We need to solve the following :
$$ \nabla^2=0, 0\leq \leq, 0\leq y\leq b $$ satisfying the boundary conditions $$ (0,y)=0, 0\leq y\leq b \\ (,0)=(,)=0, 0\leq \leq \\ _x(a,)=T\sin^3(πy/a) $$
I proceeded as following:
I'm pretty sure my steps till $Y_n(y)$ are correct. Everything beyond that was just a guess and most probably wrong. I doubt if the question itself, which was asked previously in an exam, is correct.
Help me regarding the correctness of either the question or the solution.
if you look at the non trivial boundary, i mean $u_x(a,y)$, you can see it has functionality of $y$ (as in general it should have) while $x'_n(a)$ is just a number! so this assignment $$ x'_n(a) = \frac{T \sin^3(\pi y/a)}{\sin(n \pi y/b)} $$ is either completely wrong or just mean the fraction in right hand side is a constant which doesn't seem true! so this is the start of where you should know something is wrong.
for solving this problem you should finish construction of $u(x,y)$ completely and then try to use non-trivial boundary condition. so till the point $$ x_n(x) = c_n \sinh (\frac{n\pi x}{b}) $$ everything is alright. ($c_n$ is still an unknown constant so $2c_n$ or $c_n$ doesn't matter)
so general representation of $u(x,y)$ would be: $$ u(x,y) = \sum_{n \in \mathbb{Z}} A_n u_n(x,y) =\sum_{n \in \mathbb{Z}} A_n x_n(x)y_n(y) \\ = \boxed {\sum_{n \in \mathbb{Z}} A_n \sinh (\frac{n\pi x}{b}) \sin(\frac{n\pi y}{b})}. \tag{1}\label{1} $$ therefore the problem is finding $A_n$. we also have $$ u_x(x,y) = \sum_{n \in \mathbb{Z}} A_n (\frac{n\pi}{b}) \cosh(\frac{n\pi x}{b}) \sin(\frac{n\pi y}{b}) = \frac{\pi}{b}\sum_{n \in \mathbb{Z}} n.A_n \cosh(\frac{n\pi x}{b}) \sin(\frac{n\pi y}{b}). $$ Now we should use condition $u_x(a,y)$: $$ u_x(a,y) = \frac{\pi}{b}\sum_{n \in \mathbb{Z}} n.A_n \cosh(\frac{n\pi a}{b}) \sin(\frac{n\pi y}{b}) = T \sin^3(\frac{\pi y}{a}) \tag{2}\label{2} $$ for solving this you should use orthoganality relation of $y_n(y)=\sin(\frac{n\pi y}{b})$ which is : $$ \int_0^b y_n(y) y_m(y) dy = \frac{b}{2} \delta_{n,m} $$ where $\delta_{n,m}$ is kroneker delta . So multiply $\eqref{2}$ by $y_m(y)=\sin(\frac{m\pi y}{b})$ and integrate it from $0$ to $b$ : $$ \frac{\pi}{b}\sum_{n \in \mathbb{Z}} n.A_n \cosh(\frac{n\pi a}{b}) (\frac{b}{2} \delta_{n,m}) = T \int_0^b \sin(\frac{m\pi y}{b}) \sin^3(\frac{\pi y}{a}) dy \\ A_m \frac{m \pi}{2} \cosh(\frac{m\pi a}{b}) = T \int_0^b \sin(\frac{m\pi y}{b}) \sin^3(\frac{\pi y}{a}) dy . $$ Hence $$ \boxed {A_n = \frac{2T}{n\pi \cosh(\frac{n\pi a}{b}) } \int_0^b \sin(\frac{n\pi y}{b}) \sin^3(\frac{\pi y}{a}) dy}. \tag{3}\label{3} $$ So $u(x,y)$ is expanding as $\eqref{1}$ with $A_n$ defined as $\eqref{3}$.
I'm pretty sure the integral $\eqref{3}$ should have explicit form but in case of $a$ be the denominator in $\sin^3$ it should be very complicated.