Finding a third solution from two other solutions for linear ODE not equal to 0 Edit: I am not asking how to solve the differential equation below. I know how to solve the equation as it is extremely easy. What I am asking is this:
For the equation $y' + y = 0$, if you already have the solutions $y_1, y_2$, then the solution $y_3 = c_1y_1 + c_2y_2$ where $c_1,c_2$ are any constants.
However, if that equation is modified to be $y' + y = 4$ for instance, then it doesn't look like you can find $y_3$ by taking any linear combination of $y_1,y_2$. It looks like you would need a specific linear combination, and I'd like to know how you would find that as I seem to be missing something.
In short, what constants $c_1, c_2$ would make $y_3 = c_1y_1 + c_2y_2$ a solution of that equation.
Also to whoever downvoted me: Seriously? Why are you people so obsessed with downvoting simple questions? I'm not asking for you to solve my homework or how to add 1+1. Why would you downvote me for trying to clarify something?
 A: The solution to a linear inhomogeneous ODE (inhomogeneous means zero is not a solution, or equivalently, there's a term with no $y$-dependence) is the sum of two parts, the complementary function $y_c$, which solves the homogeneous equation, and the particular integral $y_p$, which is one solution to the inhomogeneous equation. The complementary function is a sum $\sum_i A_i y_i$, and the constants $A_i$ are arbitrary unless determined by some set of boundary conditions, but the particular integral is fixed (at least up to adding bits of the complementary function to it).

This is exactly analogous to solving the inhomogeneous matrix equation
$$ Ax = b, $$
where the homogeneous equation $Ax=0$ has nonzero solutions: the general solution is the sum of a particular solution to $Ax=b$ and an arbitrary sum of solutions to the homogeneous equation.
A: When an equation is non-homogenous, then there is a particular solution $y_p(x)$ that ensures the left side always equals the right side. This particular solution is decided by the equation and doesn't have a constant multiplier.
In your case, $y' + y = 4$, the particular solution is pretty clear. Try $y = 4$, and the left side equals the right side. So, a solution to the DE is $y = 4$.
Of course, that's not the most general solution. Notice that $y = Ce^{-x} + 4$ is the most general solution. That extra term comes from $y' + y = 0$. Since the DE is of order 1, there is one constant.
