If we choose \(\rho=1\), then (2.3), (2.16), and (2.18) reduce to the following:
Corollary 1
If
\(\delta>0, \upsilon>0\), \(k\in\mathbb{R}, \zeta, \eta, \xi\in\mathbb{C}, \Re(\zeta)>0, \Re(\eta)>0\), and
\(\Re{(\xi )}>0\), then the solution of the equation
$$\begin{aligned} & \mathcal{N}(\tau)-\mathcal{N}_{0}E^{\xi }_{k,\zeta,\eta}( \tau)=-\delta^{\upsilon}{}_{0}D^{-\upsilon}_{\tau } \mathcal{N}(\tau) \end{aligned}$$
(3.1)
is given by
$$\begin{aligned} & \mathcal{N}(\tau)=\mathcal{N}_{0}\sum^{\infty}_{n=0}\frac{(\xi)_{n,k}}{\Gamma_{k}(n \zeta+\eta)}\tau^{n} E_{\upsilon ,n+1} \bigl(- \delta^{\upsilon}\tau^{\upsilon} \bigr). \end{aligned}$$
(3.2)
Corollary 2
If
\(\delta>0, \upsilon>0\), \(k\in\mathbb{R}, \zeta, \eta, \xi\in\mathbb{C}, \Re(\zeta)>0, \Re(\eta)>0\), and
\(\Re{(\xi)}>0\), then the solution of the equation
$$\begin{aligned} & \mathcal{N}(\tau)=\mathcal{N}_{0}E^{\xi }_{k,\zeta,\eta} \bigl(\delta^{\upsilon}\tau^{\upsilon} \bigr)-\delta^{\upsilon }{}_{0}D^{-\upsilon}_{\tau} \mathcal{N}(\tau) \end{aligned}$$
(3.3)
is given by
$$\begin{aligned} & \mathcal{N}(\tau)=\mathcal{N}_{0}\sum^{\infty}_{n=0}\frac{(\xi)_{n,k}}{\Gamma_{k}(n \zeta+\eta)} \bigl(\delta^{\upsilon}\tau ^{\upsilon} \bigr)^{n}E_{\upsilon,\upsilon n+1} \bigl(-\delta^{\upsilon} \tau^{\upsilon } \bigr). \end{aligned}$$
(3.4)
Corollary 3
If
\(\sigma>0, \delta>0, \upsilon>0\), \(k\in\mathbb {R}, \zeta, \eta, \xi\in\mathbb{C}, \Re(\zeta)>0, \Re(\eta)>0\), and
\(\Re{(\xi)}>0\), then the solution of the equation
$$\begin{aligned} & \mathcal{N}(\tau)=\mathcal{N}_{0}E^{\xi }_{k,\zeta,\eta} \bigl(\delta^{\upsilon}\tau^{\upsilon} \bigr)-\sigma^{\upsilon }{}_{0}D^{-\upsilon}_{\tau} \mathcal{N}(\tau) \end{aligned}$$
(3.5)
is given by
$$\begin{aligned} & \mathcal{N}(\tau)=\mathcal{N}_{0}\sum^{\infty}_{n=0}\frac{(\xi)_{n,k}}{\Gamma_{k}(n \zeta+\eta)} \bigl(\delta^{\upsilon}\tau ^{\upsilon} \bigr)^{n}E_{\upsilon,\upsilon n+1} \bigl(-\sigma^{\upsilon} \tau^{\upsilon } \bigr). \end{aligned}$$
(3.6)
If we choose \(k=1\), then (2.3), (2.16), and (2.18) reduce to the following:
Corollary 4
If
\(\delta>0, \upsilon>0\), \(\zeta, \eta, \xi\in \mathbb{C}, \Re(\zeta)>0, \Re(\eta)>0, \Re{(\xi)}>0\), and
\(\rho\in (0,1)\cup\mathbb{N}\), then the solution of the equation
$$\begin{aligned} & \mathcal{N}(\tau)-\mathcal{N}_{0}E^{\xi,\rho }_{\zeta,\eta}( \tau)=-\delta^{\upsilon}{}_{0}D^{-\upsilon}_{\tau } \mathcal{N}(\tau) \end{aligned}$$
(3.7)
is given by
$$\begin{aligned} & \mathcal{N}(\tau)=\mathcal{N}_{0}\sum^{\infty}_{n=0}\frac{(\xi)_{nq,k}}{\Gamma_{k}(n \zeta+\eta)}\tau^{n} E_{\upsilon ,n+1} \bigl(- \delta^{\upsilon}\tau^{\upsilon} \bigr). \end{aligned}$$
(3.8)
Corollary 5
If
\(\delta>0, \upsilon>0\), \(\zeta, \eta, \xi\in \mathbb{C}, \Re(\zeta)>0, \Re(\eta)>0, \Re{(\xi)}>0\), and
\(\rho\in (0,1)\cup\mathbb{N}\), then the solution of the equation
$$\begin{aligned} & \mathcal{N}(\tau)-\mathcal{N}_{0}E^{\xi,\rho }_{\zeta,\eta} \bigl(\delta^{\upsilon}\tau^{\upsilon} \bigr)=-\delta^{\upsilon }{}_{0}D^{-\upsilon}_{\tau} \mathcal{N}(\tau) \end{aligned}$$
(3.9)
is given by
$$\begin{aligned} & \mathcal{N}(\tau)=\mathcal{N}_{0}\sum^{\infty}_{n=0}\frac{(\xi)_{nq,k}}{\Gamma_{k}(n \zeta+\eta)} \bigl(\delta^{\upsilon}\tau ^{\upsilon} \bigr)^{n} E_{\upsilon,n+1} \bigl(-\delta^{\upsilon} \tau^{\upsilon } \bigr). \end{aligned}$$
(3.10)
Corollary 6
If
\(\sigma>0, \delta>0, \upsilon>0\), \(\zeta, \eta, \xi\in\mathbb{C}, \Re(\zeta)>0, \Re(\eta)>0, \Re{(\xi)}>0\), and
\(\rho \in(0,1)\cup\mathbb{N}\), then the solution of the equation
$$\begin{aligned} & \mathcal{N}(\tau)-\mathcal{N}_{0}E^{\xi,\rho }_{\zeta,\eta} \bigl(\delta^{\upsilon}\tau^{\upsilon} \bigr)=-\sigma^{\upsilon }{}_{0}D^{-\upsilon}_{\tau} \mathcal{N}(\tau) \end{aligned}$$
(3.11)
is given by
$$\begin{aligned} & \mathcal{N}(\tau)=\mathcal{N}_{0}\sum^{\infty}_{n=0}\frac{(\xi)_{nq,k}}{\Gamma_{k}(n \zeta+\eta)} \bigl(\delta^{\upsilon}\tau ^{\upsilon} \bigr)^{n}E_{\upsilon,n+1} \bigl(-\sigma^{\upsilon} \tau^{\upsilon } \bigr). \end{aligned}$$
(3.12)
If we choose \(\rho=1\) and \(k=1\), then (2.3), (2.16), and (2.18) reduce to the following:
Corollary 7
If
\(\delta>0, \upsilon>0\), \(\zeta, \eta, \xi\in \mathbb{C}, \Re(\zeta)>0, \Re(\eta)>0\), and
\(\Re{(\xi)}>0\), then the solution of the equation
$$\begin{aligned} & \mathcal{N}(\tau)-\mathcal{N}_{0}E^{\xi }_{\zeta,\eta}( \tau)=-\delta^{\upsilon}{}_{0}D^{-\upsilon}_{\tau } \mathcal{N}(\tau) \end{aligned}$$
(3.13)
is given by
$$\begin{aligned} & \mathcal{N}(\tau)=\mathcal{N}_{0}\sum^{\infty}_{n=0}\frac{(\xi)_{n}}{\Gamma(n \zeta+\eta)}\tau^{n} E_{\upsilon ,n+1} \bigl(- \delta^{\upsilon}\tau^{\upsilon} \bigr). \end{aligned}$$
(3.14)
Corollary 8
If
\(\delta>0, \upsilon>0\), \(\zeta, \eta, \xi\in \mathbb{C}, \Re(\zeta)>0, \Re(\eta)>0\), and
\(\Re{(\xi)}>0\), then the solution of the equation
$$\begin{aligned} & \mathcal{N}(\tau)=\mathcal{N}_{0}E^{\xi }_{\zeta,\eta} \bigl(\delta^{\upsilon}\tau^{\upsilon} \bigr)-\delta^{\upsilon }{}_{0}D^{-\upsilon}_{\tau} \mathcal{N}(\tau) \end{aligned}$$
(3.15)
is given by
$$\begin{aligned} & \mathcal{N}(\tau)=\mathcal{N}_{0}\sum^{\infty}_{n=0}\frac{(\xi)_{n}}{\Gamma(n \zeta+\eta)} \bigl(\delta^{\upsilon}\tau ^{\upsilon} \bigr)^{n}E_{\upsilon,\upsilon n+1} \bigl(-\delta^{\upsilon} \tau^{\upsilon } \bigr). \end{aligned}$$
(3.16)
Corollary 9
If
\(\sigma>0, \delta>0, \upsilon>0\), \(\zeta, \eta, \xi\in\mathbb{C}, \Re(\zeta)>0, \Re(\eta)>0\), and
\(\Re{(\xi)}>0\), then the solution of the equation
$$\begin{aligned} & \mathcal{N}(\tau)=\mathcal{N}_{0}E^{\xi }_{\zeta,\eta} \bigl(\delta^{\upsilon}\tau^{\upsilon} \bigr)-\sigma^{\upsilon }{}_{0}D^{-\upsilon}_{\tau} \mathcal{N}(\tau) \end{aligned}$$
(3.17)
is given by
$$\begin{aligned} & \mathcal{N}(\tau)=\mathcal{N}_{0}\sum^{\infty}_{n=0}\frac{(\xi)_{n}}{\Gamma(n \zeta+\eta)} \bigl(\delta^{\upsilon}\tau ^{\upsilon} \bigr)^{n}E_{\upsilon,\upsilon n+1} \bigl(-\sigma^{\upsilon} \tau^{\upsilon } \bigr). \end{aligned}$$
(3.18)
If we choose \(\rho=1, k=1\), and \(\xi=1\), then (2.3), (2.16), and (2.18) reduce to the following:
Corollary 10
If
\(\delta>0, \upsilon>0,\zeta, \eta\in\mathbb{C}, \Re(\zeta)>0, \Re(\eta)>0\), then the solution of the equation
$$\begin{aligned} & \mathcal{N}(\tau)-\mathcal{N}_{0}E_{\zeta ,\eta}(\tau)=- \delta^{\upsilon}{}_{0}D^{-\upsilon}_{\tau}\mathcal {N}( \tau) \end{aligned}$$
(3.19)
is given by
$$\begin{aligned} & \mathcal{N}(\tau)=\mathcal{N}_{0}\sum^{\infty}_{n=0}\frac{\tau^{n}}{\Gamma(n \zeta+\eta)} E_{\upsilon,n+1} \bigl(-\delta ^{\upsilon} \tau^{\upsilon} \bigr). \end{aligned}$$
(3.20)
Corollary 11
If
\(\delta>0, \upsilon>0,\zeta, \eta\in\mathbb{C}, \Re(\zeta)>0, \Re(\eta)>0\), then the solution of the equation
$$\begin{aligned} & \mathcal{N}(\tau)=\mathcal{N}_{0}E_{\zeta ,\eta} \bigl( \delta^{\upsilon}\tau^{\upsilon} \bigr)-\delta^{\upsilon }{}_{0}D^{-\upsilon}_{\tau} \mathcal{N}(\tau) \end{aligned}$$
(3.21)
is given by
$$\begin{aligned} & \mathcal{N}(\tau)=\mathcal{N}_{0}\sum^{\infty}_{n=0}\frac{(\delta^{\upsilon}\tau^{\upsilon})^{n}}{\Gamma(n \zeta+\eta )}E_{\upsilon,\upsilon n+1} \bigl(-\delta^{\upsilon} \tau^{\upsilon } \bigr). \end{aligned}$$
(3.22)
Corollary 12
If
\(\sigma>0, \delta>0, \upsilon>0,\zeta, \eta\in \mathbb{C}, \Re(\zeta)>0, \Re(\eta)>0\), then the solution of the equation
$$\begin{aligned} & \mathcal{N}(\tau)=\mathcal{N}_{0}E_{\zeta ,\eta} \bigl( \delta^{\upsilon}\tau^{\upsilon} \bigr)-\sigma^{\upsilon }{}_{0}D^{-\upsilon}_{\tau} \mathcal{N}(\tau) \end{aligned}$$
(3.23)
is given by
$$\begin{aligned} & \mathcal{N}(\tau)=\mathcal{N}_{0}\sum^{\infty}_{n=0}\frac{(\delta^{\upsilon}\tau^{\upsilon})^{n}}{\Gamma(n \zeta+\eta )}E_{\upsilon,\upsilon n+1} \bigl(-\sigma^{\upsilon} \tau^{\upsilon } \bigr). \end{aligned}$$
(3.24)
If we choose \(\rho=1, k=1, \xi=1\), and \(\eta=1\), then (2.3), (2.16), and (2.18) reduce to the following:
Corollary 13
If
\(\delta>0, \upsilon>0,\zeta\in\mathbb{C}\), \(\Re (\zeta)>0\), then the solution of the equation
$$\begin{aligned} & \mathcal{N}(\tau)-\mathcal{N}_{0}E_{\zeta }(\tau)=- \delta^{\upsilon}{}_{0}D^{-\upsilon}_{\tau}\mathcal{N}( \tau ) \end{aligned}$$
(3.25)
is given by
$$\begin{aligned} & \mathcal{N}(\tau)=\mathcal{N}_{0}\sum^{\infty}_{n=0}\frac{\tau^{n}}{\Gamma(n \zeta+1)} E_{\upsilon,n+1} \bigl(-\delta ^{\upsilon} \tau^{\upsilon} \bigr). \end{aligned}$$
(3.26)
Corollary 14
If
\(\delta>0, \upsilon>0,\zeta\in\mathbb{C}, \Re (\zeta)>0\), then the solution of the equation
$$\begin{aligned} & \mathcal{N}(\tau)=\mathcal{N}_{0}E_{\zeta } \bigl( \delta^{\upsilon}\tau^{\upsilon} \bigr)-\delta^{\upsilon}{}_{0}D^{-\upsilon }_{\tau} \mathcal{N}(\tau) \end{aligned}$$
(3.27)
is given by
$$\begin{aligned} & \mathcal{N}(\tau)=\mathcal{N}_{0}\sum^{\infty}_{n=0} \frac{(\delta ^{\upsilon}\tau^{\upsilon})^{n}}{\Gamma(n \zeta+1)}E_{\upsilon,\upsilon n+1} \bigl(-\delta^{\upsilon} \tau^{\upsilon } \bigr). \end{aligned}$$
(3.28)
Corollary 15
If
\(\sigma>0, \delta>0, \upsilon>0,\zeta\in\mathbb {C}; \Re(\zeta)>0\), then the solution of the equation
$$\begin{aligned} & \mathcal{N}(\tau)=\mathcal{N}_{0}E_{\zeta } \bigl( \delta^{\upsilon}\tau^{\upsilon} \bigr)-\sigma^{\upsilon}{}_{0}D^{-\upsilon }_{\tau} \mathcal{N}(\tau) \end{aligned}$$
(3.29)
is given by
$$\begin{aligned} & \mathcal{N}(\tau)=\mathcal{N}_{0}\sum^{\infty}_{n=0}\frac{(\delta^{\upsilon}\tau^{\upsilon})^{n}}{\Gamma(n \zeta +1)}E_{\upsilon,\upsilon n+1} \bigl(-\sigma^{\upsilon} \tau^{\upsilon } \bigr). \end{aligned}$$
(3.30)
If we choose \(\rho=1, k=1, \xi=1, \zeta=0\), and \(\eta=1\), then (2.3), (2.16), and (2.18) reduce to the following:
Corollary 16
If
\(\delta>0, \upsilon>0\), then the solution of the equation
$$\begin{aligned} & \mathcal{N}(\tau)-\mathcal{N}_{0}e^{\tau}=- \delta^{\upsilon}{}_{0}D^{-\upsilon}_{\tau}\mathcal{N}( \tau ) \end{aligned}$$
(3.31)
is given by
$$\begin{aligned} & \mathcal{N}(\tau)=\mathcal{N}_{0}e^{\tau}E_{\upsilon,n+1} \bigl(-\delta^{\upsilon}\tau^{\upsilon} \bigr). \end{aligned}$$
(3.32)
Corollary 17
If
\(\delta>0, \upsilon>0\), then the solution of the equation
$$\begin{aligned} & \mathcal{N}(\tau)=\mathcal{N}_{0}e^{(\delta ^{\upsilon}\tau^{\upsilon})}- \delta^{\upsilon}{}_{0}D^{-\upsilon}_{\tau }\mathcal{N}( \tau) \end{aligned}$$
(3.33)
is given by
$$\begin{aligned} & \mathcal{N}(\tau)=\mathcal{N}_{0}e^{(\delta ^{\upsilon}\tau^{\upsilon})^{n}} E_{\upsilon,\upsilon n+1} \bigl(-\delta ^{\upsilon}\tau^{\upsilon} \bigr). \end{aligned}$$
(3.34)
Corollary 18
If
\(\sigma>0, \delta>0, \upsilon>0\), then the solution of the equation
$$\begin{aligned} & \mathcal{N}(\tau)=\mathcal{N}_{0}e^{(\delta ^{\upsilon}\tau^{\upsilon})}- \sigma^{\upsilon}{}_{0}D^{-\upsilon}_{\tau }\mathcal{N}( \tau) \end{aligned}$$
(3.35)
is given by
$$\begin{aligned} & \mathcal{N}(\tau)=\mathcal{N}_{0}e^{(\delta ^{\upsilon}\tau^{\upsilon})^{n} }E_{\upsilon,\upsilon n+1} \bigl(-\sigma ^{\upsilon}\tau^{\upsilon} \bigr). \end{aligned}$$
(3.36)