Stuck on solving this PDE of 2nd Order I'm trying to solve this second order PDE:
$$-u_{xx}+2u_{xy}+3u_{yy}-(1/3)u_x+u_y=0$$
I've got up to $48V_{st}+4V_t=0$ and this implies $48V_s+4V=f(s)$. I'm stuck on how to find the $g(t)$ solution. Could I have some help, please?
 A: Consider this answer for finding separable solutions. We assume that $V$ is separable, i.e.,
$$V = V(s,t) = S(s)T(t)\tag{1}$$
putting this into your last equation, $48{V_s} +4 V = f(s)$, we find
$$48S'(s)T(t) +4 S(s)T(t) = f(s)\tag{2}$$
divide by $S(s)T(t)$ to get
$$48{{S'(s)} \over {S(s)}} + 4 = {{f(s)} \over {S(s)}}{1 \over {T(t)}}\tag{3}$$
the point at this step is that $(3)$ can only hold when the right hand side is a function of $s$ only. This will lead to
$${1 \over {T(t)}} = k\,\,\,\,\, \to \,\,\,\,48{{S'(s)} \over {S(s)}} + 4 = k{{f(s)} \over {S(s)}}\,\,\,\,\,\, \to \,\,\,\,\,\,48S'(s) +4 S(s) = kf(s)\tag{4}$$
and finally the separated parts must satisfy the following simultaneously
$$\left\{ \matrix{
  S'(s) + {1 \over {12}}S(s) = {k \over {48}}f(s) \hfill \cr 
  T(t) = {1 \over k} \hfill \cr}  \right.\tag{5}$$
Hence, you just need to solve the linear first order ODE for $S(s)$. As a recall, consider the following
$$\eqalign{
  & y'(x) + P(x)y(x) = Q(x)  \cr 
  & y(x) = {e^{ - A(x)}}\left( {y(a) + \int_a^x {Q(t){e^{A(t)}}dt} } \right)  \cr 
  & A(x) = \int_a^x {P(t)dt}  \cr}\tag{6} $$
where $a$ is an arbitrary real number. Then in your ODE
$$\eqalign{
  & S \equiv y,\,\,\,\,\,\,\,\,\,s \equiv x  \cr 
  & P(s) = {1 \over {12}}  \cr 
  & Q(s) = {k \over {48}}f(s)  \cr 
  & A(s) = \int_a^s {P(t)dt}  = \int_a^s {{1 \over {12}}dt}  = {1 \over {12}}(s - a) \cr}\tag{7} $$
and hence
$$S(s) = {e^{ - {1 \over {12}}(s - a)}}\left( {S(a) + {k \over {48}}\int_a^s {f(t){e^{{1 \over {12}}(t - a)}}dt} } \right)\tag{8}$$
A: *

*$$-{{D}_{x}^{2}}+2 {D_y}\, {D_x}+3 {{D}_{y}^{2}}-\frac{{D_x}}{3}+{D_y}=
\frac{\left( 3 {D_y}-{D_x}\right) \, \left( 3 {D_y}+3 {D_x}+1\right) }{3}
$$

*Solution of $\;3u_y-u_x=0\;$ is $\;u_1=f(3x+y)$

*Solution of $\;3u_y+3u_x+u=0\;$ is $\;u_2=e^{-\frac y3}g(x-y)$

*Then general solution of PDE $\quad-u_{xx}+2u_{xy}+3u_{yy}-(1/3)u_x+u_y=0$
$$u=u_1+u_2=f(3x+y)+e^{-\frac y3}g(x-y)$$

