A new general integral transform for solving integral equations

Graphical abstract


Introduction
Integral transforms are precious for the simplification that they bring about, most often in dealing with differential equations subject to specific boundary conditions. Appropriate choice of integral transforms helps to convert differential equations and integral equations into terms of an algebraic equation that can be solved easily. The solution achieved is, of course, the transform of the solution of the original differential equation, and it is necessary to invert this transform to complete the operation [13,23,30,37,38]. Tables are available for the common transformations, those list many functions and their transforms.
In this work, a new integral transform is introduced and is used to obtain analytical solution of higher order ODES with constant and variable coefficient and integral equations. The paper is arranged as follows.
In Section 'A new integral transform', we introduce a general integral transform in the class of Laplace transform. In Section 'Relation between the new transform and other transforms', we compare the given integral transform with those existing integral transform in the class of Laplace transform. Then this integral transform is applied to ODEs and integral equations in Section 'Solving IVP and Integral equations by new transform'. Finally, some conclusions are summarized in Section 'Conclusion'.

A new integral transform
In this section, we present a general integral transform which is cover most or even all type of integral transforms in the family of Laplace transform.
Definition 1. Let f t ð Þ be a integrable function defined for t P 0; p s ð Þ -0 and q s ð Þ are positive real functions, we define the general integral transform T s ð Þ of f t ð Þ by the formula provided the integral exists for some q s ð Þ. Thus, we can obtain the integral transform of any general function. In Table 2, we provided new integral transform of some basic functions. It must be mentioned that the new integral transform (1) for those f t ð Þ are not continuously differentiable contains terms with negative or fractional powers of q s ð Þ. Let for all t P 0, the function f t ð Þ is piecewise continuous and satisfies jf t ð Þj 6 Me kt , then T s ð Þ exists for all q s ð Þ > k. Since Proof. (I). In view of (1) we have Table 1 Definition of some integral transforms Laplace Transform

Function
New integral transform Proof.
T It is easy to show that all of those integral transforms are a special cases of Eq. (1). In the next section, we can compare these transforms with classical Laplace transform. Also we get idea, that how can we choose one of these transforms and how can we define a suitable functions for p s ð Þ and q s ð Þ in the equation (1).

Solving IVP and Integral equations by new transform
In this section, we apply this new integral transform for solving high order IVP with constant coefficient and variable coefficient as well as. Also we apply it to obtain exact solution of few type of integral equations and FDE.

Solving IVP with constant coefficient
Consider the following IVP: Now we apply new integral transform to both side of equation (5).
In view of Theorem 1, we have By substituting the initial conditions in the equation (7) we have where h s ð Þ ¼ q n s ð Þ þ a 1 q nÀ1 s ð Þ þ Á Á Á þ a n À Á ; G s ð Þ ¼ T g x ð Þ f g and W s ð Þ ¼ p s ð Þ P nÀ1 k¼0 q nÀ1Àk s ð Þy k þ a 1 P nÀ2 k¼0 q nÀ2Àk s ð Þy k þ Á Á Á þ y 0 . From Finally, by apply the inverse transform T À1 on both sides of the above equation, we obtain the exact solution: Example 1. Consider the following third-order ODE y 000 þ y 00 À 6y ¼ 0; ð11Þ y 0 ð Þ ¼ 1; y 0 0 ð Þ ¼ 0; y 00 0 ð Þ ¼ 5: By applying T on both side of (11) we have T y 000 t ð Þ þ y 00 t ð Þ À 6y t Now in view of Table 1, by applying T À1 on both side of (18) we find the exact solution as Example 2. Consider the following third-order ODE By applying T on both sides of (15) we have by replacing the initial conditions in above equation, we have Now in view of Table 1, by applying T À1 on both sides of (17) we find the exact solution as Remark 1. All the above mentioned integral transforms for ODEs with constant coefficient give same solution.
Remark 2. In view of (10), if we choose the Laplace transform for ODEs with constant coefficient, the volume of calculation be minimum.

Solving IVP with variable coefficient
Now consider the following type of IVP problems It is clear that m i ¼ 0; 1; 2; Á Á Á ; n so we must find the integral transform of T a i t ð Þy l ð Þ t ð Þ È É . According to value of a i t ð Þ and order of derivative y l ð Þ t ð Þ; l ¼ 0; 1; Á Á Á ; n À 1, we have different cases. In follow, we present few theorems which are useful to solve these type of IVP.
Theorem 3. Let p s ð Þ and q s ð Þ are differentiable and q 0 s ð Þ -0, then from the above equation we have now we take derivative on both sides of Eq. (21) respect to s, it leads after simplification we have . Now by taking derivative form both sides of this equation we have Following the same procedure we can proof (III Example 6. Consider the following fractional integral equation y t ð Þ ¼ g t ð Þ þ I a y t ð Þ; a 2 R þ ; where I a is the well known Riemann-Liouville fractional integral operator [8,9,[25][26][27]29,32,33,38]. It is defined by I a y t where G s ð Þ is T g t ð Þ f g.

Conclusion
In this paper, we introduce a general integral transform. After that we compare some integral transforms with this new integral transform. It has shown that the new integral transform cover those exiting transforms such as Laplace, Elzaki and Sumudu transforms for different value of p s ð Þ and q s ð Þ. We used this new transform for solving ODE, integral equations and fractional integral equations. Few examples have been presented to illustrate the efficiency of this integral transform.

Compliance with Ethics Requirements
This article does not contain any studies with human or animal subjects.

Funding
Not applicable.

Declaration of Competing Interest
The author declare that they have no conflict of interest.