<?xml version="1.0" encoding="UTF-8"?><!DOCTYPE article PUBLIC "-//NLM//DTD JATS (Z39.96) Journal Publishing DTD v1.3 20210610//EN" "https://jats.nlm.nih.gov/publishing/1.3/JATS-journalpublishing1-3.dtd"><article xml:lang="en" xmlns:xlink="http://www.w3.org/1999/xlink" xmlns:ali="http://www.niso.org/schemas/ali/1.0/" dtd-version="1.3" article-type="research-article"><front><journal-meta><journal-id journal-id-type="issn">2460-0245</journal-id><journal-title-group><journal-title>Journal of the Indonesian Mathematical Society</journal-title><abbrev-journal-title>JIMS</abbrev-journal-title></journal-title-group><issn pub-type="epub">2460-0245</issn><issn pub-type="ppub">2086-8952</issn><publisher><publisher-name>IndoMS</publisher-name></publisher></journal-meta><article-meta><article-id pub-id-type="doi">10.22342/jims.v32i1.1883</article-id><article-categories></article-categories><title-group><article-title>Solutions of a Generalization of Linear Volterra Integro-Differential Equations</article-title></title-group><contrib-group><contrib contrib-type="author"><name><surname>Inpoonjai</surname><given-names>Phaisatcha</given-names></name><address><country country="TH">Thailand</country><email>phaisat@rmutl.ac.th</email></address><xref ref-type="aff" rid="AFF-1"></xref><xref ref-type="corresp" rid="cor-0"></xref></contrib></contrib-group><contrib-group><contrib contrib-type="editor"><name><surname>Wijayanti</surname><given-names>Indah Emilia</given-names></name><address><country country="ID">Indonesia</country><email>ind_wijayanti@ugm.ac.id</email></address><xref ref-type="aff" rid="EDITOR-AFF-1"></xref></contrib></contrib-group><aff id="AFF-1"><institution content-type="dept">Faculty of Science and Agricultural Technology</institution><country>Rajamangala University of Technology Lanna Chiangrai</country></aff><aff id="EDITOR-AFF-1"><institution-wrap><institution>Universitas Gadjah Mada</institution><institution-id institution-id-type="ror">https://ror.org/03ke6d638</institution-id></institution-wrap><country country="ID">Indonesia</country></aff><author-notes><corresp id="cor-0">Corresponding author: Phaisatcha Inpoonjai. Email: <email>phaisat@rmutl.ac.th</email></corresp></author-notes><pub-date date-type="pub" iso-8601-date="2026-01-05" publication-format="electronic"><day>05</day><month>01</month><year>2026</year></pub-date><pub-date date-type="collection" iso-8601-date="2026-01-05" publication-format="electronic"><day>05</day><month>01</month><year>2026</year></pub-date><volume>32</volume><issue>1</issue><issue-title>MARCH</issue-title><fpage>1</fpage><lpage>21</lpage><history><date date-type="received" iso-8601-date="2024-12-25"><day>25</day><month>12</month><year>2024</year></date><date date-type="accepted" iso-8601-date="2025-11-16"><day>16</day><month>11</month><year>2025</year></date></history><permissions><copyright-statement>Copyright (c) 2026 Journal of the Indonesian Mathematical Society</copyright-statement><copyright-year>2026</copyright-year><copyright-holder>Journal of the Indonesian Mathematical Society</copyright-holder><license xlink:href="https://creativecommons.org/licenses/by-nc-nd/4.0/"><ali:license_ref xmlns:ali="http://www.niso.org/schemas/ali/1.0/">https://creativecommons.org/licenses/by-nc-nd/4.0/</ali:license_ref><license-p>This work is licensed under a Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License.</license-p></license></permissions><self-uri xlink:href="https://jims-a.org/index.php/jimsa/article/view/1883" xlink:title="1883"></self-uri><abstract><p>In this paper, we combine linear Volterra integro-differential equations of first and second kinds to be a generalization. Then, we use Laplace transform to solve an analytical solution on a convolution kernel and apply Laguerre polynomials to approximate a solution on a non-convolution kernel of this generalization.</p></abstract><kwd-group><kwd>Volterra integro-differential equation</kwd><kwd>Laplace transform</kwd><kwd>Laguerre polynomial</kwd></kwd-group><custom-meta-group><custom-meta><meta-name>File created by JATS Editor</meta-name><meta-value>https://jatseditor.com</meta-value></custom-meta><custom-meta><meta-name>issue-created-year</meta-name><meta-value>2026</meta-value></custom-meta></custom-meta-group></article-meta></front><body><sec id="sec-1"><title>1. INTRODUCTION</title><p>In this study, we consider the following linear <italic>Volterra integro-diferential equations</italic> (or only VIDEs) with initial conditions. The linear VIDEs of first kind are given by</p><disp-formula id="equation-1"><tex-math id="math-1"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \begin{array}{l} \int_ {0} ^ {x} k _ {1} (x, t) u ^ {(n)} (t) d t = f (x) u (x) + g (x) + \int_ {0} ^ {x} k _ {2} (x, t) u (t) d t, x \in [ 0, T ], \\ u (0) = a _ {0}, u ^ {'} (0) = a _ {1},..., u ^ {(n - 1)} (0) = a _ {n - 1}, \end{array}\tag{1} \end{document} ]]></tex-math></disp-formula><p>where <inline-formula><tex-math id="math-2"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle f ( x ) , g ( x ) , k _ { 1 } ( x , t ) \end{document} ]]></tex-math></inline-formula> and <inline-formula><tex-math id="math-3"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle k _ { 2 } ( x , t ) \end{document} ]]></tex-math></inline-formula> are suficiently smooth functions. The linear VIDEs of second kind are expressed by</p><disp-formula id="equation-2"><tex-math id="math-4"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \begin{array}{l} u ^ {(n)} (x) = h (x) u (x) + k (x) + \int_ {0} ^ {x} k _ {3} (x, t) u (t) d t, x \in [ 0, T ], \\ u (0) = b _ {0}, u ^ {'} (0) = b _ {1},..., u ^ {(n - 1)} (0) = b _ {n - 1}, \end{array}\tag{2} \end{document} ]]></tex-math></disp-formula><p>where <inline-formula><tex-math id="math-5"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle h ( x ) , k ( x ) \end{document} ]]></tex-math></inline-formula> and <inline-formula><tex-math id="math-6"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle k _ { 3 } ( x , t ) \end{document} ]]></tex-math></inline-formula> are suficiently smooth functions. The functions <inline-formula><tex-math id="math-7"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle k _ { 1 } ( x , t ) , k _ { 2 } ( x , t ) \end{document} ]]></tex-math></inline-formula> and <inline-formula><tex-math id="math-8"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle k _ { 3 } ( x , t ) \end{document} ]]></tex-math></inline-formula> are called <italic>kernels</italic> of the linear VIDEs.</p><p>Let <inline-formula><tex-math id="math-9"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle F ( x ) \end{document} ]]></tex-math></inline-formula> be defined and piecewise continuous function for all positive values of <inline-formula><tex-math id="math-10"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle x \end{document} ]]></tex-math></inline-formula> and be of exponential order. The<italic> Laplace</italic><italic>transform</italic> of <inline-formula><tex-math id="math-11"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle F ( x ) \end{document} ]]></tex-math></inline-formula> is defined by</p><disp-formula id="equation-3"><tex-math id="math-12"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle L \Bigl \{F (x) \Bigr \} = \int_ {0} ^ {\infty} F (x) e ^ {- s x} d x = G (s), \end{document} ]]></tex-math></disp-formula><p>where <inline-formula><tex-math id="math-13"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle F ( x ) \end{document} ]]></tex-math></inline-formula> is said to be the <italic>inverse</italic><italic>Laplace transform</italic> of <inline-formula><tex-math id="math-14"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle G ( s ) \end{document} ]]></tex-math></inline-formula> , denoted by <inline-formula><tex-math id="math-15"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle F ( x ) = \end{document} ]]></tex-math></inline-formula><inline-formula><tex-math id="math-16"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle L ^ { - 1 } { \Big \{ } G ( s ) { \Big \} } \end{document} ]]></tex-math></inline-formula> . Let us recall some useful results on the Laplace transform that shall be used in the next hereinafter.</p><p>(1) The Laplace transform of some functions:</p><disp-formula id="equation-4"><tex-math id="math-17"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \begin{array}{ll}L\{1\}=\frac{1}{s},\ s>0, &L\{x^n\}=\frac{n!}{s^{n+1}},\ s>0,\\L\{e^{ax}\}=\frac{1}{s-a},\ s>a, &L\{\sin ax\}=\frac{a}{s^2+a^2},\ s>0,\\L\{\cos ax\}=\frac{s}{s^2+a^2},\ s>0, &L\{\sinh ax\}=\frac{a}{s^2-a^2},\ s>|a|,\\L\{\cosh ax\}=\frac{s}{s^2-a^2},\ s>|a|.\end{array} \end{document} ]]></tex-math></disp-formula><p>(2) If L<sup>n</sup>F (x)<sup>o</sup> = G(s) then <inline-formula><tex-math id="math-18"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle L \Big \{ F ^ { ' } ( x ) \Big \} = s G ( s ) - F ( 0 ) \end{document} ]]></tex-math></inline-formula></p><p>(3) If L<sup>n</sup>F (x)<sup>o</sup> = G(s) then <inline-formula><tex-math id="math-19"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle L \Big \{ F ^ { ( n ) } ( x ) \Big \} = s ^ { n } G ( s ) - s ^ { n - 1 } F ( 0 ) - s ^ { n - 2 } F ^ { ' } ( 0 ) - \end{document} ]]></tex-math></inline-formula> · · · − F <sup>(n−1)</sup>(0).</p><p>(4) If <inline-formula><tex-math id="math-20"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle L \left\{ F ( x ) \right\} = G ( s ) \end{document} ]]></tex-math></inline-formula> then <inline-formula><tex-math id="math-21"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle L { \Big \{ } x ^ { n } F ( x ) { \Big \} } = ( - 1 ) ^ { n } { \frac { d ^ { n } } { d s ^ { n } } } { \Big [ } G ( s ) { \Big ] } \end{document} ]]></tex-math></inline-formula></p><p>(5) The <italic>convolution</italic> of two functions <inline-formula><tex-math id="math-22"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle F ( x ) \end{document} ]]></tex-math></inline-formula> and <inline-formula><tex-math id="math-23"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle H ( x ) \end{document} ]]></tex-math></inline-formula> , denoted by <inline-formula><tex-math id="math-24"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle F ( x ) * H ( x ) \end{document} ]]></tex-math></inline-formula> is defined by <inline-formula><tex-math id="math-25"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle F ( x ) * H ( x ) = \int _ { 0 } ^ { x } F ( t ) H ( x - t ) d t = \int _ { 0 } ^ { x } F ( x - t ) H ( t ) d t \end{document} ]]></tex-math></inline-formula> . Let <inline-formula><tex-math id="math-26"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle L \Big \{ F ( x ) \Big \} = G ( s ) \end{document} ]]></tex-math></inline-formula> and <inline-formula><tex-math id="math-27"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle L \Big \{ H ( x ) \Big \} = I ( s ) \end{document} ]]></tex-math></inline-formula> . Then the <italic>convolution theorem</italic> says that <inline-formula><tex-math id="math-28"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle L { \Big \{ } F ( x ) * H ( x ) { \Big \} } = L { \Big \{ } F ( x ) { \Big \} } L { \Big \{ } H ( x ) { \Big \} } = G ( s ) I ( s ) \end{document} ]]></tex-math></inline-formula></p><p>(6) The inverse Laplace transform of some functions:</p><disp-formula id="equation-5"><tex-math id="math-29"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \begin{array}{ll}L^{-1}\left\{\dfrac{1}{s}\right\}=1,&L^{-1}\left\{\dfrac{1}{s^{n+1}}\right\}=\dfrac{1}{n!}x^{n},\\L^{-1}\left\{\dfrac{1}{s-a}\right\}=e^{ax},&L^{-1}\left\{\dfrac{1}{s^{2}+a^{2}}\right\}=\dfrac{\sin ax}{a},\\L^{-1}\left\{\dfrac{s}{s^{2}+a^{2}}\right\}=\cos ax,&L^{-1}\left\{\dfrac{1}{s^{2}-a^{2}}\right\}=\dfrac{\sinh ax}{a},\\L^{-1}\left\{\dfrac{s}{s^{2}-a^{2}}\right\}=\cosh ax.&\end{array} \end{document} ]]></tex-math></disp-formula><p>The <italic>Laguerre polynomial</italic> is a polynomial function given by</p><disp-formula id="equation-6"><tex-math id="math-30"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle L _ {\chi} (x) = \sum_ {k = 0} ^ {\chi} \binom {\chi} {k} \frac {(- 1) ^ {k} x ^ {k}}{k !}, \binom {\chi} {k} = \frac {\chi !}{k ! (\chi - k) !}, \end{document} ]]></tex-math></disp-formula><p>where <inline-formula><tex-math id="math-31"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \chi \end{document} ]]></tex-math></inline-formula> and <inline-formula><tex-math id="math-32"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle k \end{document} ]]></tex-math></inline-formula> are called the <italic>degree</italic> and the <italic>index</italic> of the Laguerre polynomial, respectively. Some important results on the Laguerre polynomials that shall be referred in the next as follows:</p><disp-formula id="equation-7"><tex-math id="math-33"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \begin{align*}(1) \, L_{\chi}'(x) &= \frac{d}{dx} \left[ \sum_{k=0}^{\chi} \binom{\chi}{k} \frac{(-1)^k x^k}{k!} \right] = \sum_{k=1}^{\chi} \binom{\chi}{k} \frac{(-1)^k x^{k-1}}{(k-1)!}, \\(2) \, L_{\chi}''(x) &= \frac{d^2}{dx^2} \left[ \sum_{k=0}^{\chi} \binom{\chi}{k} \frac{(-1)^k x^k}{k!} \right] = \sum_{k=2}^{\chi} \binom{\chi}{k} \frac{(-1)^k x^{k-2}}{(k-2)!}, \\(3) \, L_{\chi}'''(x) &= \frac{d^3}{dx^3} \left[ \sum_{k=0}^{\chi} \binom{\chi}{k} \frac{(-1)^k x^k}{k!} \right] = \sum_{k=3}^{\chi} \binom{\chi}{k} \frac{(-1)^k x^{k-3}}{(k-3)!}, \\(4) \, L_{\chi}^{(n)}(x) &= \frac{d^n}{dx^n} \left[ \sum_{k=0}^{\chi} \binom{\chi}{k} \frac{(-1)^k x^k}{k!} \right] = \sum_{k=n}^{\chi} \binom{\chi}{k} \frac{(-1)^k x^{k-n}}{(k-n)!}, \quad n \le \chi.\end{align*} \end{document} ]]></tex-math></disp-formula><p>The Volterra integro-diferential equations are typically mathematical models in many areas of science and engineering. Solutions of these equations play vital roles in a number of processes and phenomena such as nuclear reactors, circuit analyses, wave propagation, glass forming processes, nano-hydrodynamics, visco elasticity, biological populations, etc. Therefore, there are many researchers who have been interested in the VIDEs and founded numerous methods to solve the analytical and numerical solutions of VIDEs up to the present as follows. Estimated solutions of nonlinear VIDEs of a fractional order were investigated applying the Laplace transform and the Adomian polynomials by Yang and Hou <xref ref-type="bibr" rid="BIBR-1">[1]</xref>. Moreover, the Legendre polynomial approximation was used to find numerical solutions of nonlinear VIDEs of second kind by Gachpazan, Erfanian and Beiglo <xref ref-type="bibr" rid="BIBR-2">[2]</xref>. In addition, analytical solutions of linear VIDEs of second kind were solved using the Kamal transform by Aggarwal and Gupta <xref ref-type="bibr" rid="BIBR-3">[3]</xref>. The modified Adomian decomposition method was utilized to explain exact solutions of linear VIDEs of second kind by Okai, Ilejimi and Ibrahim <xref ref-type="bibr" rid="BIBR-4">[4]</xref>. Furthermore, approximate solutions of nonlinear VIDEs involving delay were found taking a new higher order method by Jhinga, Patade and Gejji <xref ref-type="bibr" rid="BIBR-5">[5]</xref>. Other than those findings, the Sadik transform was applied to figure out exact solutions of first kind VIDEs on convolution type kernels by Aggarwal, Vyas and Sharma <xref ref-type="bibr" rid="BIBR-6">[6]</xref>. Numerical solutions of linear VIDEs were estimated using the Laguerre and Touchard polynomials by Abdullah and Ali <xref ref-type="bibr" rid="BIBR-7">[7]</xref>. So far, some asymptotic behavior of exact solutions of nonlinear VIDEs has been studied by Cakir, Gunes and Duru <xref ref-type="bibr" rid="BIBR-8">[8]</xref>. The quasilinearization technique to diference scheme also has been applied to solve estimated solutions of VIDEs in <xref ref-type="bibr" rid="BIBR-8">[8]</xref>. Recently, the asymptotic behavior of the analytical solutions of the singularly perturbed nonlinear VIDEs has been established by Cakir, Cakir and Cakir <xref ref-type="bibr" rid="BIBR-9">[9]</xref>. The uniform diference scheme on the Bakhvalov-Shishkin mesh points according to the boundary layer conditions has been introduced to find numerical solutions of VIDEs as well in <xref ref-type="bibr" rid="BIBR-9">[9]</xref>. Exact solutions of the Faltung type VIDEs for first kind have been solved applying the Kushare transform by Patil, Nikam and Shinde <xref ref-type="bibr" rid="BIBR-10">[10]</xref>.</p><p>In this research, we compound linear Volterra integro-diferential equations of first and second kinds to be a general form. Then, we take the Laplace transform to find an exact solution on a convolution type and utilize the Laguerre polynomials to estimate a solution on a non-convolution type of this generalization.</p></sec><sec id="sec-2"><title>2. MAIN RESULTS</title><p>For this section, we assume that <inline-formula><tex-math id="math-34"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle p , m \leq n \end{document} ]]></tex-math></inline-formula> and start to introduce a generalization of linear Volterra integro-diferential equations expressed by</p><disp-formula id="equation-8"><tex-math id="math-35"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \begin{array}{l} \alpha u ^ {(n)} (x) + \beta \int_ {0} ^ {x} K _ {1} (x, t) u ^ {(m)} (t) d t \\ = A (x) u (x) + B (x) + \int_ {0} ^ {x} K _ {2} (x, t) u ^ {(p)} (t) d t, x \in [ 0, T ], \\ u (0) = c _ {0}, u ^ {'} (0) = c _ {1}, \dots , u ^ {(n - 1)} (0) = c _ {n - 1}, \end{array}\tag{3} \end{document} ]]></tex-math></disp-formula><p>where <inline-formula><tex-math id="math-36"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle u ( x ) \end{document} ]]></tex-math></inline-formula> is an exponentially bounded and smooth function, <inline-formula><tex-math id="math-37"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle A ( x ) , B ( x ) , K _ { 1 } ( x , t ) \end{document} ]]></tex-math></inline-formula> and <inline-formula><tex-math id="math-38"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle K _ { 2 } ( x , t ) \end{document} ]]></tex-math></inline-formula> are exponentially bounded and suficiently smooth functions and <inline-formula><tex-math id="math-39"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \alpha , \beta \end{document} ]]></tex-math></inline-formula> are real numbers. <inline-formula><tex-math id="math-40"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \mathrm { I f } \ \alpha = 0 \end{document} ]]></tex-math></inline-formula> and <inline-formula><tex-math id="math-41"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle m = 0 \end{document} ]]></tex-math></inline-formula> or <inline-formula><tex-math id="math-42"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \alpha = 0 \end{document} ]]></tex-math></inline-formula> and <inline-formula><tex-math id="math-43"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle p = 0 \end{document} ]]></tex-math></inline-formula> , then the equation (3) is the linear VIDE of first kind and we can see <xref ref-type="bibr" rid="BIBR-6">[6]</xref> and <xref ref-type="bibr" rid="BIBR-10">[10]</xref> for more vital results. If <inline-formula><tex-math id="math-44"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \beta = 0 \end{document} ]]></tex-math></inline-formula> and <inline-formula><tex-math id="math-45"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle p = 0 \end{document} ]]></tex-math></inline-formula> , then the equation (3) is the linear VIDE of second kind and we can see <xref ref-type="bibr" rid="BIBR-1">[1]</xref>, <xref ref-type="bibr" rid="BIBR-2">[2]</xref>, <xref ref-type="bibr" rid="BIBR-3">[3]</xref>, <xref ref-type="bibr" rid="BIBR-4">[4]</xref>, <xref ref-type="bibr" rid="BIBR-5">[5]</xref>, <xref ref-type="bibr" rid="BIBR-7">[7]</xref>, <xref ref-type="bibr" rid="BIBR-8">[8]</xref> and <xref ref-type="bibr" rid="BIBR-9">[9]</xref> for more comprehensive findings.</p><p>Now, we will focus on a solution of this generalization on a convolution type with a constant function <italic>A</italic>, that is, we then consider the following initial-value problem as follows:</p><disp-formula id="equation-9"><tex-math id="math-46"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \begin{array}{l} \alpha u ^ {(n)} (x) + \beta \int_ {0} ^ {x} K _ {1} (x - t) u ^ {(m)} (t) d t \\ = A u (x) + B (x) + \int_ {0} ^ {x} K _ {2} (x - t) u ^ {(p)} (t) d t, x \in [ 0, T ], \\ u (0) = c _ {0}, u ^ {'} (0) = c _ {1},..., u ^ {(n - 1)} (0) = c _ {n - 1}. \end{array}\tag{4} \end{document} ]]></tex-math></disp-formula><p>We will utilize the Laplace transform to solve the problem as the following steps: At the beginning, applying the Laplace transform to <xref ref-type="disp-formula" rid="equation-2">(4)</xref>, we get</p><disp-formula id="equation-10"><tex-math id="math-47"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \begin{array}{l} \alpha L \Bigl \{u ^ {(n)} (x) \Bigr \} + \beta L \Bigl \{\int_ {0} ^ {x} K _ {1} (x - t) u ^ {(m)} (t) d t \Bigr \} \\ = A L \Bigl \{u (x) \Bigr \} + L \Bigl \{B (x) \Bigr \} + L \Bigl \{\int_ {0} ^ {x} K _ {2} (x - t) u ^ {(p)} (t) d t \Bigr \}. \end{array}\tag{5} \end{document} ]]></tex-math></disp-formula><p>After, using the convolution theorem to (5), we then obtain</p><disp-formula id="equation-11"><tex-math id="math-48"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \begin{array}{l} \alpha L \Big \{u ^ {(n)} (x) \Big \} + \beta L \Big \{K _ {1} (x) \Big \} L \Big \{u ^ {(m)} (x) \Big \} \\ = A L \Big \{u (x) \Big \} + L \Big \{B (x) \Big \} + L \Big \{K _ {2} (x) \Big \} L \Big \{u ^ {(p)} (x) \Big \}. \end{array}\tag{6} \end{document} ]]></tex-math></disp-formula><p>Then, taking the Laplace transform of derivatives on (6) with initial conditions, we have</p><disp-formula id="equation-12"><tex-math id="math-49"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \begin{array}{l} \alpha \Big [ s ^ {n} L \Big \{u (x) \Big \} - s ^ {n - 1} c _ {0} - s ^ {n - 2} c _ {1} - \dots - c _ {n - 1} \Big ] \\ + \beta L \Big \{K _ {1} (x) \Big \} \Big [ s ^ {m} L \Big \{u (x) \Big \} - s ^ {m - 1} c _ {0} - s ^ {m - 2} c _ {1} - \dots - c _ {m - 1} \Big ] = A L \Big \{u (x) \Big \} \\ + L \Big \{B (x) \Big \} + L \Big \{K _ {2} (x) \Big \} \Big [ s ^ {p} L \Big \{u (x) \Big \} - s ^ {p - 1} c _ {0} - s ^ {p - 2} c _ {1} - \dots - c _ {p - 1} \Big ] \end{array} \end{document} ]]></tex-math></disp-formula><p>and we also obtain</p><disp-formula id="equation-13"><tex-math id="math-50"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \begin{array}{l} \left[ \alpha s ^ {n} + \beta L \Bigl \{K _ {1} (x) \Bigr \} s ^ {m} - L \Bigl \{K _ {2} (x) \Bigr \} s ^ {p} - A \right] L \Bigl \{u (x) \Bigr \} \\ = \alpha \Bigl (s ^ {n - 1} c _ {0} + s ^ {n - 2} c _ {1} + \dots + c _ {n - 1} \Bigr) \\ + \beta L \Bigl \{K _ {1} (x) \Bigr \} \Bigl (s ^ {m - 1} c _ {0} + s ^ {m - 2} c _ {1} + \dots + c _ {m - 1} \Bigr) \\ + L \Bigl \{B (x) \Bigr \} - L \Bigl \{K _ {2} (x) \Bigr \} \Bigl (s ^ {p - 1} c _ {0} + s ^ {p - 2} c _ {1} + \dots + c _ {p - 1} \Bigr). \end{array}\tag{7} \end{document} ]]></tex-math></disp-formula><p>At last, operating the inverse Laplace transform on (7), we receive the solution of initial-value problem <xref ref-type="disp-formula" rid="equation-2">(4)</xref> as follows.</p><disp-formula id="equation-14"><tex-math id="math-51"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \begin{array}{l} u (x) = L ^ {- 1} \Bigg \{\frac {\alpha \Big (s ^ {n - 1} c _ {0} + s ^ {n - 2} c _ {1} + \cdots + c _ {n - 1} \Big)}{\alpha s ^ {n} + \beta L \Big \{K _ {1} (x) \Big \} s ^ {m} - L \Big \{K _ {2} (x) \Big \} s ^ {p} - A} \Bigg \} \\ + L ^ {- 1} \Bigg \{\frac {\beta L \Big \{K _ {1} (x) \Big \} \Big (s ^ {m - 1} c _ {0} + s ^ {m - 2} c _ {1} + \cdots + c _ {m - 1} \Big)}{\alpha s ^ {n} + \beta L \Big \{K _ {1} (x) \Big \} s ^ {m} - L \Big \{K _ {2} (x) \Big \} s ^ {p} - A} \Bigg \} \\ + L ^ {- 1} \Bigg \{\frac {L \Big \{B (x) \Big \} - L \Big \{K _ {2} (x) \Big \} \Big (s ^ {p - 1} c _ {0} + s ^ {p - 2} c _ {1} + \cdots + c _ {p - 1} \Big)}{\alpha s ^ {n} + \beta L \Big \{K _ {1} (x) \Big \} s ^ {m} - L \Big \{K _ {2} (x) \Big \} s ^ {p} - A} \Bigg \}. \end{array} \end{document} ]]></tex-math></disp-formula><p><target id="anchor-c8fc6052-0f66-4052-9cf4-51b1f8381f00" target-type="reference-target"/></p><p><bold>Example 2.1</bold><italic>Solve the Volterra integro-differential problem:</italic></p><disp-formula id="equation-15"><tex-math id="math-52"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \begin{array}{l}u^{(4)}(x)+\displaystyle\int_{0}^{x}\sin(x-t)\,u''(t)\,dt=u(x)-\dfrac{1}{2}e^{x}-\dfrac{1}{2}\cos x-\dfrac{1}{2}\sin x-x^{2}+x-1+\displaystyle\int_{0}^{x}(x-t)\,u(t)\,dt,\\[1ex]u(0)=3,\quadu'(0)=1,\quadu''(0)=1,\quadu'''(0)=1.\end{array} \end{document} ]]></tex-math></disp-formula><p><italic>Solution.</italic> Firstly, applying Laplace transform to the problem, we have</p><disp-formula id="equation-16"><tex-math id="math-53"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \begin{array}{l} {L \Big \{u ^ {(4)} (x) \Big \} + L \Big \{\int_ {0} ^ {x} \sin (x - t) u ^ {\prime \prime} (t) d t \Big \} = L \Big \{u (x) \Big \}} \\ {+ L \Big \{- \frac {1}{2} e ^ {x} - \frac {1}{2} \cos x - \frac {1}{2} \sin x - x ^ {2} + x - 1 \Big \} + L \Big \{\int_ {0} ^ {x} (x - t) u (t) d t \Big \}.} \end{array} \end{document} ]]></tex-math></disp-formula><p>Secondly, using the convolution theorem, we get</p><disp-formula id="equation-17"><tex-math id="math-54"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \begin{array}{l}L\left\{u^{(4)}(x)\right\} + L\left\{\sin x\right\}L\left\{u''(x)\right\} = L\left\{u(x)\right\} \\+ L\left\{-\frac{1}{2}e^x - \frac{1}{2}\cos x - \frac{1}{2}\sin x - x^2 + x - 1\right\} + L\left\{x\right\}L\left\{u(x)\right\}.\end{array} \end{document} ]]></tex-math></disp-formula><p>Thirdly, taking the Laplace transform of derivatives, we have</p><disp-formula id="equation-18"><tex-math id="math-55"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \begin{array}{l}s^4 L\left\{u(x)\right\} - s^3 u(0) - s^2 u'(0) - s u''(0) - u'''(0) \\+ \frac{1}{s^2 + 1} \left[ s^2 L\left\{u(x)\right\} - s u(0) - u'(0) \right] = L\left\{u(x)\right\} \\- \frac{1}{2(s - 1)} - \frac{s}{2(s^2 + 1)} - \frac{1}{2(s^2 + 1)} - \frac{2}{s^3} + \frac{1}{s^2} - \frac{1}{s} + \frac{1}{s^2} L\left\{u(x)\right\}\end{array} \end{document} ]]></tex-math></disp-formula><p>and using initial conditions, we also obtain</p><disp-formula id="equation-19"><tex-math id="math-56"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \begin{array}{l} s ^ {4} L \Big \{u (x) \Big \} - 3 s ^ {3} - s ^ {2} - s - 1 + \frac {1}{s ^ {2} + 1} \Big [ s ^ {2} L \Big \{u (x) \Big \} - 3 s - 1 \Big ] \\ = L \Big \{u (x) \Big \} - \frac {1}{2 (s - 1)} - \frac {s}{2 (s ^ {2} + 1)} - \frac {1}{2 (s ^ {2} + 1)} \\ - \frac {2}{s ^ {3}} + \frac {1}{s ^ {2}} - \frac {1}{s} + \frac {1}{s ^ {2}} L \Big \{u (x) \Big \}. \end{array} \end{document} ]]></tex-math></disp-formula><p>Fourthly, rearranging the equation, we certainly receive</p><disp-formula id="equation-20"><tex-math id="math-57"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \begin{array}{l} \left(s ^ {4} + \frac {s ^ {2}}{s ^ {2} + 1} - \frac {1}{s ^ {2}} - 1\right) L \Bigl \{u (x) \Bigr \} = 3 s ^ {3} + s ^ {2} + s + 1 + \frac {3 s}{s ^ {2} + 1} \\ + \frac {1}{s ^ {2} + 1} - \frac {1}{2 (s - 1)} - \frac {s}{2 (s ^ {2} + 1)} - \frac {1}{2 (s ^ {2} + 1)} - \frac {2}{s ^ {3}} + \frac {1}{s ^ {2}} - \frac {1}{s} \end{array} \end{document} ]]></tex-math></disp-formula><p>or</p><disp-formula id="equation-21"><tex-math id="math-58"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \left[ \frac {s ^ {8} + s ^ {6} - 2 s ^ {2} - 1}{s ^ {2} (s ^ {2} + 1)} \right] L \Bigl \{u (x) \Bigr \} = \frac {6 s ^ {9} - 4 s ^ {8} + 6 s ^ {7} - 4 s ^ {6} - 1 2 s ^ {3} + 8 s ^ {2} - 6 s + 4}{2 s ^ {3} (s - 1) (s ^ {2} + 1)}. \end{document} ]]></tex-math></disp-formula><p>Thus, we get</p><disp-formula id="equation-22"><tex-math id="math-59"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle L \Bigl \{u (x) \Bigr \} = \frac {2 (3 s - 2) (s ^ {8} + s ^ {6} - 2 s ^ {2} - 1)}{2 s (s - 1) (s ^ {8} + s ^ {6} - 2 s ^ {2} - 1)} = \frac {3 s - 2}{s (s - 1)}. \end{document} ]]></tex-math></disp-formula><p>Finally, taking the inverse Laplace transform of the equation, we suddenly have an analytical solution</p><disp-formula id="equation-23"><tex-math id="math-60"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \begin{array}{l} u (x) = L ^ {- 1} \Bigl \{\frac {3 s - 2}{s (s - 1)} \Bigr \} = L ^ {- 1} \Bigl \{\frac {s}{s (s - 1)} \Bigr \} + L ^ {- 1} \Bigl \{\frac {2 s - 2}{s (s - 1)} \Bigr \} \\ = L ^ {- 1} \Bigl \{\frac {1}{s - 1} \Bigr \} + 2 L ^ {- 1} \Bigl \{\frac {1}{s} \Bigr \} = e ^ {x} + 2. \end{array} \end{document} ]]></tex-math></disp-formula><p><target id="anchor-82353b1f-8e19-47b4-b3db-59949c43ad6e" target-type="reference-target"/></p><p><bold>Example 2.2.</bold><italic>Solve the Volterra integro-diferential problem:</italic></p><disp-formula id="equation-24"><tex-math id="math-61"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \begin{array}{l} u ^ {\prime \prime \prime} (x) + \int_ {0} ^ {x} (x - t) ^ {2} u ^ {\prime \prime} (t) d t = u (x) + \frac {9}{2} \sinh (x) - \frac {1}{2} x \cosh (x) - \frac {1}{2} x \sinh (x) - x - 1 + \\ \int_ {0} ^ {x} e ^ {x - t} u ^ {\prime} (t) d t, u (0) = 1, u ^ {\prime} (0) = - 1, u ^ {\prime \prime} (0) = 1. \end{array} \end{document} ]]></tex-math></disp-formula><p><italic>Solution</italic>. The first one, applying Laplace transform to the problem, we have</p><disp-formula id="equation-25"><tex-math id="math-62"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \begin{array}{l}\displaystyle L\left\{u'''(x)\right\} + L\left\{\int_{0}^{x} (x-t)^2 u''(t) dt\right\} = L\left\{u(x)\right\} \\\displaystyle + L\left\{\frac{9}{2}\sinh(x) - \frac{1}{2}x\cosh(x) - \frac{1}{2}x\sinh(x) - x - 1\right\} + L\left\{\int_{0}^{x} e^{x-t} u'(t) dt\right\}.\end{array} \end{document} ]]></tex-math></disp-formula><p>The second one, using the convolution theorem, we get</p><disp-formula id="equation-26"><tex-math id="math-63"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \begin{array}{l}\displaystyle L\left\{u'''(x)\right\} + L\left\{x^2\right\} L\left\{u''(x)\right\} = L\left\{u(x)\right\} \\\displaystyle = L\left\{\frac{9}{2}\sinh(x) - \frac{1}{2}x\cosh(x) - \frac{1}{2}x\sinh(x) - x - 1\right\} + L\left\{e^x\right\} L\left\{u'(x)\right\}.\end{array} \end{document} ]]></tex-math></disp-formula><p>The third one, taking the Laplace transform of derivatives, we have</p><disp-formula id="equation-27"><tex-math id="math-64"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \begin{array}{l}\displaystyle s^3 L\left\{u(x)\right\} - s^2 u(0) - s u'(0) - u''(0) + \frac{2}{s^3} \left[ s^2 L\left\{u(x)\right\} - s u(0) - u'(0) \right] \\\displaystyle = L\left\{u(x)\right\} + \frac{9}{2(s^2 - 1)} - \frac{(s^2 + 1)}{2(s^2 - 1)^2} - \frac{s}{(s^2 - 1)^2} - \frac{1}{s^2} - \frac{1}{s} \\\displaystyle + \frac{1}{s - 1} \left[ s L\left\{u(x)\right\} - u(0) \right]\end{array} \end{document} ]]></tex-math></disp-formula><p>and using initial conditions, we also obtain</p><disp-formula id="equation-28"><tex-math id="math-65"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \begin{array}{r l}& s ^ {3} L \Bigl \{u (x) \Bigr \} - s ^ {2} + s - 1 + \frac {2}{s ^ {3}} \Bigl [ s ^ {2} L \Bigl \{u (x) \Bigr \} - s + 1 \Bigr ] \\& = L \Bigl \{u (x) \Bigr \} + \frac {9}{2 (s ^ {2} - 1)} - \frac {(s ^ {2} + 1)}{2 (s ^ {2} - 1) ^ {2}} - \frac {s}{(s ^ {2} - 1) ^ {2}} - \frac {1}{s ^ {2}} - \frac {1}{s} \\& + \frac {1}{s - 1} \Bigl [ s L \Bigl \{u (x) \Bigr \} - 1 \Bigr ].\end{array} \end{document} ]]></tex-math></disp-formula><p>The fourth one, rearranging the equation, we certainly receive</p><disp-formula id="equation-29"><tex-math id="math-66"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \begin{array}{l} {\left(s ^ {3} + \frac {2}{s} - \frac {s}{s - 1} - 1\right) L \Big \{u (x) \Big \} = s ^ {2} - s + 1 + \frac {2}{s ^ {2}} - \frac {2}{s ^ {3}}} \\ {+ \frac {9}{2 (s ^ {2} - 1)} - \frac {(s ^ {2} + 1)}{2 (s ^ {2} - 1) ^ {2}} - \frac {s}{(s ^ {2} - 1) ^ {2}} - \frac {1}{s ^ {2}} - \frac {1}{s} - \frac {1}{s - 1}} \end{array} \end{document} ]]></tex-math></disp-formula><p>or</p><disp-formula id="equation-30"><tex-math id="math-67"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \begin{array}{l} \left[ \frac {s ^ {5} - s ^ {4} - 2 s ^ {2} + 3 s - 2}{s (s - 1)} \right] L \Bigl \{u (x) \Bigr \} \\ = \frac {2 s ^ {1 0} - 4 s ^ {9} + 2 s ^ {7} + 6 s ^ {6} - 8 s ^ {5} - 8 s ^ {4} + 1 6 s ^ {3} - 4 s ^ {2} - 6 s + 4}{2 s ^ {3} (s - 1) (s ^ {2} - 1) ^ {2}}. \end{array} \end{document} ]]></tex-math></disp-formula><p>Hence, we get</p><disp-formula id="equation-31"><tex-math id="math-68"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle L \Bigl \{u (x) \Bigr \} = \frac {2 (s ^ {2} - 1) (s ^ {3} - s ^ {2} + 1) (s ^ {5} - s ^ {4} - 2 s ^ {2} + 3 s - 2)}{2 s ^ {2} (s ^ {2} - 1) ^ {2} (s ^ {5} - s ^ {4} - 2 s ^ {2} + 3 s - 2)} = \frac {s ^ {3} - s ^ {2} + 1}{s ^ {2} (s ^ {2} - 1)}. \end{document} ]]></tex-math></disp-formula><p>The last one, taking the inverse Laplace transform of the equation, we suddenly have an exact solution</p><disp-formula id="equation-32"><tex-math id="math-69"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \begin{array}{l} u (x) = L ^ {- 1} \Bigl \{\frac {s ^ {3} - s ^ {2} + 1}{s ^ {2} (s ^ {2} - 1)} \Bigr \} = L ^ {- 1} \Bigl \{\frac {s ^ {3}}{s ^ {2} (s ^ {2} - 1)} \Bigr \} - L ^ {- 1} \Bigl \{\frac {s ^ {2} - 1}{s ^ {2} (s ^ {2} - 1)} \Bigr \} \\ = L ^ {- 1} \Bigl \{\frac {s}{s ^ {2} - 1} \Bigr \} - L ^ {- 1} \Bigl \{\frac {1}{s ^ {2}} \Bigr \} = \cosh x - x. \end{array} \end{document} ]]></tex-math></disp-formula><p>Here, we will emphasize on an approximate solution of this generalization for a non-convolution type given by</p><disp-formula id="equation-33"><tex-math id="math-70"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \begin{array}{l} \alpha u ^ {(n)} (x) + \beta \int_ {0} ^ {x} K _ {1} (x, t) u ^ {(m)} (t) d t \\ = A (x) u (x) + B (x) + \int_ {0} ^ {x} K _ {2} (x, t) u ^ {(p)} (t) d t, x \in [ 0, T ], \\ u (0) = c _ {0}, u ^ {'} (0) = c _ {1},..., u ^ {(n - 1)} (0) = c _ {n - 1}. \end{array}\tag{8} \end{document} ]]></tex-math></disp-formula><p>The approximation using the Laguerre polynomials is below: To start with Supposing that the function <inline-formula><tex-math id="math-71"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle u _ { \chi } ( x ) \end{document} ]]></tex-math></inline-formula> is an estimated solution of the equation (8) defined by</p><disp-formula id="equation-34"><tex-math id="math-72"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle u _ {\chi} (x) = \sum_ {k = 0} ^ {\chi} d _ {k} L _ {k} (x) = d _ {0} L _ {0} (x) + d _ {1} L _ {1} (x) + d _ {2} L _ {2} (x) + \dots + d _ {\chi} L _ {\chi} (x),\tag{9} \end{document} ]]></tex-math></disp-formula><p>where <inline-formula><tex-math id="math-73"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle m , n , p \leq \chi , L _ { k } ( x ) \end{document} ]]></tex-math></inline-formula> are the Laguerre polynomials and <inline-formula><tex-math id="math-74"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle d _ { k } \end{document} ]]></tex-math></inline-formula> are unknown constants, <inline-formula><tex-math id="math-75"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle k = 0 , 1 , . . . , \chi . \end{document} ]]></tex-math></inline-formula> . Then, writing equation (9) as a dot product, we have</p><disp-formula id="equation-35"><tex-math id="math-76"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle u _ {\chi} (x) = \left[ \begin{array}{c c c c c} L _ {0} (x) & L _ {1} (x) & L _ {2} (x) & ... & L _ {\chi} (x) \end{array} \right] \cdot \left[ \begin{array}{c} d _ {0} \\ d _ {1} \\ d _ {2} \\ \vdots \\ d _ {\chi} \end{array} \right].\tag{10} \end{document} ]]></tex-math></disp-formula><p>Next, rearranging the equation (10) in a matrix formula, we also have</p><disp-formula id="equation-36"><tex-math id="math-77"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle u _ {\chi} (x) = \left[ \begin{array}{c c c c c c} 1 & x & x ^ {2} & x ^ {3} & \ldots & x ^ {\chi} \end{array} \right] \left[ \begin{array}{c c c c c} b _ {0 0} & b _ {0 1} & b _ {0 2} & \dots & b _ {0 \chi} \\ 0 & b _ {1 1} & b _ {1 2} & \dots & b _ {1 \chi} \\ 0 & 0 & b _ {2 2} & \dots & b _ {2 \chi} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \dots & b _ {\chi \chi} \end{array} \right] \left[ \begin{array}{c} d _ {0} \\ d _ {1} \\ d _ {2} \\ \vdots \\ d _ {\chi} \end{array} \right] \end{document} ]]></tex-math></disp-formula><p>where <inline-formula><tex-math id="math-78"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle b _ { i j } \end{document} ]]></tex-math></inline-formula> are known constants. After that, finding the derivatives of <inline-formula><tex-math id="math-79"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle u _ { \chi } ( x ) \end{document} ]]></tex-math></inline-formula> , we have as follows:</p><disp-formula id="equation-37"><tex-math id="math-80"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle u _ {\chi} ^ {'} (x) = \left[ \begin{array}{l l l l l l} 0 & 1! & 2 x & 3 x ^ {2} & \dots & \chi x ^ {\chi - 1} \end{array} \right] \left[ \begin{array}{l l l l l} b _ {0 0} & b _ {0 1} & b _ {0 2} & \dots & b _ {0 \chi} \\ 0 & b _ {1 1} & b _ {1 2} & \dots & b _ {1 \chi} \\ 0 & 0 & b _ {2 2} & \dots & b _ {2 \chi} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \dots & b _ {\chi \chi} \end{array} \right] \left[ \begin{array}{l} d _ {0} \\ d _ {1} \\ d _ {2} \\ \vdots \\ d _ {\chi} \end{array} \right] \end{document} ]]></tex-math></disp-formula><disp-formula id="equation-38"><tex-math id="math-81"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle u _ {\chi} ^ {\prime \prime} (x) = \left[ \begin{array}{l l l l l l} 0 & 0 & 2! & 6 x & \ldots & \chi (\chi - 1) x ^ {\chi - 2} \end{array} \right] \left[ \begin{array}{c c c c c} b _ {0 0} & b _ {0 1} & b _ {0 2} & \dots & b _ {0 \chi} \\ 0 & b _ {1 1} & b _ {1 2} & \dots & b _ {1 \chi} \\ 0 & 0 & b _ {2 2} & \dots & b _ {2 \chi} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \dots & b _ {\chi \chi} \end{array} \right] \left[ \begin{array}{c} d _ {0} \\ d _ {1} \\ d _ {2} \\ \vdots \\ d _ {\chi} \end{array} \right] \end{document} ]]></tex-math></disp-formula><disp-formula id="equation-39"><tex-math id="math-82"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle u _ {\chi} ^ {\prime \prime \prime} (x) = \left[ \begin{array}{l l l l l l} 0 & 0 & 0 & 3! & \ldots & \chi (\chi - 1) (\chi - 2) x ^ {\chi - 3} \end{array} \right] \left[ \begin{array}{c c c c c} b _ {0 0} & b _ {0 1} & b _ {0 2} & \dots & b _ {0 \chi} \\ 0 & b _ {1 1} & b _ {1 2} & \dots & b _ {1 \chi} \\ 0 & 0 & b _ {2 2} & \dots & b _ {2 \chi} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \dots & b _ {\chi \chi} \end{array} \right] \left[ \begin{array}{c} d _ {0} \\ d _ {1} \\ d _ {2} \\ \vdots \\ d _ {\chi} \end{array} \right] \end{document} ]]></tex-math></disp-formula><p>and</p><disp-formula id="equation-40"><tex-math id="math-83"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \begin{array}{r c l} u _ {\chi} ^ {(n)} (x) & = & \left[ \begin{array}{c c c c c c} 0 & 0 & 0 & \ldots & n! & \ldots & \chi (\chi - 1) \ldots (\chi - n + 1) x ^ {\chi - n} \end{array} \right] \\ & & \left[ \begin{array}{c c c c c} b _ {0 0} & b _ {0 1} & b _ {0 2} & \dots & b _ {0 \chi} \\ 0 & b _ {1 1} & b _ {1 2} & \dots & b _ {1 \chi} \\ 0 & 0 & b _ {2 2} & \dots & b _ {2 \chi} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \dots & b _ {\chi \chi} \end{array} \right] \left[ \begin{array}{c} d _ {0} \\ d _ {1} \\ d _ {2} \\ \vdots \\ d _ {\chi} \end{array} \right] \end{array}\tag{11} \end{document} ]]></tex-math></disp-formula><p>Then, substituting the equation (11) into the equation (8), we receive</p><disp-formula id="equation-41"><tex-math id="math-84"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \begin{array}{l}\alpha \Big [ 0 0 0 \dots n! \dots \chi (\chi - 1) \dots (\chi - n + 1) x ^ {\chi - n} \Big ]\\\left[\begin{array}{c c c c c}b _ {0 0}&b _ {0 1}&b _ {0 2}&\dots&b _ {0 \chi}\\0&b _ {1 1}&b _ {1 2}&\dots&b _ {1 \chi}\\0&0&b _ {2 2}&\dots&b _ {2 \chi}\\\vdots&\vdots&\vdots&\ddots&\vdots\\0&0&0&\dots&b _ {\chi \chi}\end{array}\right].\left[\begin{array}{c}d _ {0}\\d _ {1}\\d _ {2}\\\vdots\\d _ {\chi}\end{array}\right]\\+ \beta \int_ {0} ^ {x} K _ {1} (x, t) \Big \{\Big [ 0 0 0 \dots m! \dots \chi (\chi - 1) \dots (\chi - m + 1) t ^ {\chi - m} \Big ]\\\left[\begin{array}{c c c c c}b _ {0 0}&b _ {0 1}&b _ {0 2}&\dots&b _ {0 \chi}\\0&b _ {1 1}&b _ {1 2}&\dots&b _ {1 \chi}\\0&0&b _ {2 2}&\cdot \cdot \cdot&b _ {2 \chi}\\\vdots&\vdots&\vdots&\ddots&\vdots\\0&0&0&\dots&b _ {\chi \chi}\end{array}\right].\left[ \right.\begin{array}{c}d _ {0}\\d _ {1}\\d _ {2}\\\vdots\\d _ {\chi}\end{array}\bigg ] \Big \} d t\\= A (x) \Big [ 1 x x ^ {2} x ^ {3} \dots x ^ {\chi} \Big ]\left[\begin{array}{c c c c c}b _ {0 0}&b _ {0 1}&b _ {0 2}&\dots&b _ {0 \chi}\\0&b _ {1 1}&b _ {1 2}&\dots&b _ {1 \chi}\\0&0&b _ {2 2}&\dots&b _ {1 \chi}\\\vdots&\vdots&\vdots&\ddots&\vdots\\0&0&0&\dots&b _ {\chi \chi}\end{array}\right].\left[\begin{array}{c}d _ {0}\\d _ {1}\\d _ {2}\\\vdots\\d _ {\chi}\end{array}\right]\\+ B (x) + \int_ {0} ^ {x} K _ {2} (x, t) \Bigl \{\Big [ 0 0 0 \dots p! \dots \chi (\chi - 1)... (\chi - p + 1) t ^ {\chi - p} \Bigr ]\\\end{array} \end{document} ]]></tex-math></disp-formula><disp-formula id="equation-42"><tex-math id="math-85"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \left[\begin{array}{c c c c c}b _ {0 0} & b _ {0 1} & b _ {0 2} & \dots & b _ {0 \chi} \\0 & b _ {1 1} & b _ {1 2} & \dots & b _ {1 \chi} \\0 & 0 & b _ {2 2} & \dots & b _ {2 \chi} \\\vdots & \vdots & \vdots & \ddots & \vdots \\0 & 0 & 0 & \dots & b _ {\chi \chi}\end{array}\right]\cdot\left[\begin{array}{c}d _ {0} \\d _ {1} \\d _ {2} \\\vdots \\d _ {\chi}\end{array}\right]\Big \} d t.\tag{12} \end{document} ]]></tex-math></disp-formula><p>Simplifying and integrating the equation <xref ref-type="disp-formula" rid="equation-3">(12)</xref>, we then have the new equation with unknown constants <inline-formula><tex-math id="math-86"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle d _ { 0 } , d _ { 1 } , . . . , d _ { \chi } \end{document} ]]></tex-math></inline-formula> . In order to determine <inline-formula><tex-math id="math-87"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle d _ { 0 } , d _ { 1 } , . . . , d _ { \chi } \end{document} ]]></tex-math></inline-formula> , using <italic>n</italic> initial conditions and selecting <inline-formula><tex-math id="math-88"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle x _ { i } \in [ 0 , T ] , i = 1 , 2 , . . . , \chi { - n + 1 } \end{document} ]]></tex-math></inline-formula> , with substituting in the new equation, we get a system of linear algrbraic equations of <inline-formula><tex-math id="math-89"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \chi + 1 \end{document} ]]></tex-math></inline-formula> unknown constants. Solving this system by a program, we have the values of the unknown constants, that is, the numerical solution of the initial-value problem (8) is obtained.</p><p>In order to guarantee the convergence of this method, we will verify as follows. Let <inline-formula><tex-math id="math-90"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle u ( x ) \end{document} ]]></tex-math></inline-formula> be an analytical solution of initial-value problem (8) that has derivatives of all orders at <inline-formula><tex-math id="math-91"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle x = 0 \end{document} ]]></tex-math></inline-formula> . Then, the Taylor series of <inline-formula><tex-math id="math-92"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle u ( x ) \operatorname { a t } x = 0 \end{document} ]]></tex-math></inline-formula> is defined by</p><disp-formula id="equation-43"><tex-math id="math-93"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle u (x) = u (0) + u ^ {\prime} (0) x + \frac {1}{2 !} u ^ {\prime \prime} (0) x ^ {2} + \dots + \frac {1}{\chi !} u ^ {(\chi)} (0) x ^ {\chi} + \dots . \end{document} ]]></tex-math></disp-formula><p>Thus, by the definition and process to find <inline-formula><tex-math id="math-94"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle u _ { \chi } ( x ) \end{document} ]]></tex-math></inline-formula> , we obtain that</p><disp-formula id="equation-44"><tex-math id="math-95"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle | u (x) - u _ {\chi} (x) | \leq | \frac {1}{(\chi + 1) !} u ^ {(\chi + 1)} (0) x ^ {\chi + 1} | + | \frac {1}{(\chi + 2) !} u ^ {(\chi + 2)} (0) x ^ {\chi + 2} | + \dots . \end{document} ]]></tex-math></disp-formula><p>Here, it is suficient to show that <inline-formula><tex-math id="math-96"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \scriptstyle { \frac { 1 } { \chi ! } } x ^ { \chi } \end{document} ]]></tex-math></inline-formula> converges to 0 as <inline-formula><tex-math id="math-97"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \chi \infty \end{document} ]]></tex-math></inline-formula> to confirm that <inline-formula><tex-math id="math-98"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle | u ( x ) - u _ { \chi } ( x ) | \end{document} ]]></tex-math></inline-formula> converges to 0 as <inline-formula><tex-math id="math-99"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \chi \to \infty \end{document} ]]></tex-math></inline-formula> . Since <inline-formula><tex-math id="math-100"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \begin{array} { r } { e \bigl ( \frac { \chi } { e } \bigr ) ^ { \chi } \leq \chi ! \leq e \bigl ( \frac { \chi + 1 } { e } \bigr ) ^ { \chi + 1 } } \end{array} \end{document} ]]></tex-math></inline-formula> , we get <inline-formula><tex-math id="math-101"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \begin{array} { r } { \frac { 1 } { \chi + 1 } \big ( \frac { e x } { \chi + 1 } \big ) ^ { \chi } \leq \frac { 1 } { \chi ! } x ^ { \chi } \leq \frac { 1 } { e } \big ( \frac { e x } { \chi } \big ) ^ { \chi } } \end{array} \end{document} ]]></tex-math></inline-formula> . It is easy to determine that <inline-formula><tex-math id="math-102"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \textstyle { \frac { 1 } { \chi + 1 } } \left( { \frac { e x } { \chi + 1 } } \right) ^ { \chi } \end{document} ]]></tex-math></inline-formula> and <inline-formula><tex-math id="math-103"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \textstyle { \frac { 1 } { e } } \left( { \frac { e x } { \chi } } \right) ^ { \chi } \end{document} ]]></tex-math></inline-formula> converge to 0 as as <inline-formula><tex-math id="math-104"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \chi \infty \end{document} ]]></tex-math></inline-formula> . This means that <inline-formula><tex-math id="math-105"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \scriptstyle { \frac { 1 } { \chi ! } } x ^ { \chi } \end{document} ]]></tex-math></inline-formula> converges to 0 as <inline-formula><tex-math id="math-106"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \chi \infty \end{document} ]]></tex-math></inline-formula> to confirm the convergence.<target id="anchor-054f64a5-e909-41e5-88fb-5d707eafb899" target-type="reference-target"/></p><p><bold>Example 2.3.</bold><italic>Estimate a solution of the linear Volterra integro-diferential problem using </italic><inline-formula><tex-math id="math-107"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle u _ { 6 } ( x ) \colon u ^ { ( 5 ) } ( x ) + \int _ { 0 } ^ { x } x t u ^ { \prime \prime \prime } ( t ) d t = x u ( x ) - { \frac { 3 } { 4 } } x ^ { 5 } - x ^ { 4 } + { \frac { 9 } { 2 } } x ^ { 3 } + 3 x ^ { 2 } - 2 x e ^ { x } + \end{document} ]]></tex-math></inline-formula><inline-formula><tex-math id="math-108"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle 2 e ^ { x } + \int _ { 0 } ^ { x } x t u ^ { ' } ( t ) d t , u ( 0 ) = 2 , u ^ { ' } ( 0 ) = - 1 , u ^ { ' \prime } ( 0 ) = 2 , u ^ { ' \prime \prime } ( 0 ) = 8 , u ^ { ( 4 ) } ( 0 ) = 2 , 0 \leq u ^ { ( 8 ) } ( 0 ) = 1 \end{document} ]]></tex-math></inline-formula><inline-formula><tex-math id="math-109"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle x \le 2 \end{document} ]]></tex-math></inline-formula> . <italic>An exact solution is</italic><inline-formula><tex-math id="math-110"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle u ( x ) = x ^ { 3 } - 3 x + 2 e ^ { x } \end{document} ]]></tex-math></inline-formula></p><p><italic>Solution</italic>. First, suppose that a function <inline-formula><tex-math id="math-111"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle u _ { 6 } ( x ) \end{document} ]]></tex-math></inline-formula> is an approximate solution of this problem, that is,</p><disp-formula id="equation-45"><tex-math id="math-112"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \begin{array}{l} u _ {6} (x) = d _ {0} L _ {0} (x) + d _ {1} L _ {1} (x) + d _ {2} L _ {2} (x) + \dots + d _ {6} L _ {6} (x) \\ = d _ {0} (1) + d _ {1} (- x + 1) + d _ {2} \big [ \frac {1}{2} (x ^ {2} - 4 x + 2) \big ] + d _ {3} \big [ \frac {1}{6} (- x ^ {3} + 9 x ^ {2} - 1 8 x + 6) \big ] \\ + d _ {4} \big [ \frac {1}{2 4} (x ^ {4} - 1 6 x ^ {3} + 7 2 x ^ {2} - 9 6 x + 2 4) \big ] \\ + d _ {5} \big [ \frac {1}{1 2 0} (- x ^ {5} + 2 5 x ^ {4} - 2 0 0 x ^ {3} + 6 0 0 x ^ {2} - 6 0 0 x + 1 2 0) \big ] \\ + d _ {6} \big [ \frac {1}{7 2 0} (x ^ {6} - 3 6 x ^ {5} + 4 5 0 x ^ {4} - 2 4 0 0 x ^ {3} + 5 4 0 0 x ^ {2} - 4 3 2 0 x + 7 2 0) \big ]. \end{array} \end{document} ]]></tex-math></disp-formula><p>Second, finding derivatives of <inline-formula><tex-math id="math-113"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle u _ { 6 } ( x ) \end{document} ]]></tex-math></inline-formula> , we have as follows:</p><disp-formula id="equation-46"><tex-math id="math-114"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \begin{array}{l} u _ {6} ^ {'} (x) = d _ {1} (- 1) + d _ {2} \big [ \frac {1}{2} (2 x - 4) \big ] + d _ {3} \big [ \frac {1}{6} (- 3 x ^ {2} + 1 8 x - 1 8) \big ] \\ + d _ {4} \big [ \frac {1}{2 4} (4 x ^ {3} - 4 8 x ^ {2} + 1 4 4 x - 9 6) \big ] \\ + d _ {5} \big [ \frac {1}{1 2 0} (- 5 x ^ {4} + 1 0 0 x ^ {3} - 6 0 0 x ^ {2} + 1 2 0 0 x - 6 0 0) \big ] \\ + d _ {6} \big [ \frac {1}{7 2 0} (6 x ^ {5} - 1 8 0 x ^ {4} + 1 8 0 0 x ^ {3} - 7 2 0 0 x ^ {2} + 1 0 8 0 0 x - 4 3 2 0) \big ], \end{array} \end{document} ]]></tex-math></disp-formula><disp-formula id="equation-47"><tex-math id="math-115"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \begin{array}{l}u _ {6} ^ {''} (x)= d _ {2} \big [ \frac {1}{2} (2) \big ]+ d _ {3} \big [ \frac {1}{6} (- 6 x + 1 8) \big ]+ d _ {4} \big [ \frac {1}{2 4} (1 2 x ^ {2} - 9 6 x + 1 4 4) \big ] \\+ d _ {5} \big [ \frac {1}{1 2 0} (- 2 0 x ^ {3} + 3 0 0 x ^ {2} - 1 2 0 0 x + 1 2 0 0) \big ] \\+ d _ {6} \big [ \frac {1}{7 2 0} (3 0 x ^ {4} - 7 2 0 x ^ {3} + 5 4 0 0 x ^ {2} - 1 4 4 0 0 x + 1 0 8 0 0) \big ], \\[1ex]u _ {6} ^ {'''} (x)= d _ {3} \big [ \frac {1}{6} (- 6) \big ]+ d _ {4} \big [ \frac {1}{2 4} (2 4 x - 9 6) \big ] \\+ d _ {5} \big [ \frac {1}{1 2 0} (- 6 0 x ^ {2} + 6 0 0 x - 1 2 0 0) \big ] \\+ d _ {6} \big [ \frac {1}{7 2 0} (1 2 0 x ^ {3} - 2 1 6 0 x ^ {2} + 1 0 8 0 0 x - 1 4 4 0 0) \big ], \\[1ex]u _ {6} ^ {(4)} (x)= d _ {4} \big [ \frac {1}{2 4} (2 4) \big ]+ d _ {5} \big [ \frac {1}{1 2 0} (- 1 2 0 x + 6 0 0) \big ] \\+ d _ {6} \big [ \frac {1}{7 2 0} (3 6 0 x ^ {2} - 4 3 2 0 x + 1 0 8 0 0) \big ].\end{array} \end{document} ]]></tex-math></disp-formula><p>and</p><disp-formula id="equation-48"><tex-math id="math-116"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle u _ {6} ^ {(5)} (x) = d _ {5} \big [ \frac {1}{1 2 0} (- 1 2 0) \big ] + d _ {6} \big [ \frac {1}{7 2 0} (7 2 0 x - 4 3 2 0) \big ]. \end{document} ]]></tex-math></disp-formula><p>Third, substituting the derivatives into the problem, we receive</p><disp-formula id="equation-49"><tex-math id="math-117"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \begin{array}{l} d _ {5} \left[ \frac {1}{1 2 0} (- 1 2 0) \right] + d _ {6} \left[ \frac {1}{7 2 0} (7 2 0 x - 4 3 2 0) \right] \\ + \int_ {0} ^ {x} x t \left\{d _ {3} \left[ \frac {1}{6} (- 6) \right] + d _ {4} \left[ \frac {1}{2 4} (2 4 t - 9 6) \right] + d _ {5} \left[ \frac {1}{1 2 0} (- 6 0 t ^ {2} + 6 0 0 t - 1 2 0 0) \right] \right. \\ + d _ {6} \left[ \frac {1}{7 2 0} (1 2 0 t ^ {3} - 2 1 6 0 t ^ {2} + 1 0 8 0 0 t - 1 4 4 0 0) \right] \Bigg \} d t \\ = x \left\{d _ {0} (1) + d _ {1} (- x + 1) + d _ {2} \left[ \frac {1}{2} (x ^ {2} - 4 x + 2) \right] \right. \\ + d _ {3} \left[ \frac {1}{6} (- x ^ {3} + 9 x ^ {2} - 1 8 x + 6) \right] + d _ {4} \left[ \frac {1}{2 4} (x ^ {4} - 1 6 x ^ {3} + 7 2 x ^ {2} - 9 6 x + 2 4) \right] \\ + d _ {5} \left[ \frac {1}{1 2 0} (- x ^ {5} + 2 5 x ^ {4} - 2 0 0 x ^ {3} + 6 0 0 x ^ {2} - 6 0 0 x + 1 2 0) \right] \\ + d _ {6} \left[ \frac {1}{7 2 0} (x ^ {6} - 3 6 x ^ {5} + 4 5 0 x ^ {4} - 2 4 0 0 x ^ {3} + 5 4 0 0 x ^ {2} - 4 3 2 0 x + 7 2 0) \right] \Bigg \} \\ - \frac {3}{4} x ^ {5} - x ^ {4} + \frac {9}{2} x ^ {3} + 3 x ^ {2} - 2 x e ^ {x} + 2 e ^ {x} \end{array} \end{document} ]]></tex-math></disp-formula><disp-formula id="equation-50"><tex-math id="math-118"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \begin{array}{l} + \int_ {0} ^ {x} x t \Big \{d _ {1} (- 1) + d _ {2} \big [ \frac {1}{2} (2 t - 4) \big ] + d _ {3} \big [ \frac {1}{6} (- 3 t ^ {2} + 1 8 t - 1 8) \big ] \\ + d _ {4} \big [ \frac {1}{2 4} (4 t ^ {3} - 4 8 t ^ {2} + 1 4 4 t - 9 6) \big ] \\ + d _ {5} \big [ \frac {1}{1 2 0} (- 5 t ^ {4} + 1 0 0 t ^ {3} - 6 0 0 t ^ {2} + 1 2 0 0 t - 6 0 0) \big ] \\ + d _ {6} \big [ \frac {1}{7 2 0} (6 t ^ {5} - 1 8 0 t ^ {4} + 1 8 0 0 t ^ {3} - 7 2 0 0 t ^ {2} + 1 0 8 0 0 t - 4 3 2 0) \big ] \Big \} d t. \end{array} \end{document} ]]></tex-math></disp-formula><p>Fourth, simplifying and integrating the equation, we obtain the new equation. Selecting <inline-formula><tex-math id="math-119"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle x _ { 1 } = 0 . 2 5 \end{document} ]]></tex-math></inline-formula> and <inline-formula><tex-math id="math-120"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle x _ { 2 } = 0 . 5 \end{document} ]]></tex-math></inline-formula> to substitute in the new equation with using 5 initial conditions, we get the following system:</p><disp-formula id="equation-51"><tex-math id="math-121"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \begin{array}{l} d _ {0} + d _ {1} + d _ {2} + d _ {3} + d _ {4} + d _ {5} + d _ {6} = 2, \\ - d _ {1} - 2 d _ {2} - 3 d _ {3} - 4 d _ {4} - 5 d _ {5} - 6 d _ {6} = - 1, \\ d _ {2} + 3 d _ {3} + 6 d _ {4} + 1 0 d _ {5} + 1 5 d _ {6} = 2, \\ - d _ {3} - 4 d _ {4} - 1 0 d _ {5} - 2 0 d _ {6} = 8, \\ d _ {4} + 5 d _ {5} + 1 5 d _ {6} = 2, \\ - 0. 2 5 d _ {0} - 0. 1 7 9 6 8 7 d _ {1} - 0. 1 1 8 4 8 9 d _ {2} - 0. 0 7 3 4 4 5 d _ {3} - 0. 0 5 0 3 4 1 d _ {4} \\ - 1. 0 5 3 8 3 1 d _ {5} - 5. 8 3 7 5 4 2 d _ {6} = 2. 1 7 9 2 1 1, \\ - 0. 5 d _ {0} - 0. 1 8 7 5 d _ {1} + 0. 0 4 1 6 6 6 d _ {2} + 0. 1 3 9 3 2 2 d _ {3} + 0. 0 7 6 3 0 2 d _ {4} \\ - 1. 1 6 1 2 7 3 d _ {5} - 6. 0 7 5 2 1 1 d _ {6} = 2. 8 7 5 2 8 3. \end{array} \end{document} ]]></tex-math></disp-formula><p>At last, solving the system by a program, we have</p><disp-formula id="equation-52"><tex-math id="math-122"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \begin{array}{l} d _ {0} = 1 7. 7 5 7 0 1 5, d _ {1} = - 6 1. 7 0 3 7 0 8, d _ {2} = 1 0 0. 1 6 3 3 1 3, d _ {3} = - 9 2. 7 5 6 4 7 5, \\ d _ {4} = 5 4. 9 7 1 3 9 9, d _ {5} = - 1 9. 3 5 0 1 7 7, d _ {6} = 2. 9 1 8 6 3 2. \end{array} \end{document} ]]></tex-math></disp-formula><p>Therefore, the numerical solution is</p><disp-formula id="equation-53"><tex-math id="math-123"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \begin{array}{l} u _ {6} (x) = 1 7. 7 5 7 0 1 5 - 6 1. 7 0 3 7 0 8 (- x + 1) + 1 0 0. 1 6 3 3 1 3 \big [ \frac {1}{2} (x ^ {2} - 4 x + 2) \big ] \\ - 9 2. 7 5 6 4 7 5 \big [ \frac {1}{6} (- x ^ {3} + 9 x ^ {2} - 1 8 x + 6) \big ] \\ + 5 4. 9 7 1 3 9 9 \big [ \frac {1}{2 4} (x ^ {4} - 1 6 x ^ {3} + 7 2 x ^ {2} - 9 6 x + 2 4) \big ] \\ - 1 9. 3 5 0 1 7 7 \big [ \frac {1}{1 2 0} (- x ^ {5} + 2 5 x ^ {4} - 2 0 0 x ^ {3} + 6 0 0 x ^ {2} - 6 0 0 x + 1 2 0) \big ] \\ + 2. 9 1 8 6 3 2 \big [ \frac {1}{7 2 0} (x ^ {6} - 3 6 x ^ {5} + 4 5 0 x ^ {4} - 2 4 0 0 x ^ {3} + 5 4 0 0 x ^ {2} - 4 3 2 0 x + 7 2 0) \big ]. \end{array} \end{document} ]]></tex-math></disp-formula><table-wrap id="table-1"><label>Table 1</label><caption><p>Values of exact and approximate u _ { 6 } ( x ) solutions for example <xref ref-type="custom" custom-type="reference-target" rid="anchor-054f64a5-e909-41e5-88fb-5d707eafb899">2.3.</xref></p></caption><table><colgroup><col></col><col></col><col></col><col></col></colgroup><thead><tr><th scope="col">x</th><th scope="col">Exact solution</th><th scope="col">Approx. solution</th><th scope="col">Absolute error</th></tr></thead><tbody><tr><td>0.00</td><td>2.000000</td><td>1.999999</td><td>0.000001</td></tr><tr><td>0.20</td><td>1.850806</td><td>1.850805</td><td>0.000001</td></tr><tr><td>0.40</td><td>1.847649</td><td>1.847640</td><td>0.000009</td></tr><tr><td>0.60</td><td>2.060238</td><td>2.060181</td><td>0.000057</td></tr><tr><td>0.80</td><td>2.563082</td><td>2.562883</td><td>0.000199</td></tr><tr><td>1.00</td><td>3.436564</td><td>3.436040</td><td>0.000523</td></tr><tr><td>1.20</td><td>4.768234</td><td>4.767025</td><td>0.001209</td></tr><tr><td>1.40</td><td>6.654400</td><td>6.651716</td><td>0.002684</td></tr><tr><td>1.60</td><td>9.202065</td><td>9.196116</td><td>0.005949</td></tr><tr><td>1.80</td><td>12.531295</td><td>12.518153</td><td>0.013142</td></tr><tr><td>2.00</td><td>16.778112</td><td>16.749669</td><td>0.028443</td></tr></tbody></table></table-wrap><fig id="figure-1"><label>Figure 1</label><caption><p>Graphs of exact and approximate u6(x) solutions for example <xref ref-type="custom" custom-type="reference-target" rid="anchor-054f64a5-e909-41e5-88fb-5d707eafb899">2.3.</xref></p></caption><graphic xlink:href="https://jims-a.org/index.php/jimsa/article/download/1883/562/13990" mime-subtype="jpeg" mimetype="image"><alt-text>Figure 1</alt-text></graphic></fig><p><target id="anchor-a4ae94f0-de73-4f35-b7ef-0cc4c545565d" target-type="reference-target"/></p><p><bold>Example 2.4.</bold><italic>Approximate a solution of the linear Volterra integro-diferential problem using</italic><inline-formula><tex-math id="math-124"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle u _ { 4 } ( x ) \end{document} ]]></tex-math></inline-formula><italic>and</italic><inline-formula><tex-math id="math-125"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle u _ { 5 } ( x ) \colon u ^ { ^ { \prime \prime \prime } } ( x ) + \int _ { 0 } ^ { x } x \sin ( t ) u ^ { ^ { \prime } } ( t ) d t = - 2 x ^ { 2 } u ( x ) - \cos ( x ) - \end{document} ]]></tex-math></inline-formula> 4x cos <inline-formula><tex-math id="math-126"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \begin{array} { l } { { \langle x \rangle - x \sin ( x ) + 4 x ^ { 3 } + 2 x ^ { 2 } + 4 x + \int _ { 0 } ^ { x } x \cos ( t ) u ( t ) d t , u ( 0 ) = 1 , u ^ { ' } ( 0 ) = 3 , u ^ { ' \prime } ( 0 ) = 1 , u ^ { ' \prime } ( 0 ) = 1 , u ^ { ' \prime } ( 0 ) = 1 } } \end{array} \end{document} ]]></tex-math></inline-formula><inline-formula><tex-math id="math-127"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle 0 , 0 \leq x \leq \pi \end{document} ]]></tex-math></inline-formula> . <italic>An exact solution is</italic><inline-formula><tex-math id="math-128"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle u ( x ) = \sin ( x ) + 2 x + 1 \end{document} ]]></tex-math></inline-formula></p><p><italic>Solution</italic>. Firstly, let <inline-formula><tex-math id="math-129"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle u _ { 4 } ( x ) \end{document} ]]></tex-math></inline-formula> is an approximate solution of this problem, that is,</p><disp-formula id="equation-54"><tex-math id="math-130"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \begin{array}{l} u _ {4} (x) = d _ {0} L _ {0} (x) + d _ {1} L _ {1} (x) + d _ {2} L _ {2} (x) + d _ {3} L _ {3} (x) + d _ {4} L _ {4} (x) \\ = d _ {0} (1) + d _ {1} (- x + 1) + d _ {2} \big [ \frac {1}{2} (x ^ {2} - 4 x + 2) \big ] + d _ {3} \big [ \frac {1}{6} (- x ^ {3} + 9 x ^ {2} - 1 8 x + 6) \big ] \\ + d _ {4} \big [ \frac {1}{2 4} (x ^ {4} - 1 6 x ^ {3} + 7 2 x ^ {2} - 9 6 x + 2 4) \big ]. \end{array} \end{document} ]]></tex-math></disp-formula><p>Secondly, finding derivatives of <inline-formula><tex-math id="math-131"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle u _ { 4 } ( x ) \end{document} ]]></tex-math></inline-formula> , we have as follows:</p><disp-formula id="equation-55"><tex-math id="math-132"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \begin{aligned}u_{4}'(x) &= d_{1}(-1)+ d_{2}\left[\frac{1}{2}(2x-4)\right]+ d_{3}\left[\frac{1}{6}(-3x^{2}+18x-18)\right] \\&\quad + d_{4}\left[\frac{1}{24}(4x^{3}-48x^{2}+144x-96)\right], \\[6pt]u_{4}''(x) &= d_{2}\left[\frac{1}{2}(2)\right]+ d_{3}\left[\frac{1}{6}(-6x+18)\right]+ d_{4}\left[\frac{1}{24}(12x^{2}-96x+144)\right].\end{aligned} \end{document} ]]></tex-math></disp-formula><p>and</p><disp-formula id="equation-56"><tex-math id="math-133"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle u _ {4} ^ {\prime \prime \prime} (x) = d _ {3} \left[ \frac {1}{6} (- 6) \right] + d _ {4} \left[ \frac {1}{2 4} (2 4 x - 9 6) \right]. \end{document} ]]></tex-math></disp-formula><p>Thirdly, substituting the derivatives into the problem, we receive</p><disp-formula id="equation-57"><tex-math id="math-134"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \begin{array}{l} d _ {3} \big [ \frac {1}{6} (- 6) \big ] + d _ {4} \big [ \frac {1}{2 4} (2 4 x - 9 6) \big ] \\ + \int_ {0} ^ {x} x \sin (t) \Big \{d _ {1} (- 1) + d _ {2} \big [ \frac {1}{2} (2 t - 4) \big ] + d _ {3} \big [ \frac {1}{6} (- 3 t ^ {2} + 1 8 t - 1 8) \big ] \\ + d _ {4} \big [ \frac {1}{2 4} (4 t ^ {3} - 4 8 t ^ {2} + 1 4 4 t - 9 6) \big ] \Big \} d t \\ = - 2 x ^ {2} \Big \{d _ {0} (1) + d _ {1} (- x + 1) + d _ {2} \big [ \frac {1}{2} (x ^ {2} - 4 x + 2) \big ] \\ + d _ {3} \big [ \frac {1}{6} (- x ^ {3} + 9 x ^ {2} - 1 8 x + 6) \big ] + d _ {4} \big [ \frac {1}{2 4} (x ^ {4} - 1 6 x ^ {3} + 7 2 x ^ {2} - 9 6 x + 2 4) \big ] \Big \} \\ - \cos (x) - 4 x \cos (x) - x \sin (x) + 4 x ^ {3} + 2 x ^ {2} + 4 x \\ + \int_ {0} ^ {x} x \cos (t) \Big \{d _ {0} (1) + d _ {1} (- t + 1) + d _ {2} \big [ \frac {1}{2} (t ^ {2} - 4 t + 2) \big ] \\ + d _ {3} \big [ \frac {1}{6} (- t ^ {3} + 9 t ^ {2} - 1 8 t + 6) \big ] + d _ {4} \big [ \frac {1}{2 4} (t ^ {4} - 1 6 t ^ {3} + 7 2 t ^ {2} - 9 6 t + 2 4) \big ] \Big \} d t. \end{array} \end{document} ]]></tex-math></disp-formula><p>Fourthly, simplifying and integrating the equation, we obtain the new equation. Selecting <inline-formula><tex-math id="math-135"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle x _ { 1 } = 0 . 2 5 \end{document} ]]></tex-math></inline-formula> and <inline-formula><tex-math id="math-136"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle x _ { 2 } = 0 . 5 \end{document} ]]></tex-math></inline-formula> to substitute in the new equation with using 3 initial conditions, we get the following system:</p><disp-formula id="equation-58"><tex-math id="math-137"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \begin{array}{l} d _ {0} + d _ {1} + d _ {2} + d _ {3} + d _ {4} = 1, \\ - d _ {1} - 2 d _ {2} - 3 d _ {3} - 4 d _ {4} = 3, \\ d _ {2} + 3 d _ {3} + 6 d _ {4} = 0, \\ 0. 0 6 3 1 4 9 d _ {0} + 0. 0 3 1 8 1 7 d _ {1} + 0. 0 0 5 0 4 8 d _ {2} - 1. 0 1 7 5 6 7 d _ {3} \\ - 3. 7 8 6 4 0 8 d _ {4} = - 0. 8 1 2 1 7 5, \\ 0. 2 6 0 2 8 7 d _ {0} + 0. 0 0 7 7 2 6 d _ {1} - 0. 1 7 1 6 6 4 d _ {2} - 1. 2 9 0 9 0 5 d _ {3} \\ - 3. 8 6 1 3 2 7 d _ {4} = 0. 1 2 7 5 3 9. \end{array} \end{document} ]]></tex-math></disp-formula><p>Finally, solving the system by a program, we get</p><disp-formula id="equation-59"><tex-math id="math-138"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle d _ {0} = 3. 3 0 6 0 7 1, d _ {1} = - 1. 2 8 4 8 3 8, d _ {2} = - 0. 9 8 1 9 1 3, d _ {3} = - 0. 4 0 5 9 4 4, d _ {4} = 0. 3 6 6 6 2 4. \end{document} ]]></tex-math></disp-formula><p>Therefore, the numerical solution is</p><disp-formula id="equation-60"><tex-math id="math-139"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \begin{array}{l} u _ {4} (x) = 3. 3 0 6 0 7 1 - 1. 2 8 4 8 3 8 (- x + 1) - 0. 9 8 1 9 1 3 \big [ \frac {1}{2} (x ^ {2} - 4 x + 2) \big ] \\ - 0. 4 0 5 9 4 4 \big [ \frac {1}{6} (- x ^ {3} + 9 x ^ {2} - 1 8 x + 6) \big ] \\ + 0. 3 6 6 6 2 4 \big [ \frac {1}{2 4} (x ^ {4} - 1 6 x ^ {3} + 7 2 x ^ {2} - 9 6 x + 2 4) \big ]. \end{array} \end{document} ]]></tex-math></disp-formula><table-wrap id="table-2"><label>Table 2</label><caption><p>Values of exact and approximate u _ { 4 } ( x ) solutions for example <xref ref-type="custom" custom-type="reference-target" rid="anchor-a4ae94f0-de73-4f35-b7ef-0cc4c545565d">2.4.</xref></p></caption><table><colgroup><col></col><col></col><col></col><col></col></colgroup><thead><tr><th scope="col">x</th><th scope="col">Exact solution</th><th scope="col">Approx. solution</th><th scope="col">Absolute error</th></tr></thead><tbody><tr><td>0.00</td><td>1.000000</td><td>1.000000</td><td>0.000000</td></tr><tr><td>0.31</td><td>1.937336</td><td>1.937146</td><td>0.000190</td></tr><tr><td>0.63</td><td>2.844422</td><td>2.843491</td><td>0.000931</td></tr><tr><td>0.94</td><td>3.693973</td><td>3.691509</td><td>0.002464</td></tr><tr><td>1.26</td><td>4.464331</td><td>4.457244</td><td>0.007087</td></tr><tr><td>1.57</td><td>5.141593</td><td>5.120311</td><td>0.021282</td></tr><tr><td>1.88</td><td>5.720968</td><td>5.663897</td><td>0.057071</td></tr><tr><td>2.20</td><td>6.207247</td><td>6.074761</td><td>0.132486</td></tr><tr><td>2.51</td><td>6.614333</td><td>6.343231</td><td>0.271102</td></tr><tr><td>2.83</td><td>6.963884</td><td>6.463209</td><td>0.500675</td></tr><tr><td>3.14</td><td>7.283185</td><td>6.432166</td><td>0.851019</td></tr></tbody></table></table-wrap><fig id="figure-2"><label>Figure 2</label><caption><p>Graphs of exact and approximate u4(x) solutions for example <xref ref-type="custom" custom-type="reference-target" rid="anchor-a4ae94f0-de73-4f35-b7ef-0cc4c545565d">2.4.</xref></p></caption><graphic xlink:href="https://jims-a.org/index.php/jimsa/article/download/1883/562/13991" mime-subtype="jpeg" mimetype="image"><alt-text>Figure 2</alt-text></graphic></fig><p>For the approximate solution <inline-formula><tex-math id="math-140"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle u _ { 5 } ( x ) \end{document} ]]></tex-math></inline-formula> , the first one, let <inline-formula><tex-math id="math-141"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle u _ { 5 } ( x ) \end{document} ]]></tex-math></inline-formula> be an approximate solution of this problem, that is,</p><disp-formula id="equation-61"><tex-math id="math-142"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \begin{array}{l} u _ {5} (x) = d _ {0} L _ {0} (x) + d _ {1} L _ {1} (x) + d _ {2} L _ {2} (x) + d _ {3} L _ {3} (x) + d _ {4} L _ {4} (x) + d _ {5} L _ {5} (x) \\ = d _ {0} (1) + d _ {1} (- x + 1) + d _ {2} \big [ \frac {1}{2} (x ^ {2} - 4 x + 2) \big ] + d _ {3} \big [ \frac {1}{6} (- x ^ {3} + 9 x ^ {2} - 1 8 x + 6) \big ] \\ + d _ {4} \big [ \frac {1}{2 4} (x ^ {4} - 1 6 x ^ {3} + 7 2 x ^ {2} - 9 6 x + 2 4) \big ] \\ + d _ {5} \big [ \frac {1}{1 2 0} (- x ^ {5} + 2 5 x ^ {4} - 2 0 0 x ^ {3} + 6 0 0 x ^ {2} - 6 0 0 x + 1 2 0) \big ]. \end{array} \end{document} ]]></tex-math></disp-formula><p>The second one, finding derivatives of <inline-formula><tex-math id="math-143"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle u _ { 5 } ( x ) \end{document} ]]></tex-math></inline-formula> , we have as follows:</p><disp-formula id="equation-62"><tex-math id="math-144"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \begin{aligned}u_{5}'(x) &= d_{1}(-1)+ d_{2}\left[\frac{1}{2}(2x-4)\right]+ d_{3}\left[\frac{1}{6}(-3x^{2}+18x-18)\right] \\&\quad + d_{4}\left[\frac{1}{24}(4x^{3}-48x^{2}+144x-96)\right] \\&\quad + d_{5}\left[\frac{1}{120}(-5x^{4}+100x^{3}-600x^{2}+1200x-600)\right], \\[6pt]u_{5}''(x) &= d_{2}\left[\frac{1}{2}(2)\right]+ d_{3}\left[\frac{1}{6}(-6x+18)\right]+ d_{4}\left[\frac{1}{24}(12x^{2}-96x+144)\right] \\&\quad + d_{5}\left[\frac{1}{120}(-20x^{3}+300x^{2}-1200x+1200)\right].\end{aligned} \end{document} ]]></tex-math></disp-formula><p>and</p><disp-formula id="equation-63"><tex-math id="math-145"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle u _ {5} ^ {\prime \prime \prime} (x) = d _ {3} \bigl [ \frac {1}{6} (- 6) \bigr ] + d _ {4} \bigl [ \frac {1}{2 4} (2 4 x - 9 6) \bigr ] + d _ {5} \bigl [ \frac {1}{1 2 0} (- 6 0 x ^ {2} + 6 0 0 x - 1 2 0 0) \bigr ]. \end{document} ]]></tex-math></disp-formula><p>The third one, substituting the derivatives into the problem, we receive</p><disp-formula id="equation-64"><tex-math id="math-146"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \begin{array}{l} d _ {3} \left[ \frac {1}{6} (- 6) \right] + d _ {4} \left[ \frac {1}{2 4} (2 4 x - 9 6) \right] + d _ {5} \left[ \frac {1}{1 2 0} (- 6 0 x ^ {2} + 6 0 0 x - 1 2 0 0) \right] \\ + \int_ {0} ^ {x} x \sin (t) \left\{d _ {1} (- 1) + d _ {2} \left[ \frac {1}{2} (2 t - 4) \right] + d _ {3} \left[ \frac {1}{6} (- 3 t ^ {2} + 1 8 t - 1 8) \right] \right. \\ + d _ {4} \left[ \frac {1}{2 4} (4 t ^ {3} - 4 8 t ^ {2} + 1 4 4 t - 9 6) \right] \\ + d _ {5} \left[ \frac {1}{1 2 0} (- 5 t ^ {4} + 1 0 0 t ^ {3} - 6 0 0 t ^ {2} + 1 2 0 0 t - 6 0 0) \right] \Bigg \} d t \\ = - 2 x ^ {2} \left\{d _ {0} (1) + d _ {1} (- x + 1) + d _ {2} \left[ \frac {1}{2} (x ^ {2} - 4 x + 2) \right] \right. \\ + d _ {3} \left[ \frac {1}{6} (- x ^ {3} + 9 x ^ {2} - 1 8 x + 6) \right] + d _ {4} \left[ \frac {1}{2 4} (x ^ {4} - 1 6 x ^ {3} + 7 2 x ^ {2} - 9 6 x + 2 4) \right] \\ + d _ {5} \left[ \frac {1}{1 2 0} (- x ^ {5} + 2 5 x ^ {4} - 2 0 0 x ^ {3} + 6 0 0 x ^ {2} - 6 0 0 x + 1 2 0) \right] \Bigg \} \\ - \cos (x) - 4 x \cos (x) - x \sin (x) + 4 x ^ {3} + 2 x ^ {2} + 4 x \\ + \int_ {0} ^ {x} x \cos (t) \left\{d _ {0} (1) + d _ {1} (- t + 1) + d _ {2} \left[ \frac {1}{2} (t ^ {2} - 4 t + 2) \right] \right. \\ + d _ {3} \left[ \frac {1}{6} (- t ^ {3} + 9 t ^ {2} - 1 8 t + 6) \right] + d _ {4} \left[ \frac {1}{2 4} (t ^ {4} - 1 6 t ^ {3} + 7 2 t ^ {2} - 9 6 t + 2 4) \right] \\ + d _ {5} \left[ \frac {1}{1 2 0} (- t ^ {5} + 2 5 t ^ {4} - 2 0 0 t ^ {3} + 6 0 0 t ^ {2} - 6 0 0 t + 1 2 0) \right] \Bigg \} d t. \end{array} \end{document} ]]></tex-math></disp-formula><p>The fourth one, simplifying and integrating the equation, we obtain the new equation. Selecting <inline-formula><tex-math id="math-147"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle x _ { 1 } = 0 . 2 5 , x _ { 2 } = 0 . 5 \end{document} ]]></tex-math></inline-formula> and <inline-formula><tex-math id="math-148"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle x _ { 3 } = 0 . 7 5 \end{document} ]]></tex-math></inline-formula> to substitute in the new equation with using 3 initial conditions, we get the following system:</p><disp-formula id="equation-65"><tex-math id="math-149"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \begin{array}{l} d _ {0} + d _ {1} + d _ {2} + d _ {3} + d _ {4} + d _ {5} = 1, \\ - d _ {1} - 2 d _ {2} - 3 d _ {3} - 4 d _ {4} - 5 d _ {5} = 3, \\ d _ {2} + 3 d _ {3} + 6 d _ {4} + 1 0 d _ {5} = 0, \\ 0. 0 6 3 1 4 9 d _ {0} + 0. 0 3 1 8 1 7 d _ {1} + 0. 0 0 5 0 4 8 d _ {2} - 1. 0 1 7 5 6 7 d _ {3} \\ - 3. 7 8 6 4 0 8 d _ {4} - 8. 8 3 3 0 8 1 d _ {5} = - 0. 8 1 2 1 7 5, \\ 0. 2 6 0 2 8 7 d _ {0} + 0. 0 0 7 7 2 6 d _ {1} - 0. 1 7 1 6 6 4 d _ {2} - 1. 2 9 0 9 0 5 d _ {3} \\ - 3. 8 6 1 3 2 7 d _ {4} - 8. 0 1 7 7 4 6 d _ {5} = 0. 1 2 7 5 3 9, \\ 0. 6 1 3 7 7 1 d _ {0} - 0. 2 4 9 0 2 3 d _ {1} - 0. 7 3 9 8 8 7 d _ {2} - 1. 9 5 7 6 7 1 d _ {3} \\ - 4. 2 3 2 0 4 7 d _ {4} - 7. 4 0 7 6 9 7 d _ {5} = 2. 3 7 4 5 1 5. \end{array} \end{document} ]]></tex-math></disp-formula><p>The last one, solving the system by a program, we get</p><disp-formula id="equation-66"><tex-math id="math-150"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \begin{array}{l} d _ {0} = 3. 9 0 6 5 7 9, d _ {1} = - 4. 5 0 6 7 0 4, d _ {2} = 5. 9 5 5 4 2 4, d _ {3} = - 7. 8 9 1 8 4 0, \\ d _ {4} = 4. 4 1 1 3 2 8, d _ {5} = - 0. 8 7 4 7 8 7. \end{array} \end{document} ]]></tex-math></disp-formula><p>Therefore, the numerical solution is</p><disp-formula id="equation-67"><tex-math id="math-151"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \begin{array}{l} u _ {5} (x) = 3. 9 0 6 5 7 9 - 4. 5 0 6 7 0 4 (- x + 1) + 5. 9 5 5 4 2 4 \big [ \frac {1}{2} (x ^ {2} - 4 x + 2) \big ] \\ - 7. 8 9 1 8 4 0 \big [ \frac {1}{6} (- x ^ {3} + 9 x ^ {2} - 1 8 x + 6) \big ] \\ + 4. 4 1 1 3 2 8 \big [ \frac {1}{2 4} (x ^ {4} - 1 6 x ^ {3} + 7 2 x ^ {2} - 9 6 x + 2 4) \big ] \\ - 0. 8 7 4 7 8 7 \big [ \frac {1}{1 2 0} (- x ^ {5} + 2 5 x ^ {4} - 2 0 0 x ^ {3} + 6 0 0 x ^ {2} - 6 0 0 x + 1 2 0) \big ]. \end{array} \end{document} ]]></tex-math></disp-formula><table-wrap id="table-3"><label>Table 3</label><caption><p>Values of exact and approximate u _ { 5 } ( x ) solutions for the example <xref ref-type="custom" custom-type="reference-target" rid="anchor-a4ae94f0-de73-4f35-b7ef-0cc4c545565d">2.4.</xref></p></caption><table><colgroup><col></col><col></col><col></col><col></col></colgroup><thead><tr><th scope="col">x</th><th scope="col">Exact solution</th><th scope="col">Approx. solution</th><th scope="col">Absolute error</th></tr></thead><tbody><tr><td>0.00</td><td>1.000000</td><td>1.000000</td><td>0.000000</td></tr><tr><td>0.31</td><td>1.937336</td><td>1.937318</td><td>0.000017</td></tr><tr><td>0.63</td><td>2.844422</td><td>2.844339</td><td>0.000084</td></tr><tr><td>0.94</td><td>3.693973</td><td>3.693774</td><td>0.000199</td></tr><tr><td>1.26</td><td>4.464331</td><td>4.464054</td><td>0.000276</td></tr><tr><td>1.57</td><td>5.141593</td><td>5.142007</td><td>0.000414</td></tr><tr><td>1.88</td><td>5.720968</td><td>5.725529</td><td>0.004562</td></tr><tr><td>2.20</td><td>6.207247</td><td>6.226270</td><td>0.019024</td></tr><tr><td>2.51</td><td>6.614333</td><td>6.672304</td><td>0.057971</td></tr><tr><td>2.83</td><td>6.963884</td><td>7.110808</td><td>0.146924</td></tr><tr><td>3.14</td><td>7.283185</td><td>7.610740</td><td>0.327555</td></tr></tbody></table></table-wrap><fig id="figure-3"><label>Figure 3.</label><caption><p>Graphs of exact and approximate u5(x) solutions forexample <xref ref-type="custom" custom-type="reference-target" rid="anchor-a4ae94f0-de73-4f35-b7ef-0cc4c545565d">2.4.</xref></p></caption><graphic xlink:href="https://jims-a.org/index.php/jimsa/article/download/1883/562/13992" mime-subtype="jpeg" mimetype="image"><alt-text>Figure 3.</alt-text></graphic></fig><p>To demonstrate the practical applicability of the proposed method, we consider a classical population model with memory efects, where the current growth rate depends not only on the present population but also on its past values. Let <inline-formula><tex-math id="math-152"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle u ( t ) \end{document} ]]></tex-math></inline-formula> denote the population size at time <inline-formula><tex-math id="math-153"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle t \geq 0 \end{document} ]]></tex-math></inline-formula> . The population growth with memory is defined by</p><disp-formula id="equation-68"><tex-math id="math-154"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle u ^ {\prime} (t) = r u (t) + \int_ {0} ^ {t} K (t - s) u (s) d s, u (0) = u _ {0}, \end{document} ]]></tex-math></disp-formula><p>where <inline-formula><tex-math id="math-155"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle r \in \mathbb { R } \end{document} ]]></tex-math></inline-formula> is the intrinsic growth rate and <inline-formula><tex-math id="math-156"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle K \end{document} ]]></tex-math></inline-formula> is a memory kernel. For the growth model of Drosophila, we get <inline-formula><tex-math id="math-157"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle r = 0 . 2 5 , u _ { 0 } = 5 0 \end{document} ]]></tex-math></inline-formula> and the exponentially decaying kernel <inline-formula><tex-math id="math-158"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle K ( \tau ) = 0 . 1 5 e ^ { - \bar { 0 } . 1 \tau } \end{document} ]]></tex-math></inline-formula> and see more in <xref ref-type="bibr" rid="BIBR-11">[11]</xref>. Thus, the growth model is in the form</p><disp-formula id="equation-69"><tex-math id="math-159"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle u ^ {\prime} (t) = 0. 2 5 u (t) + \int_ {0} ^ {t} 0. 1 5 e ^ {- 0. 1 (t - s)} u (s) d s, u (0) = 5 0. \end{document} ]]></tex-math></disp-formula><p>This kernel is smooth, exponentially bounded and therefore satisfies the Laplacetransform conditions and other regularity assumptions used in our analysis. <inline-formula><tex-math id="math-160"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \mathrm { B y } \end{document} ]]></tex-math></inline-formula> Laplace transform, the exact solution of the problem is <inline-formula><tex-math id="math-161"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle u ( t ) = { \frac { 6 0 0 } { 1 7 } } e ^ { 0 . 5 t } + { \frac { 2 5 0 } { 1 7 } } e ^ { - 0 . 3 5 t } \end{document} ]]></tex-math></inline-formula> Then, applying the proposed Laguerre method with <inline-formula><tex-math id="math-162"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle u _ { 4 } ( t ) \end{document} ]]></tex-math></inline-formula> and <inline-formula><tex-math id="math-163"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle u _ { 6 } ( t ) \end{document} ]]></tex-math></inline-formula> on <inline-formula><tex-math id="math-164"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle t \in [0,2] \end{document} ]]></tex-math></inline-formula> we obatian the results as follows.</p><disp-formula id="equation-70"><tex-math id="math-165"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \begin{array}{l} u _ {4} (t) = 8 0. 0 6 9 3 0 7 8 8 - 5 8. 0 2 2 4 3 1 6 9 (- t + 1) + 4 1. 8 2 8 8 1 7 7 4 \big [ \frac {1}{2} (t ^ {2} - 4 t + 2) \big ] \\ - 1 7. 3 7 9 5 2 1 4 6 \big [ \frac {1}{6} (- t ^ {3} + 9 t ^ {2} - 1 8 t + 6) \big ] \\ + 3. 5 0 3 8 2 7 5 3 \big [ \frac {1}{2 4} (t ^ {4} - 1 6 t ^ {3} + 7 2 t ^ {2} - 9 6 t + 2 4) \big ] \end{array} \end{document} ]]></tex-math></disp-formula><p>and</p><disp-formula id="equation-71"><tex-math id="math-166"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \begin{array}{l} {u _ {6} (t) = 8 0. 8 1 3 8 3 6 3 8 - 6 2. 4 3 6 8 4 4 5 5 (- t + 1) + 5 2. 3 9 0 9 7 2 7 9 \big [ \frac {1}{2} (t ^ {2} - 4 t + 2) \big ]} \\ {- 3 0. 1 5 6 4 7 5 7 6 \big [ \frac {1}{6} (- t ^ {3} + 9 t ^ {2} - 1 8 t + 6) \big ]} \\ {+ 1 1. 4 0 6 8 3 8 0 9 \big [ \frac {1}{2 4} (t ^ {4} - 1 6 t ^ {3} + 7 2 t ^ {2} - 9 6 t + 2 4) \big ]} \\ {- 2. 1 0 5 6 5 1 0 6 \big [ \frac {1}{1 2 0} (- t ^ {5} + 2 5 t ^ {4} - 2 0 0 t ^ {3} + 6 0 0 t ^ {2} - 6 0 0 t + 1 2 0) \big ]} \\ {+ 0. 0 8 7 3 2 4 1 2 \big [ \frac {1}{7 2 0} (t ^ {6} - 3 6 t ^ {5} + 4 5 0 t ^ {4} - 2 4 0 0 t ^ {3} + 5 4 0 0 t ^ {2} - 4 3 2 0 t + 7 2 0) \big ].} \end{array} \end{document} ]]></tex-math></disp-formula><table-wrap id="table-4"><label>Table 4</label><caption><p>Values of exact and approximate u4( t ) solutions with absolute errors.</p></caption><table><colgroup><col></col><col></col><col></col><col></col></colgroup><thead><tr><th scope="col">t</th><th scope="col">Exact solution</th><th scope="col">Approx. solution</th><th scope="col">Absolute error</th></tr></thead><tbody><tr><td>0.00</td><td>50.000000</td><td>50.000000</td><td>0.000000</td></tr><tr><td>0.25</td><td>53.467282</td><td>53.466132</td><td>0.001150</td></tr><tr><td>0.50</td><td>57.663500</td><td>57.662390</td><td>0.001110</td></tr><tr><td>0.75</td><td>62.663320</td><td>62.661870</td><td>0.001450</td></tr><tr><td>1.00</td><td>68.553223</td><td>68.551354</td><td>0.001868</td></tr><tr><td>1.25</td><td>75.432924</td><td>75.431314</td><td>0.001610</td></tr><tr><td>1.50</td><td>83.416991</td><td>83.415904</td><td>0.001087</td></tr><tr><td>1.75</td><td>92.636690</td><td>92.632969</td><td>0.003721</td></tr><tr><td>2.00</td><td>103.242084</td><td>103.224038</td><td>0.018046</td></tr></tbody></table></table-wrap><table-wrap id="table-5"><label>Table 5</label><caption><p>Values of exact and approximate u6( t ) solutions with absolute errors.</p></caption><table><colgroup><col></col><col></col><col></col><col></col></colgroup><thead><tr><th scope="col">t</th><th scope="col">Exact solution</th><th scope="col">Approx. solution</th><th scope="col">Absolute error</th></tr></thead><tbody><tr><td>0.00</td><td>50.000000</td><td>50.000000</td><td>0.000000</td></tr><tr><td>0.25</td><td>53.467282</td><td>53.467436</td><td>0.000154</td></tr><tr><td>0.50</td><td>57.663500</td><td>57.663728</td><td>0.000228</td></tr><tr><td>0.75</td><td>62.663320</td><td>62.663634</td><td>0.000314</td></tr><tr><td>1.00</td><td>68.553223</td><td>68.553593</td><td>0.000370</td></tr><tr><td>1.25</td><td>75.432924</td><td>75.433329</td><td>0.000405</td></tr><tr><td>1.50</td><td>83.416991</td><td>83.417464</td><td>0.000473</td></tr><tr><td>1.75</td><td>92.636690</td><td>92.637166</td><td>0.000476</td></tr><tr><td>2.00</td><td>103.242084</td><td>103.241802</td><td>0.000287</td></tr></tbody></table></table-wrap></sec><sec id="sec-3"><title>3. CONCLUDING REMARKS</title><p>In this paper, a generalization of linear VIDEs has been introduced already. In general, all results show that the Laplace transform has been efective to solve analytical solutions of the generalization on convolution type kernels repeatedly and the Laguerre polynomials have been successful to figure out numerical solutions of the generalization on non-convolution type kernels several times. However, the Kushare transform, Sadik transform and Kamal transform are other methods that can be analytically solved on convolution types of this generalization similarly. Moreover, the main advantage of this analytical method is the fact that it gives the exact solutions in just few processes and uses very less computational work. We also suggest that this numerical method can be applicable to singularly perturbed linear VIDEs, which are one case of this generalization, to obtain accurately approximate solutions.</p></sec></body><back><ack><title>Acknowledgement.</title><p>The author would like to thank the academic referees for the careful reading and helpful comments for improving this paper.</p></ack><ref-list><title>References</title><ref id="BIBR-1"><element-citation publication-type="journal"><article-title>Numerical solution of integro-diferential equations of fractional order by laplace decomposition method</article-title><source>Wseas Transactions on Mathematics</source><volume>12</volume><issue>12</issue><person-group person-group-type="author"><name><surname>Yang</surname><given-names>C.</given-names></name><name><surname>Hou</surname><given-names>J.</given-names></name></person-group><year>2013</year><page-range>1173-1183,</page-range><ext-link 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