Let $u_{xx}+u_{yy}=0$ for $(0,2\pi)\times(0,2\pi)$ with the conditions $u(0,y)=u(2\pi,y),$ $u_x(0,y)=u_x(2\pi,y)$ for $y\in(0,2\pi)$, $u(x,0)=0, u(x,2\pi)=u_0(x)$ for $x\in(0,2\pi).$ Find a solution using the Fourier method.
My approach: We are looking for a solution in the form $\sum_{k=1}^\infty X_k(x)Y_k(y),$ plugging into the equation $$\sum_{k=1}^\infty X_k''(x)Y_k(y)+X_k(x)Y_k''(y)=0.$$ From this we get $$-\frac{X_k''}{X_k}=\frac{Y_k''}{Y_k}=\lambda.$$ Now we solve for $X_k$,we have $\lambda\geq0$, because $$\int_0^{2\pi}\lambda X_k^2=\int_0^{2\pi}-X_k''X_k=[-X_kX_k]_0^{2\pi}+\int_0^{2\pi}X_k'^2$$ For $\lambda=0$ we have $X_k''=0$ thus $X_k(x)=Ax+B$, from conditions $X_k(0)=X_k(2\pi)$ we get $A=0$. For $\lambda>0$ let $\mu>0,\lambda=\mu^2$ $X_k(x)=A\cos(\mu x)+B\sin(\mu x)$, plugging the conditions yields $$A=A\cos(\mu 2\pi)+B\sin(\mu 2\pi)$$ $$B=-Asin(\mu 2\pi)+B\cos(\mu 2\pi)$$ I'm not really sure how to get the coefficients from these equations and how to continue from here. Any help is welcome.