Modified heat equation - solving PDE Conisder the following two PDEs:
$(1)$:$\;$$\displaystyle\frac{\partial \omega_1'}{\partial t}+A x_2 \frac{\partial \omega_1'}{\partial x_1}=\nu\left(\frac{\partial^2 \omega_1'}{\partial x_2^2}+\frac{\partial^2 \omega_1'}{\partial x_3^2}\right)$
and
$(2)$:$\;$
$
\begin{cases}
\displaystyle\frac{\partial \omega_1'}{\partial t}+A  \frac{\partial u_3'}{\partial x_1}+A x_2 \frac{\partial \omega_1'}{\partial x_1}=\nu\left(\frac{\partial^2 \omega_1'}{\partial x_2^2}+\frac{\partial^2 \omega_1'}{\partial x_3^2}\right)\\ \\
\displaystyle\omega_1'=\frac{\partial u_3'}{\partial x_2}-\frac{\partial u_2'}{\partial x_3}\\ \\
\displaystyle\frac{\partial u_2'}{\partial x_2}+\frac{\partial u_3'}{\partial x_3}=0
\end{cases}
$
The first PDE $(1)$ highly resembles the heat equation, aswell as the first equation in $(2)$.

How can I solve for $\omega_1'$ in both cases?

I know how to solve the heat equation using the method of separation of variables but I do not know how to apply it on these PDEs.
I am very grateful for any advice.
 A: $$\displaystyle\frac{\partial \omega_1'}{\partial t}+A x_2 \frac{\partial \omega_1'}{\partial x_1}=\nu\left(\frac{\partial^2 \omega_1'}{\partial x_2^2}+\frac{\partial^2 \omega_1'}{\partial x_3^2}\right)$$
They are four variables : $x_1\:,x_2\:,x_3\:,t.$ The function has to be separated into four functions :
$$\omega_1'=f(x_1)g(x_2)h(x_3)v(t)$$
$$fghv'+Ax_2f'hgv=\nu(fg''hv+fgh''v)$$
$$\frac{v'}{v}+Ax_2\frac{f'}{f}-\nu\frac{g''}{g}-\nu\frac{h''}{h}=0$$
A function of one variable can be equal to a function of a different variable any values of the variables only if both functions are equal to a common constant. Thus :
$$\begin{cases}
\frac{v'}{v}=a \qquad\text{function of}\quad t\quad\text{only.}\\
\frac{f'}{f}=b \qquad\text{function of}\quad x_1\quad\text{only.}\\
\frac{h''}{h}=c \qquad\text{function of}\quad x_3\quad\text{only.}\\
a+bAx_2-\nu\frac{g''}{g}-c\nu=0 \qquad\text{function of}\quad x_2\quad\text{only.}
\end{cases}$$
$g$ is obtained in solving the ODE : $\quad \nu g''=(a+bAx_2-c\nu)g\quad$ which is an Airy ODE.
$a,b,c$ are arbitrary real or complex constants.
$$\begin{cases}
v=c_1e^{at}\\
f=c_2e^{bx_1}\\
h=c_3e^{\pm\sqrt{c}\:x_3}\\
g=c_4\text{Ai}\left(z\right)+c_5\text{Bi}\left(z\right)\quad ;\quad z=\frac{a-c\nu+bAx_2}{(bA)^{2/3}\nu^{1/3}}
\end{cases}$$
$\text{Ai}(z)$ and $\text{Bi}(z)$ are the Airy functions.
$c_1,c_2,c_3,c_4,c_5$ are are arbitrary constants.
Independant solutions of the PDE :
$$\omega_1'=C_1e^{at}e^{bx_1}e^{\pm\sqrt{c}\:x_3}\text{Ai}\left(\frac{a-c\nu+bAx_2}{(bA)^{2/3}\nu^{1/3}}\right)+C_2e^{at}e^{bx_1}e^{\pm\sqrt{c}\:x_3}\text{Bi}\left(\frac{a-c\nu+bAx_2}{(bA)^{2/3}\nu^{1/3}}\right)$$
Any linear combination of those independant solutions with different arbitrary constants $C_1,C_2,a,b,c$ is solution of the PDE.
HINT :
In the case of the system of three functions $\omega_1',u'_1,u'_2$ proceed on the same manner with :
$$\begin{cases}
\omega'_1=f_1(x_1)g_1(x_2)h_1(x_3)v_1(t)  \\
u'_2=f_2(x_1)g_2(x_2)h_2(x_3)v_2(t)  \\
u'_3=f_3(x_1)g_3(x_2)h_3(x_3)v_3(t)  \\
\end{cases}$$
Separate the functions of different variables as shown above and find the 12 independant functions each one of one variable only.
I let for you this awfully booring task. Good luck !
