(s-\beta u)^{2}+\alpha^{2}u^{2}}.$ (3.14)

Proof. Applying the Definition 1 of the Shehu transform and integration by
parts, we get

$\displaystyle{\mathbb{S}}\left[\exp\left(\beta t\right)\cos(\alpha t)\right]$
$\displaystyle=$ $\displaystyle\int_{0}^{\infty}\exp\left(-\frac{(s-\beta
u)}{u}t\right)\cos(\alpha t)dt$ $\displaystyle=-\frac{u}{s-\beta
u}\lim_{\gamma\rightarrow\infty}\left[\exp\left(-\frac{(s-\beta
u)}{u}t\right)\cos(\alpha t)\right]^{\gamma}_{0}$ $\displaystyle+\frac{\alpha
u}{s-\beta u}\int_{0}^{\infty}\exp\left(-\frac{(s-\beta
u)}{u}t\right)\sin(\alpha t)dt$ $\displaystyle=$
$\displaystyle\frac{u}{s-\beta u}+\frac{\alpha u}{s-\beta
u}\int_{0}^{\infty}\exp\left(-\frac{(s-\beta u)}{u}t\right)\sin(\alpha t)dt$
$\displaystyle=$ $\displaystyle\frac{u}{s-\beta u}+\frac{\alpha
u^{2}}{(s-\alpha
u)^{2}}\lim_{\gamma\rightarrow\infty}\left[\exp\left(-\frac{(s-\beta
u)}{u}t\right)\sin(\alpha t)\right]^{\gamma}_{0}$
$\displaystyle-\frac{\alpha^{2}u^{2}}{(s-\beta
u)^{2}}\int_{0}^{\infty}\exp\left(-\frac{(s-\beta u)}{u}t\right)\cos(\alpha
t)dt$ $\displaystyle=$ $\displaystyle\frac{u}{s-\beta
u}-\frac{\alpha^{2}u^{2}}{(s-\beta
u)^{2}}\int_{0}^{\infty}\exp\left(-\frac{(s-\beta u)}{u}t\right)\cos(\alpha
t)dt.$

Simplifying the required integrals complete the proof of property 16\. $\Box$

######  Property 17.

Let the function $\frac{\exp\left(\beta t\right)-\exp\left(\alpha
t\right)}{\beta-\alpha}$ be in set A. Then its Shehu transform is given by

${\mathbb{S}}\left[\frac{\exp\left(\alpha
t\right)}{\beta-\alpha}\right]=\frac{u^{2}}{(s-\alpha u)(s-\beta u)}.$ (3.15)

Proof. Using the definition of Shehu transform, we get

$\displaystyle{\mathbb{S}}\left[\frac{\exp\left(\alpha
t\right)}{\beta-\alpha}\right]=\frac{u}{\beta-\alpha}\int_{0}^{\infty}\exp\left(\frac{-st}{u}\right)\left(\exp(\beta
t)-\exp\left(\alpha t\right)\right)dt$
$\displaystyle=\frac{1}{\beta-\alpha}\int_{0}^{\infty}e^{\frac{(\beta
u-s)}{u}t}dt-\frac{1}{\beta-\alpha}\int_{0}^{\infty}\exp\left(\frac{(\alpha
u-s)t}{u}\right)dt$ $\displaystyle=\frac{u}{(\beta-\alpha)(\beta
u-s)}\lim_{\gamma\rightarrow\infty}\left[\exp\left(-\frac{(s-\beta
u)t}{u}\right)\right]^{\gamma}_{0}$
$\displaystyle-\frac{u}{(\beta-\alpha)(\alpha
u-s)}\lim_{\gamma\rightarrow\infty}\left[\exp\left(-\frac{(s-\beta
u)t}{u}\right)\right]^{\gamma}_{0}$ $\displaystyle=$
$\displaystyle-\frac{u}{(\beta-\alpha)(\beta
u-s)}+\frac{u}{(\beta-\alpha)(\alpha u-s)}$ $\displaystyle=$
$\displaystyle\frac{-u(\alpha u-s)+u(\beta u-s)}{(\beta-\alpha)(\alpha
u-s)(\beta u-s)}=\frac{u^{2}}{(s-\alpha u)(s-\beta u)}.$

The proof is complete. $\Box$

######  Property 18.

Let the function $\frac{\beta\exp\left(\beta t\right)-\alpha\exp(\alpha
t)}{\beta-\alpha}$ be in set A. Then its Shehu transform is given by:

${\mathbb{S}}\left[\frac{\beta\exp\left(\beta t\right)-\alpha\exp\left(\alpha
t\right)}{\beta-\alpha}\right]=\frac{us}{(s-\alpha u)(s-\beta u)}.$ (3.16)

Proof:
Using the definition of Shehu transform, we get

$\displaystyle{\mathbb{S}}\left[\frac{\beta\exp\left(\beta
t\right)-\alpha\exp\left(\alpha
t\right)}{\beta-\alpha}\right]=\frac{1}{\beta-\alpha}\int_{0}^{\infty}\exp\left(\frac{-st}{u}\right)\left(\beta\exp\left(\beta
t\right)-\alpha\exp\left(\alpha t\right)\right)dt$
$\displaystyle=\frac{\beta}{\beta-\alpha}\int_{0}^{\infty}\exp\left(\frac{(\beta
u-s)}{u}t\right)dt-\frac{\alpha}{\beta-\alpha}\int_{0}^{\infty}\exp\left(\frac{(\alpha
u-s)t}{u}\right)dt$ $\displaystyle=\frac{u\beta}{(\beta-\alpha)(\beta
u-s)}\lim_{\gamma\rightarrow\infty}\left[\exp\left(-\frac{(s-\beta
u)t}{u}\right)\right]^{\gamma}_{0}$
$\displaystyle-\frac{u\alpha}{(\beta-\alpha)(\alpha
u-s)}\lim_{\gamma\rightarrow\infty}\left[\exp\left(-\frac{(s-\alpha
u)t}{u}\right)\right]^{\gamma}_{0}$ $\displaystyle=$
$\displaystyle-\frac{u\beta}{(\beta-\alpha)(\beta
u-s)}+\frac{u\alpha}{(\beta-\alpha)(\alpha u-s)}$ $\displaystyle=$
$\displaystyle\frac{-u\beta(\alpha u-s)+u\alpha(\beta
u-s)}{(\beta-\alpha)(\alpha u-s)(\beta u-s)}=\frac{us}{(s-\alpha u)(s-\beta
u)}.$

This ends the proof. $\Box$

More properties of the Shehu transform and their converges to the natural
transform, the Sumudu transform, and the Laplace transform are presented in
table 1. The comprehensive summary of Shehu transform properties are presented
in table 2.

##  4\. Applications

In this section, the applications of the proposed transform are presented. The
simplicity, efficiency and high accuracy of the Shehu transform are clearly
illustrated.

######  Example 1.

Consider the following first order ordinary differential equation

$\frac{dv(t)}{dt}+v(t)=0,$ (4.1)

subject to the initial condition

$v(0)=1.$ (4.2)

Applying the Shehu transform on both sides of Equ. (4.1), we get

$\frac{s}{u}V(s,u)-v(0)+V(s,u)=0.$ (4.3)

Substituting the given initial condition and simplifying, we deduce

$V(s,u)=\frac{u}{s+u}$ (4.4)

Taking the inverse Shehu transform of Equ. (4.4), yields

$v(t)=\exp(-t).$ (4.5)

######  Example 2.

Consider the following second order ordinary differential equation

$\frac{d^{2}v(t)}{dt^{2}}+\frac{dv(t)}{dt}=1$ (4.6)

subject to the initial conditions

$v(0)=0,\,\,\,\,\frac{dv(0)}{dt}=0.$ (4.7)

Applying the Shehu transform on both sides of Equ. (4.6), we obtain

$\frac{s^{2}}{u^{2}}V(s,u)-\frac{s}{u}v(0)-v^{\prime}(0)+\frac{s}{u}V(s,u)-v(0)=\frac{u}{s}.$
(4.8)

Substituting the given initial conditions and simplifying, we deduce

$V(s,u)=-\frac{u}{s}+\frac{u^{2}}{s^{2}}+\frac{u}{s+u}.$ (4.9)

Taking the inverse Shehu transform of Equ. (4.9), we get

$v(t)=-1+t+\exp(-t).$ (4.10)

######  Example 3.

Consider the following second nonhomogeneous order ordinary differential
equation

$\frac{d^{2}v(t)}{dt^{2}}-3\frac{dv(t)}{dt}+2v(t)=\exp(3t)$ (4.11)

subject to the initial conditions

$v(0)=1,\,\,\,\,\frac{dv(0)}{dt}=0.$ (4.12)

Applying the Shehu transform on both sides of Equ. (4.11), yields

$\frac{s^{2}}{u^{2}}V(s,u)-\frac{s}{u}v(0)-v^{\prime}(0)-3\left(\frac{s}{u}V(s,u)-v(0)\right)+2V(s,u)=\frac{u}{s-3u}.$
(4.13)

Substituting the given initial conditions and simplifying, we obtain

$V(s,u)=\frac{5}{2}\frac{u}{(s-u)}-2\frac{u}{s-2u}+\frac{1}{2}\frac{u}{(s-3u)}.$
(4.14)

Taking the inverse Shehu transform of Equ. (4.14), we get

$v(t)=\frac{5}{2}\exp(t)-2\exp(2t)+\frac{1}{2}\exp(3t).$ (4.15)

######  Example 4.

Consider the following ordinary differential equation

$\frac{d^{2}v(t)}{dt^{2}}+2\frac{dv(t)}{dt}+5v(t)=\exp(-t)\sin(t)$ (4.16)

subject to the initial conditions

$v(0)=0,\,\,\,\,\frac{dv(0)}{dt}=1.$ (4.17)

Applying the Shehu transform on both sides of Equ. (4.16), we get

$\frac{s^{2}}{u^{2}}V(s,u)-\frac{s}{u}v(0)-v^{\prime}(0)+2\left(\frac{s}{u}V(s,u)-\frac{s}{u}v(0)\right)+5V(s,u)=\frac{u^{2}}{(s+u)^{2}+u^{2}}$
(4.18)

Substituting the given initial conditions and simplifying, we get

$V(s,u)=\frac{1}{3}\frac{u^{2}}{((s+u)^{2}+u^{2})}+\frac{2}{3}\frac{u^{2}}{((s+u)^{2}+(2u)^{2})}$
(4.19)

Taking the inverse Shehu transform of Equ. (4.19), we get

$v(t)=\frac{1}{3}\exp(-t)\sin(t)+\frac{2}{3}\exp(-t)\sin(2t).$ (4.20)

######  Example 5.

Consider the following homogeneous partial differential equation

$\frac{\partial v(x,t)}{\partial t}=\frac{\partial^{2}v(x,t)}{\partial x^{2}}$
(4.21)

subject to the boundary and initial conditions

$v(0,t)=0,\,\,\,v(1,t)=0,\,\,\,\,v(x,0)=3sin(2\pi x).$ (4.22)

Applying the Shehu transform on both sides of Equ. (4.21), we get

$\frac{s}{u}V(x,s,u)-v(x,0)=\frac{d^{2}V(x,s,u)}{dx^{2}}.$ (4.23)

Substituting the given initial condition and simplifying, we get

$\frac{d^{2}V(x,s,u)}{dx^{2}}-\frac{s}{u}V(x,s,u)=-3sin(2\pi x).$ (4.24)

The general solution of Equ. (4.24) can be written as

$V(x,s,u)=V_{h}(x,s,u)+V_{p}(x,s,u),$ (4.25)

where $V_{h}(x,s,u)$ is the solution of the homogeneous part which is given by

$V_{h}(x,s,u)=\alpha_{1}\exp\left(\sqrt{\frac{s}{u}}x\right)+\alpha_{2}\exp\left(-\sqrt{\frac{s}{u}}x\right),$
(4.26)

and $V_{p}(x,s,u)$ is the solution of the nonhomogeneous part which is given
by

$V_{p}(x,s,u)=\beta_{1}sin(2\pi x)+\beta_{2}cos(2\pi x).$ (4.27)

Applying the boundary conditions on Equ. (4.26), we get

$\alpha_{1}+\alpha_{2}=0\,\,\,\,and\,\,\,\,\alpha_{1}\exp\left(\sqrt{\frac{s}{u}}\right)+\alpha_{2}\exp\left(-\sqrt{\frac{s}{u}}\right)=0\Rightarrow
V_{h}(x,s,u)=0,$

since $\alpha_{1}=\alpha_{2}=0$.
Using the method of undetermined coefficients on the nonhomogeneous part, we
get

$V_{p}(x,s,u)=\frac{3u}{s+4\pi^{2}u}\sin(2\pi x),$ (4.28)

Since, $\beta_{1}=\frac{3u}{s+4\pi^{2}u},$ and $\beta_{2}=0$.
Then Equ. (4.25) will become

$V(x,s,u)=\frac{3u}{s+4\pi^{2}u}\sin(2\pi x),$ (4.29)

Taking the inverse Shehu transform of Equ. (4.29), we get

$v(x,t)=3\exp(-4\pi^{2}t)\sin(2\pi x).$ (4.30)

Figure 1. 3D and 2D surfaces of the analytical solution of Equ. (4.21) in the
ranges $-1<x<1$, and $-1<t<1$, when $t=1$.

######  Example 6.

Consider the following nonhomogeneous partial differential equation

$\frac{\partial^{2}v(x,t)}{\partial
t^{2}}=\beta^{2}\frac{\partial^{2}v(x,t)}{\partial x^{2}}+\sin(\pi x)$ (4.31)

subject to the boundary and initial conditions

$v(0,t)=0,\,\,\,v(1,t)=0,\,\,\,\,v(x,0)=0,\,\,\,\frac{\partial
v(x,0)}{\partial t}=0,\,\,\,\beta^{2}=1.$ (4.32)

Applying the Shehu transform on both sides of Equ. (4.31), we get

$\frac{s^{2}}{u^{2}}V(x,s,u)-\frac{s}{u}v(x,0)-v^{\prime}(x,0)=\frac{d^{2}V(x,s,u)}{dx^{2}}+\frac{u}{s}\sin(\pi
x).$ (4.33)

Substituting the given initial condition and simplifying, we get

$\frac{d^{2}V(x,s,u)}{dx^{2}}-\frac{s^{2}}{u^{2}}V(x,s,u)=-\frac{u}{s}\sin(\pi
x).$ (4.34)

The general solution of Equ. (4.34) can be written as

$V(x,s,u)=V_{h}(x,s,u)+V_{p}(x,s,u),$ (4.35)

where $V_{h}(x,s,u)$ is the solution of the homogeneous part which is given by

$V_{h}(x,s,u)=\lambda_{1}\exp\left(\frac{s}{u}x\right)+\lambda_{2}\exp\left(-\frac{s}{u}x\right),$
(4.36)

and $V_{p}(x,s,u)$ is the solution of the nonhomogeneous part which is given
by

$V_{p}(x,s,u)=\eta_{1}\sin(\pi x)+\eta_{2}\cos(\pi x).$ (4.37)

Applying the boundary conditions on Equ. (4.36), we deduce

$\lambda_{1}+\lambda_{2}=0\,\,\,\,and\,\,\,\,\lambda_{1}\exp\left(\frac{s}{u}\right)+\lambda_{2}\exp\left(-\frac{s}{u}\right)=0\Rightarrow
V_{h}(x,s,u)=0,$

since $\lambda_{1}=\lambda_{2}=0$.
Using the method of undetermined coefficients on the nonhomogeneous part, we
get

$V_{p}(x,s,u)=\frac{1}{\pi^{2}}\left(\frac{u}{s}-\frac{us}{s^{2}+u^{2}\pi^{2}}\right)\sin(\pi
x),$ (4.38)

since,

$\eta_{1}=\frac{u^{3}}{s(s^{2}+u^{2}\pi^{2})}=\frac{1}{\pi^{2}}\left(\frac{u}{s}-\frac{us}{s^{2}+u^{2}\pi^{2}}\right),\,\,\,and\,\,\,\,\eta_{2}=0.$

Then Equ. (4.35) will becomes

$V(x,s,u)=\frac{1}{\pi^{2}}\left(\frac{u}{s}-\frac{us}{s^{2}+u^{2}\pi^{2}}\right)\sin(\pi
x).$ (4.39)

Taking the inverse Shehu transform of Equ. (4.39), we get

$v(x,t)=\frac{1}{\pi^{2}}\left(1-\cos(\pi t)\right)\sin(\pi x).$ (4.40)

Figure 2. 3D and 2D surfaces of the analytical solution of Equ. (4.31) in the
ranges $-1<x<1$, and $-1<t<1$.

##  5\. Conclusion

We introduced an efficient Laplace-type integral transform called the Shehu
transform for solving both ordinary and partial differential equations. We
presented its existence and inverse transform. We presented some useful
properties and their applications for solving ordinary and partial
differential equations. We provide a comprehensive list of the Laplace
transform, Sumudu transform, and the natural transform in table 1 to show
their mutual relationship with the Shehu transform. Finally, based on the
mathematical formulations, simplicity and the findings of the proposed
integral transform, we conclude that it is highly efficient because of the
following advantages:

  * •

It is a generalization of the Laplace and the Sumudu integral transforms.

  * •

Its visualization is easier than the Sumudu transform, the natural transform,
and the Elzaki transform.

  * •

The Laplace-type integral transform become Laplace transform when the variable
$u=1$ and the Yang integral transform when the variable $s=1$.

  * •

It can easily be applied directly to some class of ordinary and the partial
differential equations as demonstrated in the application section.

  * •

For advanced research in physical science and engineering, the proposed
integral transform can be considered a stepping-stone to the Sumudu transform,
the natural transform, the Elzaki transform, and the Laplace transform.

##  6\. Acknowledgements

The authors are highly grateful to the editor’s and the anonymous referees’
for their useful comments and suggestions in this paper. This research is
partially supported by the National Natural Science Foundations of China under
Grants No. (11571206). The first author also acknowledges the financial
support of China Scholarship Council (CSC) in Shandong University with grand
(CSC No: 2017GXZ025381).

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## Appendix

Table 1: Here we present a comprehensive list of the Shehu transform of some
special functions and their relationship with the natural transform
${\mathbb{N}}\left[v(t)\right]$, the Sumudu transform
$\textbf{S}\left[v(t)\right]$, and the Laplace transform.

S.NO. | $v(t)$ | ${\mathbb{S}}\left[v(t)\right]$ | ${\mathbb{N}}\left[v(t)\right]$ | $\textbf{S}\left[v(t)\right]$ | $\pounds\left[v(t)\right]$
---|---|---|---|---|---
1 | 1 | $\frac{u}{s}$ | $\frac{1}{s}$ | 1 | $\frac{1}{s}$
2 | $t$ | $\frac{u^{2}}{s^{2}}$ | $\frac{u}{s^{2}}$ | $u$ | $\frac{1}{s^{2}}$
3 | $\exp(\alpha(t))$ | $\frac{u}{s-\alpha u}$ | $\frac{1}{s-\alpha u}$ | $\frac{1}{1-\alpha u}$ | $\frac{1}{s-\alpha}$
4 | $\frac{\sin(\alpha t)}{\alpha}$ | $\frac{u^{2}}{s^{2}+\alpha^{2}u^{2}}$ | $\frac{u}{s^{2}+\alpha^{2}u^{2}}$ | $\frac{u}{1+\alpha^{2}u^{2}}$ | $\frac{1}{s^{2}+\alpha^{2}}$
5 | $\cos(\alpha t)$ | $\frac{us}{s^{2}+\alpha^{2}u^{2}}$ | $\frac{s}{s^{2}+\alpha^{2}u^{2}}$ | $\frac{1}{1+\alpha^{2}u^{2}}$ | $\frac{s}{s^{2}+\alpha^{2}}$
6 | $\cosh\alpha t$ | $\frac{us}{s^{2}-u^{2}}$ | $\frac{s}{s^{2}-u^{2}}$ | $\frac{1}{1-u^{2}}$ | $\frac{s}{s^{2}-1}$
7 | $\frac{t^{n}}{n!}\,\,n=0,1,2...$ | $\left(\frac{u}{s}\right)^{n+1}$ | $\frac{u^{n}}{s^{n+1}}$ | $u^{n}$ | $\frac{1}{s^{n+1}}$
8 | $\frac{t^{n}}{\Gamma(n+1)}\,\,n=0,1,2,...$ | $\left(\frac{u}{s}\right)^{n+1}$ | $\frac{u^{n}}{s^{n+1}}$ | $u^{n}$ | $\frac{1}{s^{n+1}}$
9 | $\cos(t)$ | $\frac{us}{s^{2}+u^{2}}$ | $\frac{s}{s^{2}+u^{2}}$ | $\frac{1}{1+u^{2}}$ | $\frac{1}{s^{2}+1}$
10 | $\sin(t)$ | $\frac{u^{2}}{s^{2}+u^{2}}$ | $\frac{u}{s^{2}+u^{2}}$ | $\frac{u}{1+u^{2}}$ | $\frac{1}{s^{2}+1}$
11 | $\frac{sinh(\alpha t)}{\alpha}$ | $\frac{u^{2}}{s^{2}-\alpha^{2}u^{2}}$ | $\frac{\alpha u}{s^{2}-\alpha^{2}u^{2}}$ | $\frac{\alpha u^{2}}{1-\alpha^{2}u^{2}}$ | $\frac{\alpha}{s^{2}-\alpha^{2}}$
12 | $\cosh(\alpha t)$ | $\frac{us}{s^{2}-\alpha^{2}u^{2}}$ | $\frac{s}{s^{2}-\alpha^{2}u^{2}}$ | $\frac{1}{1-\alpha^{2}u^{2}}$ | $\frac{s}{s^{2}-\alpha^{2}}$
13 | $\exp(\beta t)cosh(\alpha t)$ | $\frac{u(s-\beta u)}{(s-\beta u)^{2}-\alpha^{2}u^{2}}$ | $\frac{s-\beta u}{(s-\beta u)^{2}-\alpha^{2}u^{2}}$ | $\frac{1-\beta u}{(s-\beta u)^{2}-\alpha^{2}u^{2}}$ | $\frac{s-\beta}{(s-\beta u)^{2}-\alpha^{2}}$
14 | $\frac{\exp(\beta t)\sinh(\alpha t)}{\alpha}$ | $\frac{u^{2}}{(s-\beta u)^{2}-\alpha^{2}u^{2}}$ | $\frac{u}{(s-\beta u)^{2}-\alpha^{2}u^{2}}$ | $\frac{u}{(1-\beta u)^{2}-\alpha^{2}u^{2}}$ | $\frac{1}{(s-\beta)^{2}-\alpha^{2}}$
15 | $\frac{t\sin(\alpha t)}{2\alpha}$ | $\frac{u^{3}s}{(s^{2}+\alpha^{2}u^{2})^{2}}$ | $\frac{u^{2}s}{(s^{2}+\alpha^{2}u^{2})^{2}}$ | $\frac{u^{3}}{(1+\alpha^{2}u^{2})^{2}}$ | $\frac{s}{(s^{2}+\alpha^{2})^{2}}$
16 | $t\cos(\alpha t)$ | $\frac{u^{2}(s^{2}-\alpha^{2}u^{2})^{2}}{(s^{2}+\alpha^{2}u^{2})^{2}}$ | $\frac{u(s^{2}-\alpha^{2}u^{2})^{2}}{(s^{2}+\alpha^{2}u^{2})^{2}}$ | $\frac{u(1-\alpha^{2}u^{2})^{2}}{(1+\alpha^{2}u^{2})^{2}}$ | $\frac{(s^{2}-\alpha^{2})^{2}}{(s^{2}+\alpha^{2})^{2}}$
17 | $\frac{\sin(\alpha t)+\alpha t\cos(\alpha t)}{2\alpha}$ | $\frac{u^{2}s^{2}}{(s^{2}+\alpha^{2}u^{2})^{2}}$ | $\frac{us^{2}}{(s^{2}+\alpha^{2}u^{2})^{2}}$ | $\frac{u}{(1+\alpha^{2}u^{2})^{2}}$ | $\frac{s^{2}}{(s^{2}+\alpha^{2})^{2}}$
18 | $\cos\alpha t-\frac{\alpha t\sin(\alpha t)}{2}$ | $\frac{us^{3}}{(s^{2}+\alpha^{2}u^{2})^{2}}$ | $\frac{s^{3}}{(s^{2}+\alpha^{2}u^{2})^{2}}$ | $\frac{1}{(1+\alpha^{2}u^{2})^{2}}$ | $\frac{s^{3}}{(s^{2}+\alpha^{2})^{2}}$
19 | $\frac{\sin(\alpha t)-\alpha t\cos(\alpha t)}{2\alpha^{3}}$ | $\frac{u^{4}}{(s^{2}+\alpha^{2}u^{2})^{2}}$ | $\frac{u^{3}}{(s^{2}+\alpha^{2}u^{2})^{2}}$ | $\frac{u^{3}}{(1+\alpha^{2}u^{2})^{2}}$ | $\frac{1}{(s^{2}+\alpha^{2})^{2}}$
20 | $t\sinh(\alpha t)+t\cosh(\alpha t)$ | $\frac{u^{2}}{(s-\alpha u)^{2}}$ | $\frac{u}{(s-\alpha u)^{2}}$ | $\frac{u^{2}}{(1-\alpha u)^{2}}$ | $\frac{1}{(s-\alpha)^{2}}$
21 | $\frac{t\sinh(\alpha t)}{2\alpha}$ | $\frac{u^{3}s}{(s^{2}-\alpha^{2}u^{2})^{2}}$ | $\frac{u^{2}s}{(s^{2}-\alpha^{2}u^{2})^{2}}$ | $\frac{u^{2}}{(1-\alpha^{2}u^{2})^{2}}$ | $\frac{s}{(s^{2}-\alpha^{2})^{2}}$
22 |  $Si(\alpha t)$ (Sine integral) | $\frac{u}{s}tan^{-1}\left(\frac{\alpha u}{s}\right)$ | $\frac{1}{s}tan^{-1}\left(\frac{\alpha u}{s}\right)$ | $tan^{-1}\left(u\sqrt{\alpha^{2}}\right)$ | $\frac{1}{s}tan^{-1}\left(\frac{\alpha}{s}\right)$
23 |  $Ci(\alpha t)$ (Cosine integral) | $-\frac{u}{2s}\log\left(\frac{s^{2}+\alpha^{2}}{\alpha^{2}}\right)$ | $-\frac{1}{2s}\log\left(\frac{s^{2}+\alpha^{2}u^{2}}{\alpha^{2}u^{2}}\right)$ | $-\frac{1}{2}\log\left(\frac{\alpha^{2}u^{2}+1}{\alpha^{2}u^{2}}\right)$ | $-\frac{1}{2s}\log\left(\frac{s^{2}+\alpha^{2}}{\alpha^{2}}\right)$
24 |  $Ei(\alpha t)$ (Exp. integral) | $-\frac{u}{s}\log\left(\frac{\alpha u-s}{\alpha u}\right)$ | $-\frac{1}{s}\log\left(\frac{\alpha u-s}{\alpha u}\right)$ | $\log\left(\frac{\alpha u-1}{\alpha u}\right)$ | $-\frac{1}{s}\log\left(\frac{\alpha-s}{\alpha}\right)$
25 | $\frac{(3-\alpha^{2}t^{2})\sin(\alpha t)-3\alpha t\cos(\alpha t)}{8\alpha^{5}}$ | $\frac{u^{6}}{(s^{2}+\alpha^{2}u^{2})^{3}}$ | $\frac{u^{5}}{(s^{2}+\alpha^{2}u^{2})^{3}}$ | $\frac{u^{5}}{(1+\alpha^{2}u^{2})^{3}}$ | $\frac{1}{(s^{2}+\alpha^{2})^{3}}$
26 | $\frac{(3-\alpha^{2}t^{2})\sin(\alpha t)+5\alpha t\cos(\alpha t)}{8\alpha}$ | $\frac{u^{2}s^{4}}{(s^{2}+\alpha^{2}u^{2})^{3}}$ | $\frac{us^{4}}{(s^{2}+\alpha^{2}u^{2})^{3}}$ | $\frac{u}{(1+\alpha^{2}u^{2})^{3}}$ | $\frac{s^{4}}{(s^{2}+\alpha^{2})^{3}}$
27 | $\frac{(8-\alpha^{2}t^{2})\cos(\alpha t)-7\alpha t\sin(\alpha t)}{8}$ | $\frac{us^{5}}{(s^{2}+\alpha^{2}u^{2})^{3}}$ | $\frac{s^{5}}{(s^{2}+\alpha^{2}u^{2})^{3}}$ | $\frac{1}{(1+\alpha^{2}u^{2})^{3}}$ | $\frac{s^{5}}{(s^{2}+\alpha^{2})^{3}}$
28 | $\frac{t^{2}\sin(\alpha t)}{2\alpha}$ | $\frac{u^{4}(3s^{2}-\alpha^{2}u^{2})}{(s^{2}+\alpha^{2}u^{2})^{3}}$ | $\frac{u^{3}(3s^{2}-\alpha^{2}u^{3})}{(s^{2}+\alpha^{2}u^{2})^{3}}$ | $\frac{u^{3}(-3+\alpha^{2}u^{2})}{(1+\alpha^{2}u^{2})^{3}}$ | $\frac{(3s^{2}-\alpha^{2})}{(s^{2}+\alpha^{2})^{3}}$
29 | $\frac{t^{2}\cos(\alpha t)}{2}$ | $\frac{u^{3}(s^{3}-3\alpha^{2}u^{2}s)}{(s^{2}+\alpha^{2}u^{2})^{3}}$ | $\frac{u^{2}(s^{3}-3\alpha^{2}u^{2}s)}{(s^{2}+\alpha^{2}u^{2})^{3}}$ | $\frac{u^{2}(1-3\alpha^{2}u^{2})}{(1+\alpha^{2}u^{2})^{3}}$ | $\frac{(s^{3}-3\alpha^{2}s)}{(s^{2}+\alpha^{2})^{3}}$
30 | $\frac{t^{3}\sin(\alpha t)}{24\alpha}$ | $\frac{su^{5}(s-\alpha u)^{2}}{(s^{2}+\alpha^{2}u^{2})^{4}}$ | $\frac{su^{4}(s-\alpha u)^{2}}{(s^{2}+\alpha^{2}u^{2})^{4}}$ | $\frac{u^{4}(1-\alpha u)^{2}}{(1+\alpha^{2}u^{2})^{4}}$ | $\frac{s(s-\alpha)^{2}}{(s^{2}+\alpha^{2})^{4}}$
31 | $\frac{\exp(\alpha t)-\exp(\beta t)}{\alpha-\beta}\,\,\alpha\neq\beta$ | $\frac{u^{2}}{(s-\alpha u)(s-\beta u)}$ | $\frac{u}{(s-\alpha u)(s-\beta u)}$ | $\frac{u}{(1-\beta u)(1-\alpha u)}$ | $\frac{1}{(s-\beta)(s-\alpha)}$
32 | $\frac{\alpha\exp(\alpha t)-\beta\exp(\beta t)}{\alpha-\beta}\,\,\alpha\neq\beta$ | $\frac{us}{(s-\beta u)(s-\alpha u)}$ | $\frac{s}{(s-\beta u)(s-\alpha u)}$ | $\frac{1}{(1-\beta u)(1-\alpha u)}$ | $\frac{s}{(s-\beta)(s-\alpha)}$
33 | $I_{0}(\alpha t)$ | $\frac{u}{\sqrt{s^{2}-\alpha^{2}u^{2}}}$ | $\frac{1}{\sqrt{s^{2}-\alpha^{2}u^{2}}}$ | $\frac{1}{\sqrt{1-\alpha^{2}u^{2}}}$ | $\frac{1}{\sqrt{s^{2}-\alpha^{2}}}$
34 | $\delta(t-\alpha)$ | $u\exp\left(\frac{-\alpha s}{u}\right)$ | $\frac{1}{u}\exp\left(\frac{-\alpha s}{u}\right)$ | $\frac{1}{u}\exp\left(\frac{-\alpha}{u}\right)$ | $\exp(-\alpha s)$
35 | $J_{0}(\alpha t)$ | $\frac{u}{\sqrt{s^{2}+\alpha^{2}u^{2}}}$ | $\frac{1}{\sqrt{s^{2}+\alpha^{2}u^{2}}}$ | $\frac{1}{\sqrt{1+\alpha^{2}u^{2}}}$ | $\frac{1}{\sqrt{s^{2}+\alpha^{2}}}$

Table 2: General properties of Shehu transform

S.NO. | Definition/Property | ${\mathbb{S}}\left[v(t)\right]=V(s,u)$ | Transforms
---|---|---|---
1 | Definition | ${\mathbb{S}}\left[v(t)\right]$ | $\int_{0}^{\infty}\exp\left(\frac{-st}{u}\right)v(t)\,dt;\,\,\,\,s\,>0,\,\,u>0$
2 | Inverse | $v(t)={\mathbb{S}^{-1}}\left[V(s,u)\right]$ | $\frac{1}{2\pi i}\int_{\alpha-i\infty}^{\alpha+i\infty}\frac{1}{u}\exp\left(\frac{st}{u}\right)V(s,u)\,ds$
3 | Linearity | ${\mathbb{S}}\left[\alpha v(t)+\beta w(t)\right]$ | $\alpha{\mathbb{S}}\left[v(t)\right]+\beta{\mathbb{S}}\left[w(t)\right]$
4 | Change of scale | ${\mathbb{S}}\left[v(\alpha t)\right]$ | $\frac{u}{\alpha}V\left(\frac{s}{\alpha},u\right)$
5 | Derivatives | ${\mathbb{S}}\left[v^{\prime}(t)\right]$ | $\frac{s}{u}V(s,u)-v(0)$
|  | ${\mathbb{S}}\left[v^{\prime\prime}(t)\right]$ | $\frac{s^{2}}{u^{2}}V(s,u)-\frac{s}{u}v(0)-v^{\prime}(0)$
|  | ${\mathbb{S}}\left[v^{\prime\prime\prime}(t)\right]$ | $\frac{s^{3}}{u^{3}}V(s,u)-\frac{s^{2}}{u^{2}}v(0)-\frac{s}{u}v^{\prime}(0)-v^{\prime\prime}(0)$
|  | $\vdots$ |
|  | ${\mathbb{S}}\left[v^{(n)}(t)\right]$ | $\frac{s^{n}}{u^{n}}V(s,u)-\sum_{k=0}^{n-1}\left(\frac{s}{u}\right)^{n-(k+1)}$

Table 3: Summary of some integral transform and their definitions

S.No. | Integral transform | Definition
---|---|---
1 | Laplace transform | $\pounds[f(t)]=F(s)=\int_{0}^{\infty}\exp\left(-st\right)f(t)dt$
2 | Fourier transform | $\digamma[f(t)]=f(\omega)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\exp\left(-i\omega t\right)f(t)dt$
3 | Melling transform | $M[f(s);s]=f^{*}(s)=\int_{0}^{\infty}x^{s-1}f(x)dx$
4 | Hankel’s transform | $F_{v}(s)=H_{v}[f(r)]=\int_{0}^{\infty}rf(r)J_{v}(sr)dr,\,\,\,\,\,r\geq 0$
5 | Sumudu transform | $\textbf{S}[f(t)]=G(u)=\frac{1}{u}\int_{0}^{\infty}\exp\left(\frac{-t}{u}\right)f(t)dt$
6 | Laplace-Carson transform |  $\hat{f}_{C}(p)=p\int_{0}^{\infty}\exp\left(-pt\right)f(t)dt$, $t\geq 0$
7 | Atangana-Kilicman transform | $M_{n}(s)=M_{n}[f(x)](s)=\int_{0}^{\infty}x^{n}\exp\left(-xs\right)f(x)dx$
8 | El-zaki transform | $E[f(t)]=T(u)=u\int_{0}^{\infty}\exp\left(\frac{-t}{u}\right)f(t)dt$
9 | Yang transform | $Y[\phi(\tau)]=\phi(\omega)=\int_{0}^{\infty}\exp\left(\frac{-\tau}{\omega}\right)\phi(\tau)d\tau$
10 | natural transform | ${\mathbb{N}^{+}}[f(t)]=R(s,u)=\frac{1}{u}\int_{0}^{\infty}\exp\left(\frac{-st}{u}\right)f(t)dt,\,\,\,\,s\,>0,\,\,u>0$
11 | z-transform | $X(z)=Z\\{x[n]\\}=\sum_{n=-\infty}^{\infty}x[n]z^{-n},\,\,n\in{\mathbb{Z}},\,\,z\in{\mathbb{C}}$
12 |  ${\mathbb{M}}$-transform | $M_{\rho,m}[f(t)](u,v)=\int_{0}^{\infty}\frac{\exp\left(-ut\right)f(vt)}{\left(t^{m}+v^{m}\right)^{\rho}}dt,\,\,\rho\in{\mathbb{C}},\,\Re(\rho)\geq 0,\,m\in{\mathbb{Z}_{+}}$