Solution: The Dirichlet problem in the polar form can be written as \begin{equation}\label{eq:23Aug2024-1} \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, \end{equation} with $0 < r < 2$ and $\theta \in [0,2\pi ]$. The boundary condition will be \begin{align*} u(2,\theta ) & = 1 + 2(2\cos \theta )^2 = 1 + 4(2\cos ^2\theta) \\ & = 1 + 4 \left( 1 + \cos 2\theta \right) = 5 + 4 \cos 2\theta . \end{align*} We recall that the solution of the equation \eqref{eq:23Aug2024-1} is given by \begin{equation}\label{eq:23Aug2024-2} u(r,\theta ) = A_0 + \sum_{n=1}^{\infty} r^n \left[ A_n \cos (n \theta ) + B_n \sin (n \theta ) \right]. \end{equation} Applying the boundary condition to the above solution, we get \begin{align*} 5 + 4 \cos 2\theta = A_0 + \sum_{n=1}^{\infty} 2^n \left[ A_n \cos (n \theta ) + B_n \sin (n \theta ) \right]. \end{align*} This implies, \begin{gather*} A_0 = 5, \quad A_2 = 1, \quad A_n = 0, \quad n \neq 0,2\\ B_n = 0, \quad n \in \mathbb{N} . \end{gather*} Therefore, the solution to the given equation will be \[ u(r,\theta ) = 5 + r^2 \cos(2\theta). \] Note that $x= 0 , y = 1$ means $r = 1$ and $\theta = \frac{\pi}{2}$. Thus, \[ u(0,1) = u \left(1, \frac{\pi}{2}\right) = 5 - 1 = 4. \]