So you have that a heat flows between the two fluids, proportional to $T_1(x)-T_0(x)$ , over some time $\delta t$: while in that same time the fluid moves forward by an amount $v~\delta t$. With suitably integrated “linear” specific heat capacities $c_{0,1}$ and some shared energy transfer coefficient $\kappa$ this means
$$
c_1 [ T_1(x+v_1\delta t) - T_1(x) ] =
-\kappa [ T_1(x)-T_0(x) ]\delta t\\
c_0 [ T_0(x+v_0\delta t) - T_0(x) ] =
-\kappa [ T_0(x)-T_1(x) ]\delta t
$$
and expanding the term on the left to first order we get a matrix equation
$$
\frac{\mathrm d\phantom t}{\mathrm d x}\begin{bmatrix}T_0\\T_1\end{bmatrix}=-\kappa \begin{bmatrix}
1/(v_0 c_0)&-1/(v_0 c_0)\\
-1/(v_1 c_1)&1/(v_1 c_1)
\end{bmatrix}\begin{bmatrix}
T_0\\T_1
\end{bmatrix}.
$$
To analyze this system is to analyze this matrix. Let me simplify,
$$ \mathbf M = -\kappa \begin{bmatrix}
1/(v_0 c_0)&-1/(v_0 c_0)\\
-1/(v_1 c_1)&1/(v_1 c_1)
\end{bmatrix} = \begin{bmatrix}
-\alpha&\alpha\\
\beta&-\beta
\end{bmatrix}
$$
There is one honking obvious eigenvector,$ \begin{bmatrix}1\\1
\end{bmatrix}$. It has eigenvalue 0, and this says that when the two temperatures are the same there is no heat transfer and you are in a steady state. This matrix is, in other words, a projection!
Because the trace $-\alpha-\beta$ is the sum of the eigenvalues and we already know one eigenvalue to be zero, that actually is the other eigenvalue. The direction is then easy to solve: the other eigenvector is $\begin{bmatrix}
\alpha\\
-\beta
\end{bmatrix}.$
Once you have done this you have diagonalized the equations, and the rest is linearity. So we decompose the initial condition into the diagonalizing basis for $\mathbf M$ as
$$
\begin{bmatrix}T_0(0)\\
T_1(0)
\end{bmatrix}
=\frac{\beta T_0(0) +\alpha T_1(0)}{\alpha +\beta}
\begin{bmatrix}1\\1
\end{bmatrix}
+
\frac{T_0(0)-T_1(0)}{\alpha +\beta}
\begin{bmatrix}\alpha\\-\beta
\end{bmatrix}
$$
and then we use linearity to study how each of these two terms evolves over $x$, since we can reconstruct the full evolution as a sum of the independent evolutions for linear differential equations.
Well, we said that for the first term $\mathrm d\vec T/\mathrm dx=0$ so that term stays constant independent of $x$. But for the second term we have the equation$$
{\mathrm d\vec T\over\mathrm dx}=-(\alpha+\beta) \vec T,\\
\vec T(x) = \vec T(0) \exp(-(\alpha+\beta)x).
$$
Combining the two, yes, we have an exponential decay of temperature along the length of the heat exchanger, when run in this parallel configuration. The first term is the steady state, and the second term is the part that is exponentially decaying.
