Trying to solve a non-homogeneous differential equation, whether it is linear, Bernoulli, Euler, you solve the related homogeneous equation and then you look for a particular solution depending on the "class" of the non-homogeneous term.

Just to make an example, let's say we have this equation

$$y'' - y' - 2y = 2x e^x$$

we solve $$y'' -2y' +y =0$$ and we get solutions of the form $y_{\text{hom}}(x) = C_1 e^{2x} + C_2 e^{-x}$.

Then, since the known term is of the form $p(x)(x) e^{\mu x}$, and $\mu$ is not a solution to the characteristic polynomial, we should find the solutions through those of the form $q(x)e^{\mu x}$. If, on the other hand, $\mu$ were a solution of the characteristic polynomial, we should have looked for solutions through those of the form $q(x) x^q e^{\mu x}$ where $q$ is the multiplicity of the root $\mu$. (where $q(x)$ is the general polynomial of the same degree as $p(x)$.

I have a whole list of these "classes" of known terms, like $p(x)$, $p(x)e^{\mu x}$, $e^{\alpha x}(a \sin(\beta x) + a' \cos(\beta x))$, and the corresponding general solutions (like I said before, also with the multiplicity involved).

My question is, where do these come from? I have no idea how such classes were found or anything; my book and my professors just said these were given and we should just learn them.

Edit: from the comment I see that something's going on, unfortunately I don't have enough background to know how to work with Laurent series. I would appreciate any explanation. Thanks.

  • $\begingroup$ the "classes of known terms" you mention are just linear combinations of $x^ne^{\mu x}$. By taking derivatives wrt. $\mu$ you can reduce it just to $e^{\mu x}$. Solving a constant linear ODE with $e^{\mu x}$ at the RHS is reasonably easy and unsurprising (in fact, solve it for $e^{\mu+\epsilon}$ and expand to a Laurent series in $\epsilon$ if you you want to easily include the case when $\mu$ is a root of the char. poynomial) $\endgroup$
    – user8268
    May 16, 2011 at 10:45
  • $\begingroup$ @user8268: would you care to post your comment as an answer? (feel free to add something if you want) Even though this question didn't raise much attention, I still got some info ;) $\endgroup$
    – Andy
    May 17, 2011 at 19:51

1 Answer 1


The possible right-hand-sides for this method are, themselves, solutions of linear homogeneous differential equations with constant coefficients. User8268 provided another description of this class in a comment.

If your equation is $\mathsf Ly=f$, where $\mathsf L$ is a linear differential operator with constant coefficients, and $\mathsf Mf=0$ where $\mathsf M$ is another such operator, then your desired solution is also a solution of $\mathsf{ML}y=0$, and knowledge of solutions of such (homogeneous differential equations with constant coefficients) can then be used to show what the form of $y$ must be.

  • $\begingroup$ Always thank you very much. I have improved with a specific mathjax code your nice answer. $\endgroup$
    – Sebastiano
    Jul 9, 2020 at 11:32

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