= Solution
{c}
With $x=2z-1$, the <chain rule> gives $T_{xx}=V_{zz}/4$. Thus
$$
V_t=\frac14V_{zz},\quad0<z<1,\qquad V(0,t)=V(1,t)=0,\qquad V(z,0)=T_0.
$$
<Separation of variables> gives the <eigenfunctions> $\sin(n\pi z)$ with time factors $e^{-n^2\pi^2t/4}$. The <Fourier sine coefficients> of the initial constant are
$$
b_n=2\int_0^1T_0\sin(n\pi z)\,dz=\frac{2T_0(1-(-1)^n)}{n\pi}.
$$
Only odd modes survive, so
$$
\boxed{V(z,t)=\frac{4T_0}{\pi}\sum_{k=0}^\infty\frac{\sin((2k+1)\pi z)}{2k+1}e^{-(2k+1)^2\pi^2t/4}.}
$$
Returning to $x$, $\sin((2k+1)\pi(x+1)/2)=(-1)^k\cos((2k+1)\pi x/2)$, hence
$$
\boxed{T(x,t)=\frac{4T_0}{\pi}\sum_{k=0}^\infty\frac{(-1)^k}{2k+1}\cos\!\left(\frac{(2k+1)\pi x}{2}\right)e^{-(2k+1)^2\pi^2t/4}.}
$$
For every $t>0$ the exponential coefficients give <uniform convergence> of the differentiated series on later-time compact sets, verifying the PDE and <boundary conditions>. As $t\downarrow0$, the solution tends to $T_0$ at every interior point and in <L2 norm>, while remaining zero at the endpoints. The expression is visibly even in $x$, consistent with the corrected symmetry argument for $b=0$.
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