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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.

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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.

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  • $\begingroup$ You're the third person to not answer my question; instead, you starting talking about the subtraction aspect which I did not mention even once! 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? $\endgroup$ – The Pointer Mar 7 '17 at 8:58
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    $\begingroup$ Yes, if you do it with $x_p(t)$ only, you get $f(t)$ in the right-hand side. That's what it means for $x_p(t)$ to be a solution. And if you do it with $x_p(t)+x_c(t)$, you also get $f(t)$, just because $x_c(t)$ doesn't contribute. And that means that $x_p(t)+x_c(t)$ is also a solution. Which is the point: you get another solution to the same equation. If $x_c(t)$ had given a nonzero contribution, you wouldn't get $f(t)$, hence $x_p(t)+x_c(t)$ wouldn't have been a solution. $\endgroup$ – Hans Lundmark Mar 7 '17 at 10:20
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    $\begingroup$ No, no, no, it is definitely not true that $x_c(t)+x_p(t)$ is the same thing as $f(t)$! Maybe that's the whole source of your confusion? $\endgroup$ – Hans Lundmark Mar 7 '17 at 10:56
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    $\begingroup$ When you take $x_p(t)$ and do certain operations on it, then you get $f(t)$. And the same operations on the function $x_p(t)+x_c(t)$ also gives $f(t)$. $\endgroup$ – Hans Lundmark Mar 7 '17 at 10:57
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    $\begingroup$ And the most general function that will give $f(t)$ when you do those operations is obtained as $x_p(t)+x_c(t)$ where $x_c(t)$ is the most general function which gives zero when you do those operations on it. $\endgroup$ – Hans Lundmark Mar 7 '17 at 10:59
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If $f$ is some linear function (doesn't have to be about differential equations), $f(x)=z$ and $f(y)=z$ then $f(x)-f(y)=0=f(x-y)$. Thus $x-y$ solves the homogeneous equation. Therefore it is enough to find a particular solution and the general homogeneous solution, because the difference of two particular solutions will always be one of the homogeneous solutions. You don't have to proceed this way but it is usually easier. Just taking a particular solution will not give you all solutions unless there are no nontrivial homogeneous solutions (which never happens in ordinary differential equations).

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  • $\begingroup$ Thanks for the response. I do not understand what relevance your example has to the proof in the OP; indeed, there is no mention of subtracting. $\endgroup$ – The Pointer Mar 6 '17 at 21:44
  • $\begingroup$ @ThePointer It's another presentation of the same idea. $\endgroup$ – Ian Mar 6 '17 at 21:45
  • $\begingroup$ Can you please clarify the idea I'm having trouble with in the OP? $\endgroup$ – The Pointer Mar 6 '17 at 21:53
  • $\begingroup$ @ThePointer My last sentence addresses that issue, I think: the homogeneous solution is indeed "destroyed" when you substitute it into the equation, but it wasn't zero to begin with, so it gives you a new solution. For example, there are more solutions to $y'=1$ than just $y=x$, there is $x+C$, and that $C$ is coming from the homogeneous solution, i.e. the solutions to $y'=0$. $\endgroup$ – Ian Mar 6 '17 at 22:05
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The theory of linear, 2nd-order equations says that the general solution is a double infinity of solutions. When you solve the homogeneous equation, there are 2 arbitrary constants. Letting those constants vary over all the real numbers gives the double infinity of solutions.

The particular solution to the nonhomogenous equation is just one solution. You you need infinity-squared minus one more solutions to have the general solution. By happy coincidence*, if you add a homogeneous solution to the particular solution, you get another particular solution. Since there is a double infinity of homogeneous solutions, we can form a double infinity of non-homogeneous solutions. In fact, ALL solutions can be constructed this way, so it is, in fact, the general solution.

Why do we need the general solution? Because the particular solution has no arbitrary constants in it, which means you can't solve a corresponding IVP. With the arbitrary constants, you can solve ANY corresponding IVP.

*Yeah, I know.

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  • $\begingroup$ Thanks for the response. Unfortunately, this does not address my question. $\endgroup$ – The Pointer Mar 6 '17 at 21:52
  • $\begingroup$ @ThePointer I'm pretty sure it exactly answers the question. Are you sure you asked what you wanted? $\endgroup$ – B. Goddard Mar 6 '17 at 21:54
  • $\begingroup$ Well, I never mentioned infinities. I'm just a student trying to understand the concept I posted in the OP. I even provided the proof and specified the reasoning that I do not understand. I'm probably just too stupid to understand your answer, but none of it looks like it addresses my post. $\endgroup$ – The Pointer Mar 6 '17 at 21:57
  • $\begingroup$ The infinities are the point. You add ONE nonhom solution to each of the infinitely many hom solutions and it gives you infinitely many nonhom sols. That's why we add homs to nonhoms. And the reason it's necessary, is that it give you ALL nonhom solutions. $\endgroup$ – B. Goddard Mar 6 '17 at 22:01

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