To explain my question, let me consider the following example,

$$y''(x) + y(x) = 1.$$

We do now that the correspoding homogeneous equation has the general solution of the form

$$y_c(x) = c_1 sin(x) + c_2 cos(x),$$

and if we consider the particular solution $y_p(x) = 1$, it is claimed that we would have found the general solution of the non-homogeneous equation as

$$y(x) = c_1 sin(x) + c_2 cos(x) + 1.$$

However, notice that we could have used $y_{1} (x) = 3 sin(x) + 1$ as our particular solution, but using the particular solution $y_p(x) = 1$ has a "simple" form in the sense that it does not contain a part from particular solution of the corresponding homogeneous equation.

My question is that, for any given linear ODE, how many such "simple" solutions (i.e that does not contain a part from the corresponding homogeneous equation) are there to this linear ODE ?


Aside from that, to me, the solutions of the form

$$y(x) = y_c(x) + y_p(x)$$ covers only a subset of the solutions to a given linear ODE because I do not see how would we prove that if $y_2(x)$ is a solution of the given ODE, then $y_2(x)$ can be written of the form

$$y_2 (x) = y_c(x) + y_p(x) ? $$


It is true that there is often one or more "exceptional" solutions, as you are guessing. One way of seeing this is analyzing the long-time behavior. (In this post, I will refer to the variable $x$ as "time", for consistency with the OP notation).

For the equation $y''+y=1$, all solutions oscillate as $x\to \infty$ except one: the constant solution $y(x)=1$. Another example is the equation $y'=-y+2x$ with general solution $$ y(x)=Ce^{-x}+2x-2,$$ (see Hubbard & West "Differential equations", figure 1.1.6, for a nice picture of this family of solutions). In this case, as $x\to\infty$, all solutions collapse on the exceptional solution $y(x)=2x-2$, which is called a funnel.

  • $\begingroup$ So, in general, a linear n-th order inhomogeneous ODE might have more than one "exceptional" solution which in a sense corresponds only to the inhomogeneous part of the ODE, right ? $\endgroup$ – onurcanbektas Mar 15 '18 at 7:06
  • $\begingroup$ The problem is that there is no definition of "excepcional". We could define "excepcional" as "having a different behaviour at infinity", as I suggest, but this is still too vague to prove anything rigorously. I would content myself with some examples and move on. $\endgroup$ – Giuseppe Negro Mar 15 '18 at 23:33

I think that your context is the "method of undetermined coeffcients" (or, its modification, "the annihilator method"). This method indeed provides the existence of precisely one solution $y_p(x)$ of the nonhomogeneous equation, in whose expression no terms from the general solution of the complementary homogeneous equation occur.

Another thing is, what does "simple" mean? For the equation $$ y''(x) = \cos{x} $$ the general solution, $c_1 x + c_2$, of the complementary equation is, at least to me, simpler than $y_p(x) = - \cos{x}$.

Regarding your digression: if $y_2(x)$ is a solution of the nonhomogeneous equation, then $y_2(x) - y_p(x)$ is a solution of the complementary homogeneous equation: just plug the difference into the homogeneous equation.

  • $\begingroup$ the "simple" has nothing to do with the actual meaning of the word simple, you can say "exceptional" if you like, I just wanted to denote the kind of solution that I was talking. $\endgroup$ – onurcanbektas Mar 15 '18 at 7:08
  • $\begingroup$ Plus, you digression doesn't contradict with what I'm claiming, so I do not see your point there. $\endgroup$ – onurcanbektas Mar 15 '18 at 7:10
  • $\begingroup$ Dear Madam/Sir! This was your digression, not mine. You did not understand why $y_2(x)$ can be written as $y_c(x) + y_p(x)$, so I tried to explain it to you. I repeat: by plugging $y_2(x)$ and $y_p(x)$ into the nonhomogeneous equation and subtracting the equation for $y_p(x)$ from the equation for $y_2(x)$ one obtains that $y_2(x) - y_p(x)$ satisfies the complementary homogeneous equation. $\endgroup$ – user539887 Mar 15 '18 at 22:03

Of course, you could use $y_1(x)=3\sin(x)+1$ instead of $y_1(x)=1$.

This would give $\quad y(x)=c_1\sin(x)+c_2\cos(x)+3\sin(x)+1\quad$ instead of $\quad y(x)=c_1\sin(x)+c_2\cos(x)+1$

Note that $\quad y(x)=c_1\sin(x)+c_2\cos(x)+3\sin(x)+1 \quad=\quad (c_1+3)\sin(x)+c_2\cos(x)+1\quad$

So, you just replace $c_1$ by $c_1+3$ which is also a constant $c'_1=c_1+3$


The result is the same since $c_1$ or $c'_1$ are both arbitrary constants.

Doesn't matter the particular solution $y_1(x)$ you chose, $y_1(x)=1$ or $y_1(x)=3\sin(x)+1$, or any other, this is the same result, but expressed with different symbols for the arbitrary constant.

  • $\begingroup$ I'm really sorry, but I have to ask, can you explain what is my question ? because I cannot see any answer, even to a misunderstood question, to any part of my actual question. $\endgroup$ – onurcanbektas Mar 14 '18 at 9:48

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