Sea \( T=T(x ; y) \) es la función temperatura de una placa metálica en el punto \( (x ; y) \) y está dada por la ecuación \( \operatorname{sen}(x+T)+\operatorname{sen}(x+y)=1 \) ¿para que valor de la constante a en el punto \( \left(\pi ; \frac{\pi}{2} ; \pi\right) \) se verifica la ecuación: \[ x \frac{\partial^{2} T}{\partial x^{2}}+z \frac{\partial^{2} T}{\partial y^{2}}=a\left(\frac{\partial T}{\partial y}-\frac{\partial T}{\partial x}\right) \]

Para resolver este problema, primero necesitamos encontrar el valor de la constante \( a \) en el punto \( \left(\pi ; \frac{\pi}{2} ; \pi\right) \) que verifica la ecuación dada: \[ x \frac{\partial^{2} T}{\partial x^{2}}+z \frac{\partial^{2} T}{\partial y^{2}}=a\left(\frac{\partial T}{\partial y}-\frac{\partial T}{\partial x}\right) \] Dado que \( T = T(x, y) \), primero calculamos las derivadas parciales necesarias de \( T \) con respecto a \( x \) y \( y \). 1. **Derivadas parciales de \( T \):** \[ \frac{\partial T}{\partial x} = \frac{\partial}{\partial x} \left( \sin(x + T) + \sin(x + y) \right) \] \[ \frac{\partial T}{\partial x} = \cos(x + T) \cdot \frac{\partial T}{\partial x} + \cos(x + y) \] \[ \frac{\partial T}{\partial x} - \cos(x + T) \cdot \frac{\partial T}{\partial x} = \cos(x + y) \] \[ \left(1 - \cos(x + T)\right) \frac{\partial T}{\partial x} = \cos(x + y) \] \[ \frac{\partial T}{\partial x} = \frac{\cos(x + y)}{1 - \cos(x + T)} \] \[ \frac{\partial T}{\partial y} = \frac{\partial}{\partial y} \left( \sin(x + T) + \sin(x + y) \right) \] \[ \frac{\partial T}{\partial y} = \cos(x + T) \cdot \frac{\partial T}{\partial y} + \cos(x + y) \] \[ \frac{\partial T}{\partial y} - \cos(x + T) \cdot \frac{\partial T}{\partial y} = \cos(x + y) \] \[ \left(1 - \cos(x + T)\right) \frac{\partial T}{\partial y} = \cos(x + y) \] \[ \frac{\partial T}{\partial y} = \frac{\cos(x + y)}{1 - \cos(x + T)} \] 2. **Segundas derivadas parciales de \( T \):** \[ \frac{\partial^{2} T}{\partial x^{2}} = \frac{\partial}{\partial x} \left( \frac{\cos(x + y)}{1 - \cos(x + T)} \right) \] \[ \frac{\partial^{2} T}{\partial x^{2}} = \frac{-\sin(x + y) \cdot \frac{\partial T}{\partial x} - \cos(x + y) \cdot \frac{\partial^{2} T}{\partial x^{2}}}{(1 - \cos(x + T))^2} \] \[ \frac{\partial^{2} T}{\partial x^{2}} = \frac{-\sin(x + y) \cdot \frac{\cos(x + y)}{1 - \cos(x + T)} - \cos(x + y) \cdot \frac{\partial^{2} T}{\partial x^{2}}}{(1 - \cos(x + T))^2} \] \[ \frac{\partial^{2} T}{\partial x^{2}} = \frac{-\sin(x + y) \cdot \cos(x + y) - \cos(x + y) \cdot \frac{\partial^{2} T}{\partial x^{2}}}{(1 - \cos(x + T))^2} \] \[ \frac{\partial^{2} T}{\partial x^{2}} = \frac{-\sin(x + y) \cdot \cos(x + y)}{(1 - \cos(x + T))^2} - \frac{\cos(x + y) \cdot \frac{\partial^{2} T}{\partial x^{2}}}{(1 - \cos(x + T))^2} \] \[ \frac{\partial^{2} T}{\partial x^{2}} = \frac{-\sin(x + y) \cdot \cos(x + y) - \cos(x + y) \cdot \