Question about solutions to the differential equation 
Any two solutions to the differential equation of the first order $y’+a(x)\cdot y=f(x)$ is $y(x)=e^{A(x)}\cdot(F(x)+C)$ differ only in the constant $C$?

Is this true? Or is it that the constant $C$ only influences $F(x)$ and so that makes it an important only to that part of the solution?
 A: First, let me correct the solution to the original equation.
If we have a linear first order differential equation of the form
$$y'+a(t)y=f(t)$$
then the solution to the above problem is given by the method of integration factors:
$$y(t) = \frac{\int_{t_0}^{t} f(s)\mu(s)ds + C}{\mu(t)} $$
Here, $\mu(t)$ is the integrating factor and can be calculated using:
$$\mu(t) = e^{\int_{t_0}^{t}a(s)ds +C_0 } = e^{C_0}e^{\int_{t_0}^{t}a(s)ds}=Ke^{\int_{t_0}^{t}a(s)ds}$$
Here, $C,C_0,K$ are all constants. I'm sure you must have seen this in your ODE course, but I am just restating it for convenience.
To derive the solution, you "come up" with a function $\mu(t)$ such that when multiplying both sides of the ODE, you can easily integrate both sides.
Note that $\mu'(t)=a(t)\mu(t)$, so the original equation becomes (using the product rule):
$$\mu(t)y'+a(t)\mu(t)y = \mu(t)y'+\mu'(t)y=(\mu(t)y)'=f(t)\mu(t)$$
Integrating the above equation with respect to $t$ ant then dividing by $\mu(t)$ gives us our solution.
Here is where you should be careful-the $+C$ goes with the integral, and needs to be divided through by $\mu(t)$. It is often useful to put everything with the integral in brackets, including the arbitrary constant, and then divide.
Also, remember to multiply $f(t)$ by $\mu(t)$ and have the coefficient in front of the $y'$ term as just 1. This is called "normal" form.
So with that out of the way, I can answer the question you asked. The arbitrary constant needs to be there, and gives a family of solutions, each differing by an arbitrary constant. 
You can, however, specify the constant using an initial value (say $y(t_0)=y_0$). Specifying an initial value allows you to solve for $C$.This is true due to the existence and uniqueness theorem of ordinary differential equations.
