Show in detail how integration for the final answer was found, using the IC and BC's.

The initial and boundary conditions are given by T(0,x2)=0,x2T(x1,0)=0,x2T(x1,1)=1T(x1,x2)=c0x1+f(x2) Substitution in the temperature equation leads to the following: dx22d2f=c0(1x22). Integration and using the boundary conditions gives T(x1,x2)=23x1+43(x226x24)+c2. The initial and boundary conditions are given by T(0,x2)=0,x2T(x1,0)=0,x2T(x1,1)=1T(x1,x2)=c0x1+f(x2) Substitution in the temperature equation leads to the following: dx22d2f=c0(1x22). Integration and using the boundary conditions gives T(x1,x2)=23x1+43(x226x24)+c2