Solving $y^{(x)}(x)=ax+b$ in closed form: What function equals $ax+b$ when you take the $n$th derivative at $x=n$? (with graphs) Based on the fun of:

Conjectured simple ODE solution: $$y^{(y(x))}(x)=f(x)\mathop \implies\limits^?(y(x))!+c_0Γ(y(x))=\int\limits_{c_1}^xf(t)(x-t)^{y(x)-1} dt $$

Imagine we had a function of which its nth derivative at $x=n$ was some function:
$$\frac{d^{x}y(x)}{dx^x}=y^{(x)}(x)=ax+b\implies y(0)=b\implies y’(1)=a+b,\int y(x)dx\bigg|_{x=-1}=a-b$$
Now let’s use the fractional integral as the inverse of the xth derivative as $x$ is any constant. Therefore we can use Cauchy’s Repeated Integral formula:
$$y^{(x)}(x)=ax+b\implies y(x)+c_1=\text I^{x}_x (ax+b)$$
We can also generalize the lower integration bound as there may be no analog for the indefinite integral:
$$\,_a\text I^n_x g(x)\mathop=^\text{def}\frac 1{Γ(n)}\int_a^xg(t)(x-t)^{n-1}dt \implies y(x)+c_1=\,_{c_0}\text I^x_x (ax+b)= \frac 1{Γ(x)}\int_{c_0}^x(at+b)(x-t)^{x-1}dt \mathop=^{x\ne 0,-1}\\\implies y(x)=^?\frac{(x-c_0)^x(ac_0x+ax+bx+b)}{(x+1)!}+c_1$$
Here is an interactive graph of the conjectured answer.

The problem is that this formula only works for the trivial $a=b=0$ case.

Here is my graphical attempt for $y^{(x)}(x)=1$:
$$y(0)=1,y’(1)=1,y’’(1)=1$$
maybe we can assume a function looking like the following with all constants of integration being $0$ for simplicity. Here is the graph with these properties:

Notice how this function matches up with our conjectured result:
$$y^{(x)}(x)=1\implies y(x)=\frac{x^x}{x!}$$
Actually we can generalize this process by doing:
$$y^{f(x)}(x)=g(x)\implies y(x)=\text I_x^n g(x),n=f(x)$$
given that $f\ne y$ so that we can solve the equation.
Then just use a definition of the $n$th integral or derivative and to finish.
I have also found this solution $n\in\Bbb N$ derivatives for $ax+b=1$ with many initial conditions:
$$y’’’(x)=1,y(0)=y’(1)=y’’(2)=y’’’(3)=1\implies y^{(n)}(x)=1,y^{(n)}(n)=y^{(n-1)}(n-1)=1=…=y’’(2)=1,y’(1)=1,y(0)=1:$$
which would give the following polynomial for $n=6$:
$$y(x)= \frac1{720} (720 - 4866 x + 4560 x^2 - 1500 x^3 + 270 x^4 - 24 x^5 + x^6), y^{(6)}(6)= y^{(5)}(5)= y^{(4)}(4)= y^{(3)}(3)= y^{(2)}(2) = y’(1)=y(0)=1 $$
Therefore our conjectured solution for $n\in\Bbb N$ is:
$$y=\frac1{n!}\sum_{k=0}^n c_k(-1)^k x^k,$$
where $c_k$ are the coefficients of our solution polynomial which may relate to these Polynomial function.
So how can $y^{(x)}(x)=ax+b$ be  solved for $y(x)$ in closed form? Please correct me and give me feedback!
 A: I was able to find the following solution after a while with this interactive graph for the following differential equation. The equation at least works for all $x\in\Bbb N$:
$$y^{(x)}(x)=1$$
I was trying to solve $$y^{(x)}(x)=0$$ but many functions can be found with multiple roots which would be “trivial solutions”, so this accidental find does work. Let’s see how to derive it. All constants will be assumed to be $1$ for simplicity:
$$y^{(x)}(x)=1\implies y^{(x-1)}(x)=x+c\implies y^{(x-2)}(x)=\frac{x^2}2+x+1\implies y^{(x-3)}=\frac {x^3}{3\cdot 2}+\frac{x^2}2+x+1$$
Therefore:
$$y^{(x-k)}(x)=\sum_{n=0}^k\frac{x^n}{n!}\implies y^{(x-x)}(x)=\sum_{n=0}^x \frac{x^n}{n!}$$
which is exactly the discrete sum definition of the Regularized Gamma function:
$$e^zQ(n,z)=\sum_{m=0}^{n-1}\frac{x^m}{m!},n\in\Bbb N$$
Therefore:
$$\sum_{n=0}^{(x+1)-1} \frac{x^n}{n!}=e^x Q(x+1,x)$$
and finally one solution is:
$$y^{(x)}(x)\implies y(x)= \sum_{n=0}^{x} \frac{x^n}{n!}\ne e^x Q(x+1,x) $$
Note that the sum does work based on Desmos, but the closed form may not.
Interestingly enough, if we assume the false closed form we get a familiar Gompertz constant and Exponential Integral function:
$$\frac d{dx} e^xQ(x+1,x)|_{x=1}=2\gamma -e\text{Ei(x)}=2\gamma+\text G =1.75077…$$
The final result also implies, but probably does not use, the central Exponential Sum function:
$$e_n(x)\mathop=^\text{def} \sum_{k=0}^n \frac{x^k}{k!} =e^x Q(n+1,x)\implies \sum_{k=0}^x\frac{x^k}{k!}=e_x(x)$$
This is still not a general solution, so I will not accept it. I could probably use the same method to solve  for a more general solution. Do not forget to see the graph for proof Please correct me and give me feedback!
