PDE
$u_{t} + xtu_{x} = x^{2}, u(x,0) = \phi(x)$

Question

PDE
$u_{t} + xtu_{x} = x^{2}, u(x,0) = \phi(x)$

anonymous · Answer

No idea, lol. Book is no help at all, so all I've been able to do is take guesses at methods and ways of solving with no luck.

anonymous · Answer

@Kainui

anonymous · Answer

what have u tried so far

anonymous · Answer

Theres only two substitutions we've been given and neither seem to get me much anywhere. I may be doing the manipulation improperly, though. The first substitution is given an equation 
$u_{t} + cu_{x} + \lambda u = f(x,t)$ , let $\xi = x-ct$ and $\tau = t$. Then
$u_{t} = -cu_{\xi} + u_{\tau}$
$u_{x} = u_{\xi}$
I assume this is the proper substitution to use, just not able to get it to work. I'm guessing I'm unaware of some sort of chain rule going on, but yeah.

As for the second one, I doubt it will work because I'm not sure if this substitution is applicable where you could replace y with t, but given an equation
$a(x,y)u_{x}(x,y) + b(x,y)u_{y}(x,y) = 0$, there are the substitutions $x' = ax + by$ and $y' = bx-ay$. 
Now, I know my equation isn't homogeneous, but I figured that it may have just left the possibility of a homogeneous solution plus a particular solution. That and I wasn't sure what else I could try.