Could someone help solve this PDE? Given,
$U_t-U_{xx}-2U_x=0$
Using method of separation of variables, find ALL possible solution.
My answer :
$U(x,t)=X(x)T(t)$
$T'(t)/T(t)$=$[X''(x)+2X'(x)]/X(x)$=$\lambda$
From here I'm stuck. Could someone show me some working steps for me to proceed further.
 A: The solution to 
$$ T'(t) = \lambda T(t) $$
is simply $T = A e^{\lambda t}$, where $A$ is a parameter. 
The solution to 
$$ X''(x) + 2X'(x) - \lambda X(x) = 0$$ 
can be found by setting $r_\pm = -1 \pm \sqrt{ 1 + \lambda}$ and writing as
$$ \left(\frac{d}{dx} + r_+\right)\left( \frac{d}{dx} + r_-\right) X(x) = 0 $$
so by using the method of integrating factors we have
$$ e^{-r_+ x} \frac{d}{dx} e^{r_+ x - r_-x} \frac{d}{dx} e^{r_- x} X(x) = 0 $$
which we can solve by directly integrating and obtaining
$$ X(x) = C_+ e^{-r_+ x} + C_- e^{-r_- x} $$
where $C_\pm$ are parameters. 
A: This is exactly the same question as PDE separation of variables.
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.
