Solve differential equation $-f'(x)= a_1 f(a_2 x+a_3)$ with $f(0)=1$. How to solve the following differential equation
\begin{align}
-f'(x)= a_1 f(a_2 x+a_3),
\end{align}
where $f(0)=1$.  
I looked around I think this falls under the category of discrete delayed differential equations.  
 A: Case $1$: $a_2>0$
Let $f(x)=\int_0^\infty e^{-xt}K(t)~dt$ ,
Then $\int_0^\infty te^{-xt}K(t)~dt=a_1\int_0^\infty e^{-(a_2x+a_3)t}K(t)~dt$
$\int_0^\infty te^{-xt}K(t)~dt-a_1\int_0^\infty e^{-a_2xt}e^{-a_3t}K(t)~dt=0$
$a_2^2\int_0^\infty te^{-a_2xt}K(a_2t)~dt-a_1\int_0^\infty e^{-a_2xt}e^{-a_3t}K(t)~dt=0$
$\int_0^\infty e^{-a_2xt}(a_2^2tK(a_2t)-a_1e^{-a_3t}K(t))~dt=0$
$\therefore a_2^2tK(a_2t)-a_1e^{-a_3t}K(t)=0$
Let $\begin{cases}t_1=\log_{a_2}t\\K_1(t_1)=K(t)\end{cases}$ ,
Then $a_2^{t_1+2}K_1(t_1+1)=a_1e^{-a_3a_2^{t_1}}K_1(t_1)$
A: Edition of 06.03.19

First, if $\underline{a_1=0}$ or $\underline{a_2=0},$ then $f(x)=0.$
If $\underline{a_2=1},$ then finding of the solution in the form of
$$f(x)= \mathrm{const}\cdot e^{-kx}\tag1$$
gives
$$ke^{-kx} = a_1e^{-kx-ka_3},$$
$$ke^{ka_3}=a_1,$$
$$k=
\begin{cases}
a_1,\quad\text{if}\quad a_3=0,\\[4pt]
\dfrac{\log(a_1a_3)}{a_3 W(\log(a_1a_3))},\quad\text{otherwize},
\end{cases}\tag2
$$
where $W(t)$ is the Lambert W-function.

Assume $\underline{a_1\not=0},\quad \underline{a_2\not = 0},\quad \underline{a_2\not= 1}.$
Taking in account that
$$f'\left(\frac{a_3}{1-a_2} - x\right) = a_1f\left(\frac{a_2a_3}{1-a_2}-a_2x+a_3\right) = a_1f\left(a_2\left(\frac{a_3}{a_2(1-a_2)}-x\right)\right),$$
denote
$$r=\frac{a_3}{a_2(1-a_2)},\quad y=r-x,\quad g(y) = f\left(\dfrac y{a_1}\right).\tag3$$
Then 
$$g'(y) = \dfrac1{a_1}f'\left(\dfrac y{a_1}\right) = f\left(\dfrac{a_2}{a_1}\,y\right),$$
$$g'(y) = g(a_2y),\quad g(r) = 1.\tag4$$
Boundary condition defines factor of $g(y).$ Behavior of $g(y)$ is linked with the parameter $a_2.$
Denote $v=g(0),$ and let us calculate derivatives of $g(y):$
\begin{align}
&g(0) = v,\\[4pt]
&g'(y) = g(a_2y),\quad g'(0) = v,\\[4pt]
&g''(y) = a_2 g'(a_2 y) = a_2g(a_2^2y),\quad g''(0) = a_2v,\\[4pt]
&g'''(y) = a_2^3 g'(a_2^2y) = a_2^3 g(a_2^3y),\quad g'''(0) = a_2^3v,\dots,\\[4pt]
&g^{(n)}(y) =a_2^{\frac12n(n-1)}g\left(a_2^{\frac12n(n-1)}\right),\quad 
g^{(n)}(0) =a_2^{\frac12n(n-1)}v,\dots
\end{align}
Maclaurin series are
$$g(y) = v \left(1+y+\frac12a_2y^2+\frac16a_2^3y^3+\dots+\dfrac1{n!}a_2^{\frac12n(n-1)}y^n+\dots\right).\tag5$$
If $\underline{|a_2|>1}$ then $n! \le n^n = e^{n\log n},$ and the series $(5)$ diverge.
If $\underline{a_2 = 1}$ then the series $(5)$ converge to $ve^x,$ and then 
$$g(y) = e^{x-r}.\tag6$$
Formula $(6)$ cannot be used in the solution, but it gives a limit case for $g(y).$
If $\underline{a_2 = -1}$ then the series $(5)$ converge to $v(\sin x + \cos x),$ and then 
$$g(y)=\dfrac{\cos\left(\dfrac\pi4-y\right)}{\cos\left(\dfrac\pi4-r\right)}.\tag7$$
If $\underline{|a_2|<1}$ then the series $(5)$ converge, wherein
$$g(y) = \dfrac {\sum\limits_{n=0}^\infty \dfrac1{n!}a_2^{\frac12n(n-1)}y^n}
{\sum\limits_{n=0}^\infty \dfrac1{n!}a_2^{\frac12n(n-1)}r^n}.\tag8$$

The Wolfram Alpha plot of the numerators of $(8)$ shown above, illustrates the behavior of the function $g(y)$ for $a_2=\pm 0.9, \pm 0.6, \pm 0.3,\quad (y=-5,5).$ 
The solution of the issue equation in the case $|a_2| < 1$ is
$$f\left(x,\vec a\right) = g(a_1(r-x)).\tag9$$
A: Obviously $f(x)$ is in $C^{\infty}(\textbf{R})$. Also it hold by induction
$$
f^{(\nu)}(x)=(-a_1)^{\nu}a_2^{\nu(\nu-1)/2}f\left(g_n(x)\right)\textrm{, }\forall x\in\textbf{R}
$$
I.e if $g(x)=a_2x+a_3$, then 
$$
g_n(x)=(a_3+a_2(a_3+a_2(a_3+\ldots+a_2(a_3+a_2x\underbrace{)\ldots)))}_{n-parenthesis}.
$$
Hence
$$
f^{(\nu)}(x)=(-1)^{\nu}a_1^{\nu}a_2^{\nu(\nu-1)/2}f\left(a_3(1+a_2^1+a_2^2+\ldots+a_2^{\nu-1})+a_2^{\nu} x\right)\textrm{, }\forall \nu=1,2,\ldots.
$$
Hence when $\nu=1,2,\ldots$, we get
$$
f^{(\nu)}(-\frac{a_3}{a_2-1})=(-a_1)^{\nu}a_2^{\nu(\nu-1)/2}f\left(-\frac{a_3}{a_2-1}\right).
$$
If we set $c=\frac{a_3}{1-a_2}$,then
$$
f(x)=\sum^{\infty}_{n=0}\frac{f^{(n)}(c)}{n!}(x-c)^{n}=f(c)\left(1+\sum^{\infty}_{n=1}\frac{a_2^{n(n-1)/2}(-a_1)^{n}}{n!}(x-c)^n\right),
$$
for all $x\in\textbf{R}$, when $|a_2|<1$.
