Tough PDE on Separation of Variables I want to know on how to solve this question :
Given the PDE is : $U_t -tU -txU_x+tU_x=0$
Use separation of variables to find ALL possible solutions
Could someone help me this question because the PDE is too long and very frustrating. It took me 2 hours to try to solve but end up I cant solve it. This my first lesson in PDE. If someone can show me till the final step I would be much much happy and appreciated.
I need to get this done because my final exam is coming soon.
Please guys help me out
This question is 25 marks. ( I got this question from past years midterm paper)
 A: The trick with separation of variables is to write the solution $U$ as a product $X(x)T(t)$, so that your equation becomes
$$ X T'-t X T-t x X' T + t X' T = 0$$
Now divide through by $t X T$ so that we completely separate the dependence on $x$ and $t$ from each other:
$$\frac{T'}{t T} - 1 = (x-1) \frac{X'}{X} $$
Note that the left-hand-side only depends on $t$, while the right-hand side only depends on $x$.  Thus, they are both constant, so we can set each side equal to some constant, $-\lambda$:
$$\begin{align} \frac{T'}{t T} - 1 &=  -\lambda \\ (x-1) \frac{X'}{X} &= -\lambda \\ \end{align} $$
You may now solve each equation separately.  I imagine you have initial conditions that will determine each component $X$ and $T$, and will determine $\lambda$.
EDIT
You will now see why I chose $-\lambda$ rather than $\lambda$.  The solution of the $t$ equation is straightforward:
$$T(t) = T_0 e^{-(\lambda - 1) t^2/2} $$
The solution to the $x$ equation can be seen by rearranging terms:
$$(x-1)X'(x) + \lambda X(x) = 0$$
which may be rewritten as
$$ [(x-1)^{\lambda} X(x)]' = 0 \implies X(x) = X_0 (x-1)^{-\lambda}$$
Combining these solutions, we get
$$U(x,t) = X(x)T(t) = X_0 T_0 (x-1)^{-\lambda} e^{-(\lambda - 1) t} = K (x-1)^{-\lambda} e^{-(\lambda - 1) t^2/2}$$
You now need to specify some sort of initial conditions to determine $K$ and $\lambda$.
A: Let $U(x,t)=X(x)T(t)$ ,
Then $X(x)T'(t)-tX(x)T(t)-txX'(x)T(t)+tX'(x)T(t)=0$
$X(x)T'(t)=txX'(x)T(t)-tX'(x)T(t)+tX(x)T(t)$
$X(x)T'(t)=((x-1)X'(x)+X(x))tT(t)$
$\dfrac{(x-1)X'(x)+X(x)}{X(x)}=\dfrac{T'(t)}{tT(t)}=s$
$\begin{cases}\dfrac{(x-1)X'(x)+X(x)}{X(x)}=s\\\dfrac{T'(t)}{tT(t)}=s\end{cases}$
$\begin{cases}\dfrac{X'(x)}{X(x)}=\dfrac{s-1}{x-1}\\\dfrac{T'(t)}{T(t)}=ts\end{cases}$
$\begin{cases}X(x)=c_1(s)(x-1)^{s-1}\\T(t)=c_2(s)e^{\frac{t^2s}{2}}\end{cases}$
$\therefore U(x,t)=\int_s C(s)(x-1)^{s-1}e^{\frac{t^2s}{2}}~ds$ or $\sum\limits_s C(s)(x-1)^{s-1}e^{\frac{t^2s}{2}}$
$=e^{\frac{t^2}{2}}\int_s C(s)(x-1)^{s-1}e^{\frac{t^2(s-1)}{2}}~ds$ or $e^{\frac{t^2}{2}}\sum\limits_s C(s)(x-1)^{s-1}e^{\frac{t^2(s-1)}{2}}$
$=e^{\frac{t^2}{2}}\int_s C(s)\left((x-1)e^{\frac{t^2}{2}}\right)^{s-1}~ds$ or $e^{\frac{t^2}{2}}\sum\limits_s C(s)\left((x-1)e^{\frac{t^2}{2}}\right)^{s-1}$
$=e^{\frac{t^2}{2}}f\left((x-1)e^{\frac{t^2}{2}}+1\right)$
Compare the result with solving this PDE by using the method of characteristics:
$U_t-tU-txU_x+tU_x=0$
$U_t-t(x-1)U_x=tU$
$\dfrac{dt}{ds}=1$ , letting $t(0)=0$ , we have $t=s$
$\dfrac{dx}{ds}=-t(x-1)=-s(x-1)$ , letting $x(0)=x_0$ , we have $x=(x_0-1)e^{-\frac{s^2}{2}}+1=(x_0-1)e^{-\frac{t^2}{2}}+1$
$\dfrac{dU}{ds}=tU=sU$ , letting $U(0)=f(x_0)$ , we have $U(x,t)=f(x_0)e^{\frac{s^2}{2}}=f\left((x-1)e^{\frac{t^2}{2}}+1\right)e^{\frac{t^2}{2}}$
