Assume $n=1$ and $u(x,t)=v(\frac{x}{\sqrt{t}})$.
(a) Show that $u_t = u_{xx}$ if and only if $$v''+\frac z2 v' = 0. \tag{$*$}$$ Show that the general solution of $(*)$ is $$v(z)=c \int_0^z e^{-s^2/4} \, ds + d.$$
(b) Differentiate $u(x,t)=v(\frac x{\sqrt{t}})$ with respect to $x$ and select the constant $c$ properly, to obtain the fundamental solution $\Phi$ for $n=1$. Explain why this procedure produces the fundamental solution. (Hint: What is the initial condition for $u$?)
This problem is from PDE Evans, 2nd edition, Chapter 2 Exercise 13. My question is on part B: why does the procedure allow us to get the fundamental solution $\Phi$?
For context, we have (page 46 in the textbook):
DEFINITION. The function $$\Phi(x,t) := \begin{cases} \frac 1{(4\pi t)^{n/2}} e^{-\frac{|x|^2}{4t}} & (x \in \mathbb{R}^n, t > 0) \\ 0 & (x \in \mathbb{R}^n, t = 0)\end{cases}$$ is called the fundamental solution of the heat equation.
Firstly, I understand part (a). I also understand most of part (b). I followed the process exactly and obtained the fundamental solution $\Phi$ for $n=1$, with the constant $c$ chosen to be $c=\frac 1{(4\pi)^{1/2}}$. I get: $$u_x(x,t)=\frac{\partial}{\partial x} v(\frac x{\sqrt{t}}) = \frac c{t^{1/2}} e^{-\frac{|x|^2}{4t}}=\frac 1{(4\pi t)^{1/2}} e^{-\frac{|x|^2}{4t}} =: \Phi,$$ as required.
But the last part of the question asks why does this procedure work, which is asked for in the last two lines of the stated problem. I do not know why this works; I've tried to find the initial condition for $u$ per the given hint, that is, find $u(x,0)$ (substitute $t=0$), but to no avail.
A possible explanation: since $u(x,t)$ solves the heat equation, so does $u_x(x,t).$ And recall the initial condition for the fundamental solution is the $\delta$ distribution. And thus we have $1=u(x,0)=c\int^\infty_{-\infty} e^{\frac{s^2}{4}}\mathrm{d}s=2c\sqrt{\pi}.$ So $c=\frac{1}{\sqrt{4\pi}}$