Solving a steady state temperature distribution problem with the help of Hankel Transform

Question

Could someone please help me with the following problem. I am totally new to Hankel Transform. This question is my first assignment related to Hankel transform. I tried to solve it but couldn't get the desired result. I shall be thankful if someone could point out my mistake or help me with completing my solution. Here is the question and my attempt. Thank you. Question
The steady temperature distribution $u(r,z)$ in a semi-infinite solid with $z\geq{0}$ is governed by the system \begin{align} u_{rr}+\frac{1}{r}u_{r}+u_{zz}=-Aq(r),\qquad 0<r<\infty, \quad z>0\label{ii}\\ u(r,0)=0\label{iii} \end{align} where $A$ is a constant and $q(r)$ represents the steady heat source. Show that the solution is given by $$u(r,z)=A\int_{0}^{\infty}\tilde{q}(k)J_{0}(kr)k^{-1}(1-e^{-kz})dk,$$ where $\tilde{q}(k)$ is the zeroth-order Hankel transform of $q(r)$.
Solution:
The steady temperature distribution $u(r,z)$ in a semi-infinite solid with $z\geq{0}$ is governed by the system $$u_{rr}+\frac{1}{r}u_{r}+u_{zz}=-Aq(r),\qquad 0<r<\infty, \quad z>0$$ $$u(r,0)=0$$. As asked in question, we have to find $u(r,z)$ by the method of Hankel transform of zero-order. So let $\tilde{u}(k,z)$ be the zero-order Hankel transform of $u(r,z)$, i.e., $$\tilde{u}(k,z)=\mathcal{H}_{0}\{u(r,z)\}=\int_{0}^{\infty}ru(r,z)J_{0}(kr)dr$$ Applying zero-order Hankel transform on both sides of equation $\ref{ii}$, we have: \begin{align*} \int_{0}^{\infty}r\left(u_{rr}+\frac{1}{r}u_{r}+u_{zz}\right)J_{0}(kr)dr=\int_{0}^{\infty}r\left(-Aq(r)\right)J_{0}(kr)dr\\ \mathcal{H}_{0}\left\{\left(u_{rr}+\frac{1}{r}u_{r}\right)\right\}+\int_{0}^{\infty}ru_{zz}J_{0}(kr)dr=-A\int_{0}^{\infty}r\left(q(r)\right)J_{0}(kr)dr\\ \text{from properties of Hankel transform}\\ -k^{2}\tilde{u}(k,z)+\frac{d^{2}}{dz^{2}}\int_{0}^{\infty}ru(r,z)J_{0}(kr)dr=-A\mathcal{H}_{0}\{q(r)\}\\ -k^{2}\tilde{u}(k,z)+\frac{d^{2}}{dz^{2}}\mathcal{H}_{0}\{u(r,z)\}=-A\tilde{q}(k)\\ -k^{2}\tilde{u}(k,z)+\frac{d^{2}}{dz^{2}}{\tilde{u}(k,z)}=-A\tilde{q}(k) \end{align*} rearranging the above equation, we get \begin{align}\label{iv} \frac{d^{2}}{dz^{2}}{\tilde{u}(k,z)}-k^{2}\tilde{u}(k,z)=-A\tilde{q}(k)\\ f(D)(\tilde{u}(k,z))=\left(D^{2}-k^{2}\right)(\tilde{u}(k,z)=-A\tilde{q}(k) \notag\\ \end{align} equation \ref{iv} is 2nd order, linear, non-homogeneous ODE. The given conditions under the Hankel transform is \begin{align} \mathcal{H}_{0}\{u(r,0)\}=\mathcal{H}_{0}\{0\}\notag\\ \tilde{u}(k,0)=0\label{v} \end{align} Now, solving, $\ref{iv}$ with the help of $\ref{v}$. The associated homogeneous equation to $\ref{iv}$ is $$\frac{d^{2}}{dz^{2}}{\tilde{u}(r,z)}-k^{2}\tilde{u}(k,z)=-A\tilde{q}(k)$$ and hence the solution is $$\tilde{u}_{c}(k,z)=c_{1}e^{-kz}+c_{2}e^{kz}$$ Now, the solution of non-homogeneous part is \begin{align*} \tilde{u}_{p}(k,z)&=\frac{1}{f(D)}(-A\tilde{q}(k))\\ &=\frac{-A\tilde{q}(k)}{\left(D^{2}-k^{2}\right)}e^{0}\\ &=\frac{-A\tilde{q}(k)}{\left(0-k^{2}\right)}e^{0}\\ &=\frac{-A\tilde{q}(k)}{-k^{2}}\\ &=\frac{A\tilde{q}(k)}{k^{2}} \end{align*} Thus the solution of equation $\ref{iv}$ is \begin{align}\label{vi} \tilde{u}(k,z)=c_{1}e^{-kz}+c_{2}e^{kz}+\frac{A\tilde{q}(k)}{k^{2}} \end{align} Applying the given condition, $\tilde{u}(k,0)=0$ to above solution equation: \begin{align*} \tilde{u}(k,0)&=c_{1}e^{-k0}+c_{2}e^{k0}+\frac{A\tilde{q}(k)}{k^{2}}\\ 0&=c_{1}+c_{2}+\frac{A\tilde{q}(r)}{k^{2}}\\ \implies c_{1}&=-c_{2}-\frac{A\tilde{q}(k)}{k^{2}} \end{align*} Thus solution $\ref{vi}$ is, now, \begin{align*} \tilde{u}(k,z)&=(-c_{2}-\frac{A\tilde{q}(r)}{k^{2}})e^{-kz}+c_{2}e^{kz}+\frac{A\tilde{q}(k)}{k^{2}}\\ &=-c_{2}e^{-kz}-\frac{A\tilde{q}(r)}{k^{2}}e^{-kz}+c_{2}e^{kz}+\frac{A\tilde{q}(k)}{k^{2}}\\ \tilde{u}(k,z)&=c_{2}\left(e^{kz}-e^{-kz}\right)+\frac{A\tilde{q}(k)}{k^{2}}(1-e^{-kz})\\ \text{applying Hankel inverse on both sides}\\ u(r,z)&=\int_{0}^{\infty}kc_{2}(e^{kz}-e^{-kz})J_{0}(kr)dk+\int_{0}^{\infty}k\frac{A\tilde{q}(k)}{k^{2}}(1-e^{-kz})J_{0}(kr)dk\\ &=c_{2}\int_{0}^{\infty}k(e^{kz}-e^{-kz})J_{0}(kr)dk+A\int_{0}^{\infty}k^{-1}{\tilde{q}(k)}(1-e^{-kz})J_{0}(kr)dk\\ &=c_{2}\int_{0}^{\infty}k(e^{kz}-e^{-kz})J_{0}(kr)dk+A\int_{0}^{\infty}{\tilde{q}(k)}J_{0}(kr)k^{-1}(1-e^{-kz})dk\\ \end{align*}