Solving $V_t=V_{xx}$ on $\{(x,t):x\in\mathbb{R}\}, t>0$ and $V(x,t)=e^tf(x)$ for some function $f$. 
Find solutions to $V_t=V_{xx}$ on $\{(x,t):x\in\mathbb{R}\}, t>0$ and $V(x,t)=e^tf(x)$ for some function $f$.

So I have that $V_{xx}(x,t)=e^tf''(x)$ and $V_t(x,t)=e^tf(x)$
But from the equation I get $e^tf''(x)=e^tf(x)$?
So I should have $f''(x)=f(x)$?
I'm not really sure what to be doing from here, or if what I've done even makes sense.
But solving $y''-y=0$
I get $r^2-r=r(r-1)=0$
So I have $r=0,r=1$
So $f(x)=c_1e^{x}+c_2$ is a solution?
But this doesn't seem to work with me given equations. Since if $V(x,t)=c_1e^te^x+c_2e^t$
then $V_t=c_1e^te^x+c_2e^t\neq V_{xx}$
 A: So l take it that we are asked for solutions of
$V_t = V_{xx} \tag 1$
on the set
$\Omega = \{ (x, t): x \in \Bbb R, 0 < t \in \Bbb R \} \tag 2$
which take the form
$V(x, t) = e^tf(x), \tag 3$
where
$f: \Bbb R \to \Bbb R \tag 4$
is sufficiently differentiable.  How differentiable?  Well, it is implicitly indicated that
$V_{xx} = (e^t f(x))_{xx} = e^t f''(x) \tag 5$
exists, hence
$f(x) \in C^2(\Bbb R, \Bbb R). \tag 6$
Now it follows from (3) that
$V_t = e^tf(x), \tag 7$
and this in concert with (1) and (5) yields
$e^t f''(x) = e^tf(x), \tag 8$
whence, since
$e^t \ne 0, \tag 9$
we have
$f''(x) = f(x), \tag{10}$
which of course may be solved in the usual fashion: setting
$f(x) = e^{rx}, \tag{11}$
it is easy to see that
$f'(x) = re^{rx}, \tag{12}$
$f''(x) = r^2e^{rx}, \tag{13}$
and thus via (10), (11), and (13),
$r^2e^{rx} = e^{rx}, \tag{14}$
from which, dividing by $e^{rx} \ne 0$,
$r^2 = 1, \tag{15}$
that is,
$r = \pm 1. \tag{16}$
We conclude that $e^{\pm x}$ are two linearly independent solutions of (10), and since (10) is a second order equation, these two functions span its entire space of solutions.  Thus we may write
$f(x) = c_+ e^x + c_-e^{-x}; \tag{17}$
now combining (3) and (17) we find that
$V(x, t) = e^t(c_+ e^x + c_-e^{-x}), \tag{18}$
the requisite result.  Note that we have also shown that
$f(x) \in C^\infty(\Bbb R, \Bbb R) \tag{19}$
in the course of the derivation of (18).
A: You made a mistake when writing out the characteristic polynomial of $y''-y=0$.
The correct polynomial is $r^2 -1$. Otherwise what you have done is correct.
