$$\lambda_h \frac{\partial^2 \theta_w}{\partial x^2} + \lambda_c V \frac{\partial^2 \theta_w}{\partial y^2} =\beta_h e^{-\beta_h x} \int e^{\beta_h x} \theta_w(x,y) \, \mathrm{d}x + \beta_c e^{-\beta_c y} \int e^{\beta_c y} \theta_w(x,y) \, \mathrm{d}y$$
on a domain where $x,y$ vary between $0$ to $1$. The bc(s) are $$\frac{\partial \theta_w(0,y)}{\partial x}=\frac{\partial \theta_w(1,y)}{\partial x}=0 $$
$$\frac{\partial \theta_w(x,0)}{\partial y}=\frac{\partial \theta_w(x,1)}{\partial y}=0 $$
The two terms on the RHS come from solving the following two $$ \frac{\partial \theta_h}{\partial x} + \beta_h (\theta_h - \theta_w) = 0, \frac{\partial \theta_c}{\partial y} + \beta_c (\theta_c - \theta_w) = 0$$
which have bc(s) as $\theta_h(0,y)=1$ and $\theta_c(x,0)=0$. Then subsequently $\theta_h$ and $\theta_c$ were eliminated from the following $$\lambda_h \frac{\partial^2 \theta_w}{\partial x^2} + \lambda_c V \frac{\partial^2 \theta_w}{\partial y^2} = \frac{\partial \theta_h}{\partial x} + V\frac{\partial \theta_c}{\partial y} $$
to reach the top equation
So to start i had three coupled PDEs
\begin{eqnarray} \frac{\partial \theta_h}{\partial x} + \beta_h (\theta_h - \theta_w) &=& 0,\\ \frac{\partial \theta_c}{\partial y} + \beta_c (\theta_c - \theta_w) &=& 0,\\ \lambda_h \frac{\partial^2 \theta_w}{\partial x^2} + \lambda_c V \frac{\partial^2 \theta_w}{\partial y^2} - \frac{\partial \theta_h}{\partial x} - V\frac{\partial \theta_c}{\partial y} &=& 0 \end{eqnarray} and then i worked onto the top equation
Any advice on how to move forward with this problem ?