Differentiable function and polynomials: Proof of $\phi (x)=ce^x + q(x)$ unclear to me. I came across a proof which I can't quite understand:
If $p:\mathbb{R}\to\mathbb{R}$ is a polynomial of degree $n$ and $\phi:\mathbb{R}\to\mathbb{R}$ is a differentiable function with $\phi'=\phi+p$, there exists a polynomial $q$ of degree $n$ with a number $c\in\mathbb{R}$ with $\phi(x)=ce^x+q(x)$ for all $x\in\mathbb{R}$.
The proof is given as follows:
$$\phi'=\phi+p$$
$$\Leftrightarrow\phi'-\phi = p$$
$$\Leftrightarrow\phi'*e^{-x}-\phi*e^{-x}=p*e^{-x}$$
$$\Leftrightarrow\left(\frac{d}{dx}\right)[\phi*e^{-x}]=p*e^{-x}$$
$$\int\left(\frac{d}{dx}\right)[\phi*e^{-x}]=\int p*e^{-x}$$
$$\Leftrightarrow\phi*e^{-x}=q*e^{-x}+c$$
$$\Rightarrow\phi(x)=ce^x+q(x)$$
Isn't the notation (especially the integrals) missing something?
Why exactly are we multiplying by $e^{-x}$ in line 3? Is this a special technique?
How does the integral with $p$ in it suddenly yield $q$? And why is the arbitrary constant $c$ on the left hand side of the equation missing?
Also, the conclusion is unclear to me. How exactly do the $e^{-x}$ and $c$ disappear and how does the $ce^x$ appear?
Thank you very much in advance!
 A: 
Why exactly are we multiplying by $e^{−x}$ in line 3:

The OE $$\phi'(x)-\phi(x) = p(x)$$ is a Linear differential equation of first degree and as an method we need a magic function called Integrating factor to solve it. The method tells us to multiply this given function to the OE, both sides of it, just to make the OE an exact OE. Here we have $-1$ as the coefficient of $\phi(x)$ in OE, so according to cited approach the integrating factor is: $$\mu(x)=\exp\int(-1)dx=\text{e}^{-x}$$ Now continue from line 3 to the rest.

How does the integral with p in it suddenly yield $q$?

Do as @nullUser noted, he did the rest complete.
A: The notation is quite abusive. The author assumes the reader is familiar enough with differential equations to understand what the notation means, where the $dx$'s go and so on.
We multiply by $e^{-x}$ because we recognize the ODE $y'-y = p$ to be first order linear, which can always be solved by multiplying by a suitable integrating factor, which in this case turns out to be $e^{-x}$. See http://tutorial.math.lamar.edu/Classes/DE/Linear.aspx for examples.
The statement $\int p(x) e^{-x}dx = q(x)e^{-x}+c$ follows by induction from the fact that $\int x^k e^{-x}dx$ is solved by repeated integration by parts. Differentiating powers of $x$ and integrating $e^{-x}$ always leaves you with a term of the form $x^n e^{-x}$ up to a multiplicative constant.
To get the last line from its predecessor just multiply the equation by $e^{x}$. 
A: Do you remember the product rule from calculus ? Using that we have $${d \over dx} \left( \phi*e^{-x}\right) = {d \over dx}\phi*e^{-x}+{d \over dx}{e^{-x}}*\phi$$
Is the motivation of why we multiplied $e^{-x}$ to the equation.
The reason why $p$ sudenly seems like it turned into a $q$, is because you integrated 
$$\int p(x)*e^{-x}dx = q(x)*e^{-x}+c$$
Here, when you integrate the polynomial becomes something different.
For example using integration by parts,$$\int x*e^{-x}dx = -(1+x)e^{-x}+c$$
where $p(x) = x$ and $q(x)=-(1+x)$
Finally, the reason why it seems like $e^{-x}$ suddenly dissapeared, is because it hasn't. When we look at the very last step, all it's doing is multiplying $e^x$ on both sides. And $e^{-x}*e^x = 1$.
Does this help ?
