Undergrad Question about solving linear first order non-homogeneous ODEs

[CGandC]

TL;DR

Why solution of $y'+2xy = 8x$ is $y=4+C e^{-x^2}$ and not $y=e^{-C_1}(4+C_2 e^{-x^2})$?

A general equation for linear first order non-homogeneous ODE is: $y' + a(x)y = b(x)$.
The procedure to solve ( assuming $a(x) , b(x)$ are continuous so that the fundamental theorem of calculus could be used ) it is to multiply it by $e^{A(x)}$ ( here $A'(x) = a(x)$ ) s.t. $A(x) = \int^x { a(t)dt }$ and we get $(ye^{A(x)})'=b(x)e^{A(x)}$.
Hence $ye^{A(x)}=\int^x {b(t)e^{A(t)}dt + C}$, hence $y(x) = e^{-A(x)}(\int^x {b(t)e^{A(t)}dt } + C )$.

The question is: why when we consider$A(x) = \int^x { a(t)dt }$ we do not add another constant? , i.e., why are we not writing $A(x) = \int^x { a(t)dt } + \tilde{C}$ where $\tilde{C}$ is another constant?.

Example: Consider $y'+2xy = 8x$. I tried solving it as follows:
Here $A'(x) = 2x$ and so $A(x) = x^2 + C_1$. Multiply both sides of ODE by $e^{A(x)} = e^{x^2 + C_1 }$ and so $(ye^{x^2 +C_1} )' = 8xe^{x^2 }$, hence $ye^{x^2 +C_1} = \int^x 8te^{t^2 }dt = 4e^{x^2} + C_2$; from this we have $y=e^{-C_1}(4+C_2 e^{-x^2})$.
However, the "true" solution according to the procedure above is $y=4+C e^{-x^2}$ ( only one constant ). Why is it so? It seems to me like the constant part $e^{-C_1}$ cannot be "intermingled" into the other constant $C_2$ and that we must have $y=e^{-C_1}(4+C_2 e^{-x^2})$.

Thanks for any help!______________________

Edit: as I was typing this I just noticed that when I multiplied both side and that I've written $(ye^{x^2 +C_1} )' = 8xe^{x^2 }$ - this is wrong since it must be $(ye^{x^2 +C_1} )' = 8xe^{x^2 + C_1 }$ , this was my mistake ( and then it turns out the constant $C_1$ doesn't come up in the solution so it doesn't matter if we write $A(x) = \int^{x} a(t)dt$ or $A(x) = \int^{x} a(t)dt + C_1$ )

 

[Mark44]

So you answered your own question, right?

 

[anuttarasammyak]

Just to confirm your own answer

Example: Consider y′+2xy=8x. I tried solving it as follows:

\frac{dy}{8-2y}=xdx
\ln |y-4|=-x^2+C_1
y=4+C_2e^{-x^2}

 

[CGandC]

So you answered your own question, right?

Yes,