I am using the book "Equations of Mathematical Physics V.S. Vladimirov" to solve PDE using Separation of variables. And in his "A Collection of Problems on the Equations of Mathematical Physics (1986)" on page 249, there is following problem (20.22.1):
$$\begin{equation} \begin{cases} u_{tt} = u_{xx} + \frac{1}{x} u_x, \quad 0 < x < 1\\ |u(t, 0)| < \infty\\ u(t, 1) = sin^2x\\ u(0, x) = \frac{1}{2} [1 - \frac{J_0(2x)}{J_0(2)}]\\ u_t(0, x) = 0 \end{cases} \end{equation}$$
I have begun solving this problem by making the boundary condition homogeneous using following substitution $$u(t, x) = sin^2t + v(t, x)$$ and obtaining following non-homogeneous problem:
$$\begin{equation} \begin{cases} v_{tt} - v_{xx} - \frac{1}{x} v_x = -2cos2t, \quad 0 < x < 1\\ |v(t, 0)| < \infty\\ v(t, 1) = 0\\ v(0, x) = \frac{1}{2} [1 - \frac{J_0(2x)}{J_0(2)}]\\ v_t(0, x) = 0 \end{cases} \end{equation}$$
$$\text{From here I used separation of variables } v(t, x) = T(t)X(x)$$
$$\frac{T''}{T} = \frac{X''}{X} + \frac{1}{x} \frac{X'}{X} = \lambda$$
$$\begin{equation} \begin{cases} x^2X'' + xX' + \lambda x^2X = 0, \quad 0 < x < 1\\ |v(t, 0)| < \infty\\ v(t, 1) = 0\\ \end{cases} \end{equation}$$
$$\text{Here when } \lambda < 0 \text{ we use following substitution to turn the problem into Bessel equation } \lambda = -\mu^2$$
$$x^2X'' + xX' + \mu^2 x^2X = 0 \text{ and setting } y = \mu x \text{ we get}$$
$$y^2X''_{y} + yX'_y + y^2X = 0 \text{ we know that solution to this Bessel equation is } X = c_1J_0(\mu x) + c_2H_0(\mu x) $$
$$\text{and from boundary condition we can see that } c_2 = 0 \text{ which gives us } X = c_1J_0(\mu x)$$
$$\begin{equation} 0 = X(1) = c_1J_0(\mu) \Rightarrow J_0(\mu) = 0, \quad \mu_k = \theta_k, \quad k \geqslant 0 \end{equation}$$
Then finally we obtain following eigenvalue and eigenfunction:
$$\begin{equation} \lambda_k = -\theta^2, \quad X_k(x) = J_0(\theta_k x), \quad k \geqslant 0 \end{equation}$$
$$\begin{equation} v_k = T_k(t)X_k(x), \quad f(t) = -2cos2t \end{equation}$$
Then the solution becomes
$$\begin{equation} v(t, x) = \sum_{k = 0}^{\infty} T_k(t)X_k(x) \end{equation}$$
and from here we substitute this to equation and get
$$\begin{equation} \sum_{k = 0}^{\infty} [T_k'' - \lambda_k T_k]X_k = f(t) \end{equation}$$
and set
$$\begin{equation} f(t) = \sum_{k = 0}^{\infty} C_kX_k \end{equation}$$
$$\begin{equation} \sum_{k = 0}^{\infty} [T_k'' - \lambda_k T_k]X_k = \sum_{k = 0}^{\infty} C_kX_k \end{equation}$$
$$\begin{equation} \sum_{k = 0}^{\infty} [T_k'' - \lambda_k T_k - C_k(t)]X_k = 0 \end{equation}$$
using the initial condition we get
$$\begin{equation} \begin{cases} T''_k - \lambda_k T_k - C_k = 0\\ T_k(0) = b_k\\ T'_k(0) = 0 \end{cases} \end{equation}$$
$$\begin{equation} \begin{cases} \frac{1}{2} [1 - \frac{J_0(2x)}{J_0(2)}] = v(0, x) = \sum_{k = 0}^{\infty} T_k(0)X_k\\ 0 = v_t(0, x) = \sum_{k = 0}^{\infty} T'_k(0)X_k \end{cases} \end{equation}$$
$$\text{and here is part I got stuck. How can I find } T_k \text{ using Fourier expansion. I don't think I will be able to evaluate following integrals}$$
$$\begin{equation} C_k(t) = \frac{<-2cos2t, X_k>}{||X_k||^2} = \frac{-\int_{0}^{1} 2cos2tX_kdx}{||X_k||^2} \end{equation}$$
$$\begin{equation} T_k(0) = \int \frac{1}{2} [1 - \frac{J_0(2x)}{J_0(2)}]X_k(x)dx \end{equation}$$
I don't think I've any made errors on the way here. How can I take it from here?
