# Particular Integrals for Second Order Differential Equations with constant coefficients

My question is about finding a particular integral for a second order inhomogeneous differential equation (constant coefficients) when the inhomogeneous term is of the same form as the complementary solution to the homogeneous equation. It is clear that multiplying the complementary solution by the independent variable produces a function, Y, that is not a solution to the homogeneous equation. Should I, however, always expect Y to satisfy the inhomogeneous equation? Or is Y merely a good guess for most basic undergraduate problems that I am likely to encounter?

Thank you in anticipation of your help.

## 1 Answer

Not quite. For example, $y'' = 0$ has solutions $y = a t + b$. $y = (a t + b) t$ satisfies $y'' + a y = 2 a$.

What is true is that if $r$ is a root of order $m$ of the polynomial $P(x)$, then $q(t) e^{rt}$ is a solution of $P(D) y = 0$ for any polynomial $q$ of degree $\le m-1$, and for any polynomial $p$ of degree $d$, $P(D) y = p(t) e^{rt}$ will have solutions of the form $s(t) e^{rt}$ where $s(t)$ is a polynomial of degree $d+m$.

• Thank you for response. I am, however, still a little confused. I may have misinterpreted your notation but I think what you are saying is that if r is a root of the degree m polynomial P(x) and q(t) is a polynomial of degree less than or equal to m-1 then q(t)*e^(rt) is a solution to P(D)y=0. Yet consider P(x)=x^2-1. This has roots +/- 1. While the function y= e^t is a solution, y= t*e^t is not a solution of P(D)y=0. Could you please elaborate on your earlier point? Thank you very much in anticipation of your help. Perhaps I don't understand what is meant by "root of order m" – St Andrews Nov 1 '12 at 17:39
• $r$ is a root of order $m$ of the polynomial $P(x)$ if $(x-r)^m$ divides $P(x)$ but $(x-r)^{m+1}$ does not. For example, $+1$ and $-1$ are roots of order $1$ of $P(x) = x^2 - 1 = (x+1)(x-1)$. On the other hand, $+1$ is a root of order $2$ of $x^3-x^2-x+1 = (x-1)^2 (x+1)$. – Robert Israel Nov 1 '12 at 18:28
• So, as you say, $e^t$ is a solution of $(D^2-1) y = 0$ but $t e^t$ is not, because for $q(t) e^t$ to be a solution, the polynomial $q$ should have degree $\le 1 - 1 = 0$. On the other hand, $t e^t$ is a solution of $(D^3-D^2-D+1) y = 0$. – Robert Israel Nov 1 '12 at 18:34
• Thank you very much for your help: you have made things a lot clearer for me. – St Andrews Nov 1 '12 at 20:12