= Solution
A <material derivative> measures the rate experienced by a moving fluid particle. If its position is $\boldsymbol X(t)$ with $\dot{\boldsymbol X}=\boldsymbol u(\boldsymbol X,t)$, the <chain rule> for any scalar field $F$ gives
$$
\frac d{dt}F(\boldsymbol X(t),t)=\partial_tF+\dot{\boldsymbol X}\cdot\nabla F=\partial_tF+\boldsymbol u\cdot\nabla F.
$$
Hence the operator is $D/Dt=\partial_t+\boldsymbol u\cdot\nabla$, and the <material derivative> differs from the time derivative at a fixed spatial point.
In the steady temperature problem $\partial_t\theta=0$ and the transverse velocity vanishes. Thus at each fixed $y\in(-1,1)$,
$$
(1-y^2)\partial_x\theta=-\theta.
$$
Integrating in $x$ and applying the inlet value gives
$$
\boxed{\theta(x,y)=\theta_0\exp\left[-\frac{x}{1-y^2}\right],\qquad x>0,\ -1<y<1.}
$$
A particle's travel time from the inlet is $x/(1-y^2)$, so the formula is also the exponential decay along its <characteristic curve>. For fixed $x>0$ the temperature tends to zero near either wall, because the travel time becomes large. No value at $y=\pm1$ is imposed by an inlet condition there, since the wall velocity is zero and the stated channel is open in $y$.
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