I am working on solving the following boundary value problem over an annulus with inner radius 1 and outer radius 2.
$$\frac{\partial^2 u}{\partial r^2}+\frac{1}{r}\frac{\partial u}{\partial r}+\frac{1}{r^2}\frac{\partial^2 u}{\partial \theta^2}=0, 1<r<2, -\pi \le \theta \le \pi$$ $$u(1,\theta)=\sin{4\theta}-\cos{\theta}, -\pi \le \theta \le \pi$$ $$u(2,\theta)=\sin{\theta}, -\pi \le \theta \le \pi$$
The separation of variables $u(r,\theta)=R(r)T(\theta)$ yields the following equations for some constant $\lambda$.
$$\frac{r^2R''(r)+rR'(r)}{R(r)}=-\frac{T''(\theta)}{T(\theta)}=\lambda$$
The continuity equations $T(-\pi)=T(\pi)$ and $T'(-\pi)=T'(\pi)$ shows that $\lambda$ must be non-negative for a nontrivial solution. For $\lambda=0$, $T(\theta)=A$, and for $\lambda>0$:
$$T(\theta)=A_n\cos{\sqrt{\lambda}\theta}+B_n\sin{\sqrt{\lambda}\theta}$$
From this, assuming a nontrivial solution implies that $\sqrt{\lambda}=n$ for some positive integer $n$.
Using $\lambda=n^2$, the equation for $R(r)$ is the Cauchy-Euler equation:
$$r^2R''(r)+rR'(r)-n^2R(r)=0$$
With $\lambda=0$ leading to the solution $R_0(r)=C+D\ln{r}$, and $\lambda>0$ leading to the solution $R_n(r)=C_nr^n+D_nr^{-n}$.
Thus, the general solution for $u(r,\theta)$ is given as follows:
$$u(r,\theta)=C+D\ln{r}+\sum_{n=1}^{\infty}(C_nr^n+D_nr^{-n})(A_n\cos{n\theta}+B_n\sin{n\theta})$$
Since the set of functions $\{1, \cos{\theta}, \cos{2\theta},..., \sin{\theta}, \sin{2\theta}, ...\}$ is linearly independent, the $n\notin\{1,4\}$ terms are zero in the $r=1$ boundary condition, producing:
$$\sin{4\theta}-\cos{\theta}=(C_4+D_4)(A_4\cos{4\theta}+B_4\sin{4\theta})+(C_1+D_1)(A_1\cos{\theta}+B_1\sin{\theta})$$
And then coefficient matching yields the system:
$$0=(C_4+D_4)A_4$$ $$1=(C_4+D_4)B_4$$ $$-1=(C_1+D_1)A_1$$ $$0=(C_1+D_1)B_1$$
Thus, $A_4=0$ and $B_1=0$. Likewise, the $r=2$ boundary condition produces:
$$\sin{\theta}=(C_1 2^1+D_1 2^{-1})(A_1\cos{\theta}+B_1\sin{\theta})$$
Which yields the equation $1=(C_1 2^1+D_1 2^{-1})B_1$ in direct contradiction of $B_1=0$ found from the previous boundary condition. Since the trivial solution does not satisfy the boundary conditions, it seems like there is no solution to this problem.
Edit: It turns out that the listen solution (given in the answer below) does satisfy the boundary condition at $r=2$, my mistake. However, the issue I have with this a solution of the form
$$(\alpha_1r+\gamma_1r^{-1})\cos{\theta}+(\beta_1r+\delta_1r^{-1})\sin{\theta}$$
is not necessarily factorable into the form $(C_1r+D_1r^{-1})(A_1\cos{\theta}+B_1\sin{\theta})$ and, in fact, with the stated solution, it is not. This is where my confusion lies.
Let's redo the algebra a little bit. Extracting from your general solution the terms involving the functions that appear on the boundary, we posit a solution of the form $$ u(r,\theta) = \beta_4 r^4 \sin(4\theta) + \delta_4 r^{-4} \sin(4\theta) + \alpha_1 r \cos(\theta) + \beta_1 r \sin(\theta) \\ + \gamma_1 r^{-1} \cos(\theta) +\delta_1 r^{-1} \sin(\theta). $$ At $r=1$ the equation reads $$ \sin(4\theta) - \cos(\theta) = (\beta_4+\delta_4) \sin(4\theta) + (\alpha_1 + \gamma_2) \cos(\theta) + (\beta_1 + \delta_1) \sin(\theta), $$ which yields the equations $$ \beta_4 + \delta_4 =1 \\ \alpha_1 + \gamma_1 = -1 \\ \beta_1 + \delta_1 = 0. $$ At $r=2$ the equation reads $$ \sin(\theta) = (2^4 \beta_4 + 2^{-4} \delta_4) \sin(4\theta) +(2\alpha_1 +2^{-1} \gamma_2) \cos(\theta) +(2\beta_2 + 2^{-1} \delta_1)\sin(\theta), $$ which yields the equations $$ 2^4 \beta_4 + 2^{-4} \delta_4 = 0\\ 2 \alpha_1 + 2^{-1} \gamma_1 =0 \\ 2\beta_1 + 2^{-1} \delta_1 =1. $$ We can solve the resulting linear system with relative ease, as it decouples into three pairs of equations with two unknowns each. We get $$ \beta_4 = -1/255, \delta_4 = 256/255, \alpha_1 = 1/3, \gamma_1 = -4/3, \\ \beta_1 = 2/3, \delta_1 = -2/3. $$
Hence $$ u(r,\theta) = \frac{1}{3} r \cos(\theta) - \frac{4}{3} r^{-1} \cos(\theta) + \frac{2}{3} r \sin(\theta) - \frac{2}{3} r^{-1} \sin(\theta) \\ - \frac{1}{255} r^4 \sin(4\theta) + \frac{256}{255} r^{-4}\sin(4\theta), $$ and it's a simple matter to chack by hand that this does the job.
EDIT: Let's look again at the general solution.
You show above that a separated solution $u(r,\theta) = R(r) T(\theta)$ leads to solutions of the form $$ R(r) = r^n \text{ or } R(r) = r^{-n} $$ and $$ T(\theta) = \cos(n\theta) \text{ or } T(\theta) = \sin(n\theta) $$ when $n \ge 1$, and $$ R(r) = 1 \text{ or } R(r) = \log(r) \text{ and } T(\theta) = 1 $$ when $n=0$. From this we get four possible solutions when $n \ge 1$: $$ u_n(r,\theta)= r^n \cos(n\theta) \\ u_n(r,\theta) = r^{-n} \cos(n\theta) \\ u_n(r,\theta) = r^n \sin(n \theta) \\ u_n(r,\theta) = r^{-n} \sin(n\theta) $$ and two possible solutions when $n=1$: $$ u_0(r,\theta) = 1 \text{ and } u_0(r,\theta) = \log(r). $$ The general form of the solution is then a linear combination of these: $$ u(r,\theta) = \omega + \chi \log(r) \\ + \sum_{n=1}^\infty \left[\alpha_n r^n \cos(n\theta) + \beta_n r^n \sin(n\theta) + \gamma_n r^{-n} \cos(n\theta) + \delta_n r^{-n} \sin(n\theta) \right]. $$
If we start with this then we're led naturally to my solution. The error you're running into is due to the factored form you put the solution in. To invert we need to solve $$ C_n A_n = \alpha_n, C_n B_n = \beta_n, D_n A_n = \gamma_n, D_n B_n = \delta_n. $$ I don't think we can invert this in a nice way.