In each of the following problems, use the method of separation of variables to solve the one$-$dimensional heat equation. Also, define the steady$-$state temperature at the midpoint of the region.

a$.\frac{\text{d}\text{e}\text{l}\text{T}}{\text{d}\text{e}\text{l}\text{t}}=\frac{de{l}^{2}T}{del{x}^2},T(-,t)=T(,t),\frac{\text{d}\text{e}\text{l}\text{T}(-,t)}{\text{d}\text{e}\text{l}\text{x}}=\frac{\text{d}\text{e}\text{l}\text{T}(,t)}{\text{d}\text{e}\text{l}\text{x}}\frac{\text{d}\text{e}\text{l}\text{T}}{\text{d}\text{e}\text{l}\text{t}}=\frac{1}{2}\frac{de{l}^{2}T}{del{x}^2},T(0,t)=100,T(2,t)=50,T(x,0)=100-13x0T(x,0)=x+,-$

$d.\frac{\text{d}\text{e}\text{l}\text{T}}{\text{d}\text{e}\text{l}\text{t}}=\frac{1}{2}\frac{de{l}^{2}T}{del{x}^2},T(0,t)=100,T(2,t)=50,T(x,0)=100-13x,0\frac{\text{d}\text{e}\text{l}\text{T}}{\text{d}\text{e}\text{l}\text{t}}=\frac{de{l}^{2}T}{del{x}^2},\frac{\text{d}\text{e}\text{l}\text{T}(0,t)}{\text{d}\text{e}\text{l}\text{x}}=\frac{\text{d}\text{e}\text{l}\text{T}(,T)}{\text{d}\text{e}\text{l}\text{x}}=0,T(x,0)=cosx,0$

$c.\frac{\text{d}\text{e}\text{l}\text{T}}{\text{d}\text{e}\text{l}\text{t}}=\frac{de{l}^{2}T}{del{x}^2},T(-,t)=T(,t),\frac{\text{d}\text{e}\text{l}\text{T}(-,t)}{\text{d}\text{e}\text{l}\text{x}}=\frac{\text{d}\text{e}\text{l}\text{T}(,t)}{\text{d}\text{e}\text{l}\text{x}}$

$T(x,0)=x+,-$

$d.\frac{\text{d}\text{e}\text{l}\text{T}}{\text{d}\text{e}\text{l}\text{t}}=\frac{1}{2}\frac{de{l}^{2}T}{del{x}^2},T(0,t)=100,T(2,t)=50,T(x,0)=100-13x,0\frac{\text{d}\text{e}\text{l}\text{T}}{\text{d}\text{e}\text{l}\text{t}}=\frac{1de{l}^{2}T}{2del{x}^2},T(0,t)=T(3,t)=0,T(x,0)=sinx,0$

$b.\frac{\text{d}\text{e}\text{l}\text{T}}{\text{d}\text{e}\text{l}\text{t}}=\frac{de{l}^{2}T}{del{x}^2},\frac{\text{d}\text{e}\text{l}\text{T}(0,t)}{\text{d}\text{e}\text{l}\text{x}}=\frac{\text{d}\text{e}\text{l}\text{T}(,T)}{\text{d}\text{e}\text{l}\text{x}}=0,T(x,0)=cosx,0$

$c.\frac{\text{d}\text{e}\text{l}\text{T}}{\text{d}\text{e}\text{l}\text{t}}=\frac{de{l}^{2}T}{del{x}^2},T(-,t)=T(,t),\frac{\text{d}\text{e}\text{l}\text{T}(-,t)}{\text{d}\text{e}\text{l}\text{x}}=\frac{\text{d}\text{e}\text{l}\text{T}(,t)}{\text{d}\text{e}\text{l}\text{x}}$

$T(x,0)=x+,-$

$d.\frac{\text{d}\text{e}\text{l}\text{T}}{\text{d}\text{e}\text{l}\text{t}}=\frac{1}{2}\frac{de{l}^{2}T}{del{x}^2},T(0,t)=100,T(2,t)=50,T(x,0)=100-13x,0$