PDE separation of variables Hi could someone guide me this problem
It says ,
$ u_t - u_{xx}-2 u_x=0  $
Use the method of separation of variables to find all possible solutions. 
Could someone help me out for this problem. I'm beginner at PDE. I would be much appreciated if you able to show some partial work so that I can understand.
Thanks in advance for taking my consideration
 A: If $u(x,t) = X(x)T(t)$, then 
$$\frac{T'}{T} = \frac{X''}{X} + 2 \frac{X'}{X}$$
LHS depends on $t$ only, RHS depends on $x$ only, so both sides are equal to a constant, say $\lambda$.  You can take it from here...
A: Case $1$: $\text{Re}(t)\geq0$
Let $u(x,t)=X(x)T(t)$ ,
Then $X(x)T'(t)-X''(x)T(t)-2X'(x)T(t)=0$
$X(x)T'(t)=X''(x)T(t)+2X'(x)T(t)$
$X(x)T'(t)=(X''(x)+2X'(x))T(t)$
$\dfrac{T'(t)}{T(t)}=\dfrac{X''(x)+2X'(x)}{X(x)}=-(f(s))^2-1$
$\begin{cases}\dfrac{T'(t)}{T(t)}=-(f(s))^2-1\\X''(x)+2X'(x)+((f(s))^2+1)X(x)=0\end{cases}$
$\begin{cases}T(t)=c_3(s)e^{-t((f(s))^2+1)}\\X(x)=\begin{cases}c_1(s)e^{-x}\sin(xf(s))+c_2(s)e^{-x}\cos(xf(s))&\text{when}~f(s)\neq0\\c_1xe^{-x}+c_2e^{-x}&\text{when}~f(s)=0\end{cases}\end{cases}$
$\therefore u(x,t)=C_1xe^{-x-t}+C_2e^{-x-t}+\int_sC_3(s)e^{-x-t((f(s))^2+1)}\sin(xf(s))~ds+\int_sC_4(s)e^{-x-t((f(s))^2+1)}\cos(xf(s))~ds~\text{or}~C_1xe^{-x-t}+C_2e^{-x-t}+\sum\limits_sC_3(s)e^{-x-t((f(s))^2+1)}\sin(xf(s))+\sum\limits_sC_4(s)e^{-x-t((f(s))^2+1)}\cos(xf(s))$
Case $2$: $\text{Re}(t)\leq0$
Let $u(x,t)=X(x)T(t)$ ,
Then $X(x)T'(t)-X''(x)T(t)-2X'(x)T(t)=0$
$X(x)T'(t)=X''(x)T(t)+2X'(x)T(t)$
$X(x)T'(t)=(X''(x)+2X'(x))T(t)$
$\dfrac{T'(t)}{T(t)}=\dfrac{X''(x)+2X'(x)}{X(x)}=(f(s))^2-1$
$\begin{cases}\dfrac{T'(t)}{T(t)}=(f(s))^2-1\\X''(x)+2X'(x)+(1-(f(s))^2)X(x)=0\end{cases}$
$\begin{cases}T(t)=c_3(s)e^{t((f(s))^2-1)}\\X(x)=\begin{cases}c_1(s)e^{-x}\sinh(xf(s))+c_2(s)e^{-x}\cosh(xf(s))&\text{when}~f(s)\neq0\\c_1xe^{-x}+c_2e^{-x}&\text{when}~f(s)=0\end{cases}\end{cases}$
$\therefore u(x,t)=C_1xe^{-x-t}+C_2e^{-x-t}+\int_sC_3(s)e^{-x+t((f(s))^2-1)}\sinh(xf(s))~ds+\int_sC_4(s)e^{-x+t((f(s))^2-1)}\cosh(xf(s))~ds~\text{or}~C_1xe^{-x-t}+C_2e^{-x-t}+\sum\limits_sC_3(s)e^{-x+t((f(s))^2-1)}\sinh(xf(s))+\sum\limits_sC_4(s)e^{-x+t((f(s))^2-1)}\cosh(xf(s))$
Hence $u(x,t)=\begin{cases}C_1xe^{-x-t}+C_2e^{-x-t}+\int_sC_3(s)e^{-x-t((f(s))^2+1)}\sin(xf(s))~ds+\int_sC_4(s)e^{-x-t((f(s))^2+1)}\cos(xf(s))~ds&\text{when}~\text{Re}(t)\geq0\\C_1xe^{-x-t}+C_2e^{-x-t}+\int_sC_3(s)e^{-x+t((f(s))^2-1)}\sinh(xf(s))~ds+\int_sC_4(s)e^{-x+t((f(s))^2-1)}\cosh(xf(s))~ds&\text{when}~\text{Re}(t)\leq0\end{cases}$
or $\begin{cases}C_1xe^{-x-t}+C_2e^{-x-t}+\sum\limits_sC_3(s)e^{-x-t((f(s))^2+1)}\sin(xf(s))+\sum\limits_sC_4(s)e^{-x-t((f(s))^2+1)}\cos(xf(s))&\text{when}~\text{Re}(t)\geq0\\C_1xe^{-x-t}+C_2e^{-x-t}+\sum\limits_sC_3(s)e^{-x+t((f(s))^2-1)}\sinh(xf(s))+\sum\limits_sC_4(s)e^{-x+t((f(s))^2-1)}\cosh(xf(s))&\text{when}~\text{Re}(t)\leq0\end{cases}$
This is already the general solution of $u_t-u_{xx}-2u_x=0$ . Note that when without any I.C.s, the form of $f(s)$ can choose arbitrary, but when I.C.s are given, the form of $f(s)$ and the choice whether using the integration kernel or using the summation kernel should choose wisely in order to accommodate the I.C.s to get the most nice form of the solution, especially the number of I.C.s is more than two.
