Solution of $\frac{\partial f}{\partial t}(t,x) = 2 \frac{\partial^2f}{\partial x^2}(t,x)$ Consider the PDE
$$\frac{\partial f}{\partial t}(t,x) = 2 \frac{\partial^2f}{\partial x^2}(t,x)\tag{1} $$
with $t\ge0,\ x\in\mathbb R,\ f(0,x)=e^x$. I want to find $f(t,x)$.
I know that the heat equation
$$\frac{\partial p}{\partial t}(t,x) = \frac{1}{2}\frac{\partial^2p}{\partial x^2}(t,x)\tag{2}$$
with $t\ge0,\ x\in\mathbb R,\ p(0,x)=h(x)$ has the solution $p(t,x) =\mathbb E[h(x+W_t)]$ where $W_t$ is a Brownian motion.
I have tried things like setting $p(t,x)=f(2t,x)$, but I do not seem to be able to put $(1)$ into the form of $(2)$. How can I find $f(t,x)$ from using the general solution of the heat equation?
 A: Following the suggtestion on Cameron Williams' comment:
We set $u(t,y)=f(t,2y)$ with $y=x/2$ and $\overline h(y)=e^{2y}$. Then
$$\frac{\partial u}{\partial t}(t,y)=\frac{\partial f}{\partial t}(t,x), \quad \quad \quad \frac{\partial^2 u}{\partial y^2}(t,y)=4\frac{\partial^2 f}{\partial x^2}(t,x)$$
So
$$\frac{\partial u}{\partial t}(t,y)=\frac12\frac{\partial^2 u}{\partial y^2}(t,y) \quad \text{with}\quad u(0,y)=\overline h(y)$$
$$\iff \frac{\partial f}{\partial t}(t,x)=\frac42\frac{\partial^2 f}{\partial x^2}(t,x)=2\frac{\partial^2 f}{\partial y^2}(t,x) \quad \text{with}\quad f(0,x)=u(0,y)=\overline h(y)=h(x)$$
Now the solution of $u$ is given by $u(t,y)=\mathbb E[\overline h(y+W_t)]=e^{2y}\mathbb E[e^{2W_t}]=e^{2y+2t}$ and therefore $f(t,x)=u(t,x/2)=e^{2t+x}$.

Follow-up question: I am a relativ beginner using change of variables. So how does one actually spot this particular change of variables in the first place (i.e. what tips you off), and also, is there a way to perform this change of variables maybe quicker (less awkward than I did)?
A: $\newcommand{\+}{^{\dagger}}%
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Let's define
$\ds{\tilde{\fermi}\pars{s,x} = \int_{0}^{\infty}\fermi\pars{t,x}\expo{-st}\,\dd s}$.
Then,
$$
-\fermi\pars{0,x} + s\tilde{\fermi}\pars{s,x}
= 2\,\partiald[2]{\tilde{\fermi}\pars{s,x}}{x}\quad\imp\quad
\pars{\partiald[2]{}{x} - \half\,s}\tilde{\fermi}\pars{s,x} = -\,\half\,\expo{x}
$$
Then, $\tilde{\fermi}\pars{s,x} = A\expo{x}$ such that $A - sA/2 = -1/2\quad\imp\quad
A = -1/\bracks{2\pars{1 - s/2}} = 1/\pars{s - 2}$ which leads to:
$$
\tilde{\fermi}\pars{s,x} = {\expo{x} \over s - 2}
\quad\mbox{and}\quad
\fermi\pars{t,x} = \expo{x}\int_{\gamma - \ic\infty}^{\gamma + \ic\infty}
{\expo{st} \over s - 2}\,{\dd s \over 2\pi\ic}\quad\mbox{with}\quad\gamma > 2
$$
$$
\color{#0000ff}{\large\fermi\pars{t,x} = \expo{x\ +\ 2t}}
$$

Even more simple: Write $\fermi\pars{t,x} \equiv \expo{x}\varphi\pars{t}$ and you get
$\dot{\varphi}\pars{t} = 2\varphi\pars{t}$ with $\varphi\pars{0} = 1$. Then, $\varphi\pars{t} = \expo{2t}$

