How can I solve the following system of linear partial differential equations or simplify them to solvable form? both Z and Y depend on x and t variables.
\begin{align} \frac{\partial Y}{\partial t}+\frac{\partial Y}{\partial x}&=Z-Y \\ \frac{1}{c}\frac{\partial Z}{\partial t}&=Y-Z \end{align}
Let
$$k=\frac{1}{c}$$
We have
\begin{cases} \displaystyle Z(x,t)-Y(x,t)=\frac{\partial Y}{\partial x}+\frac{\partial Y}{\partial t}\\ \displaystyle Z(x,t)-Y(x,t)=-k\frac{\partial Z}{\partial t} \end{cases}
Trivial solutions require $Z(x,t)=Y(x,t)=\operatorname{const}.$
Separating each function for nontrivial solutions gives
\begin{cases} \displaystyle Z(x,t)=Y(x,t)+\frac{\partial Y}{\partial x}+\frac{\partial Y}{\partial t}\\ \displaystyle Y(x,t)=Z(x,t)+k\frac{\partial Z}{\partial t} \end{cases}
We can find $Z_t$ to decouple this system of equations
$$\frac{\partial Z}{\partial t}=\frac{\partial Y}{\partial t}+\frac{\partial^2 Y}{\partial t\partial x}+\frac{\partial^2 Y}{\partial t^2}$$
\begin{cases} \displaystyle Z(x,t)=Y(x,t)+\frac{\partial Y}{\partial x}+\frac{\partial Y}{\partial t}\\ \displaystyle Y(x,t)=Y(x,t)+\frac{\partial Y}{\partial x}+\frac{\partial Y}{\partial t}+k\left(\frac{\partial Y}{\partial t}+\frac{\partial^2 Y}{\partial t\partial x}+\frac{\partial^2 Y}{\partial t^2}\right) \end{cases}
Let
$$Y(x,t)=X(x)T(t)$$ $$Z(x,t)=\tilde{X}(x)\tilde{T}(t)$$
Then
\begin{cases} \tilde{X}\tilde{T}=XT+X'T+XT'\\ X'T+XT'+k\left(XT'+X'T'+XT''\right)=0 \end{cases}
Focus on the second equation and divide by $XT$
$$\frac{X'}{X}+\frac{T'}{T}+k\left(\frac{T'}{T}+\frac{X'T'}{XT}+\frac{T''}{T}\right)=0\implies (1+k)\frac{T'}{T}+k\frac{T''}{T}=-\left(1+k\frac{T'}{T}\right)\frac{X'}{X}$$
Taking partial derivatives on both sides, we have
$$0=-\left(1+k\frac{T'}{T}\right)\frac{\partial}{\partial x}\left(\frac{X'}{X}\right)$$
$$\frac{\partial}{\partial t}\left((1+k)\frac{T'}{T}+k\frac{T''}{T}\right)=-k\frac{X'}{X}\frac{\partial}{\partial t}\left(\frac{T'}{T}\right)$$
It seems possible at first that
$$k\frac{T'}{T}=-1$$
But the first equation is always satisfied since the second implies that
$$\frac{X'}{X}=\xi=\operatorname{const}$$
So it is not necessary nor consistent that $\displaystyle k\frac{T'}{T}=-1$.
We can continue with this new information
$$\xi+\frac{T'}{T}+k\frac{T'}{T}+k\xi\frac{T'}{T}+k\frac{T''}{T}=0$$
We have a first order ordinary differential equation for $X$ and a second order ordinary differential equation for $T$. Solving both automatically solves $Z(x,t)=\tilde{X}\tilde{T}$.
$$T''+\left(\frac{k\xi+k+1}{k}\right)T'+\frac{\xi}{k}T=0$$
Solving for $T$ becomes slightly less tedious if we let
$$\eta=\frac{k\xi+k+1}{k}\\ \mu=\eta^2-4\frac{\xi}{k}$$
Following the procedures of second order ODEs
$$r^2+\eta r+\frac{\xi}{k}=0$$
$$r=\frac{1}{2}\left(-\eta\pm\sqrt{\mu}\right)$$
So
$$T(t)=C_1e^{\frac{1}{2}\left(-\eta+\sqrt{\mu}\right)t}+C_2e^{-\frac{1}{2}\left(\eta+\sqrt{\mu}\right)t}\\X(x)=C_3e^{\xi x}$$
And we have our solutions
$$\boxed{Y(x,t)=e^{\xi x}\left(\tilde{C}_1e^{\frac{1}{2}\left(-\eta+\sqrt{\mu}\right)t}+\tilde{C}_2e^{-\frac{1}{2}\left(\eta+\sqrt{\mu}\right)t}\right)}$$ $$\boxed{Z(x,t)=(1+\xi)Y(x,t)+\frac{\partial Y}{\partial t}}$$
