Why Is the Solution of a Linear Nonhomogeneous Constant-Coefficient Differential Equation the Sum of a Particular and Homogeneous Solution?
My textbook gives the following proof:
Let $x_c(t)$ be a solution of the homogeneous equation and $x_p(t)$ be a solution of the particular equation.
$a\dfrac{d^2x}{dt^2} + b\dfrac{dx}{dt} + cx = a\left(
\dfrac{d^2x_c}{dt^2} + \dfrac{d^2x_p}{dt^2} \right) + b\left(
\dfrac{dx_c}{dt} + \dfrac{dx_p}{dt} \right) + c(x_c + x_p)$
$= \left( a\dfrac{d^2x_c}{dt^2} + b\dfrac{dx_c}{dt} + cx_c \right) +
\left( a\dfrac{d^2x_p}{dt^2} + b\dfrac{dx_p}{dt} + cx_p \right)$
$= 0 + f(t) = f(t) $
I don't see how this proves that a homogeneous solution and a particular solution are required to find the solution of a linear non homogeneous constant-coefficient differential equation?
In fact, the proof shows that we end up with $= 0 + f(t) = f(t) $ where $f(t)$ is what we were seeking all along. Therefore, since $\left( a\dfrac{d^2x_c}{dt^2} + b\dfrac{dx_c}{dt} + cx_c \right) = 0$, what was the point of the solution to the homogeneous equation? It seems that the homogeneous solution was redundant from the beginning, since only the particular solution, $\left( a\dfrac{d^2x_p}{dt^2} + b\dfrac{dx_p}{dt} + cx_p \right)$, contributed towards the solution of the linear non homogeneous constant-coefficient differential equation ($\left( a\dfrac{d^2x_p}{dt^2} + b\dfrac{dx_p}{dt} + cx_p \right) = f(t)$).
I would greatly appreciate it if people could please take the time to clarify concept.
EDIT
I've received 3 answers; none of which address my question or do it sufficiently. My question is very clear: How does the above calculation show that both a particular solution and a homogeneous solution are required for a general solution? I then mentioned that the homogeneous solutions ends up equating to $0$ and the particular solution equates to $f(x)$; in other words, it seems that the particular solution was the only thing required to get the general solution $f(x)$, which seemingly makes the homogeneous solution redundant? What am I misunderstanding?
Given the poor response it has received, I am going to request that a moderator delete this question.
Best Answer
If all you want is to find some solution of the ODE, well, then finding one particular solution $x_p(t)$ is enough. But that solution might not be the right one, if you have additional conditions (boundary values, etc.). Usually one needs to know all the solutions of the ODE, in order to pick the one that satisfies the extra conditions.
The computation that you have reproduced from your textbook shows that $x_p(t)+x_c(t)$ is another solution of the same ODE (if $x_c(t)\neq 0$), so this indicates a way of finding more solutions than just $x_p(t)$. (The point here is that, as you have realized, only $x_p(t)$ gives a contribution $f(t)$ to the right-hand side, and adding $x_c(t)$ doesn't spoil that.)
But that still doesn't really explain the main issue, which is that you find all the solutions in this way. This is where the subtraction argument comes in. Suppose you have happened to find one particular solution $x_{p,1}(t)$ while your friend found another one $x_{p,2}(t)$. Then the difference $x_c(t)=x_{p,2}(t)-x_{p,1}(t)$ will satisfy the homogeneous equation, since you get $\ldots=f(t)-f(t)=0$ in the right-hand side when plugging in. So two different particular solutions to your ODE can't be arbitrarily different; they can only differ by a solution to the homogeneous ODE. That's why you know all the solutions to your ODE if you know (a) one of them to begin with, and (b) all the solutions $x_c(t)$ of the homogeneous equation.