## Table of Particular Solutions

The general solution of the equation
$A \frac{d^2y}{dt^2}+B \frac{dy}{dt}+Cy=f(t)$
is the sum of two parts.
$y_c$
, called the complementary solution, is the solution to the homogeneous equation
$A \frac{d^2y}{dt^2}+B \frac{dy}{dt}+Cy=0$
.
$y_p$
, called the particular solution, is any solution to the non homogeneous equation
$A \frac{d^2y}{dt^2}+B \frac{dy}{dt}+Cy=f(t)$
.
The particular solution must be matched to the function
$f(t)$
. The table gives some examples.
 $f(t)$ $y_p$ A B $2+5$  t (or polynomial of degree n) $A+Bt$  (or polynomial of degree n) $D e^{\omega t}, \; \omega \neq \frac{-B \pm \sqrt{B^2-4AC}}{2A}$ $Ee^{ \omega t}$ $A e^{\omega t}, \; \omega = \frac{-B \pm \sqrt{B^2-4AC}}{2A}$ $(C_1+C_2t)e^{ \omega t}$ $Dsin \omega t, \; \omega \neq \frac{-B \pm \sqrt{B^2-4AC}}{2A}$ $Esin \omega t +F cos \omega t$ $Dcos \omega t, \; \omega \neq \frac{-B \pm \sqrt{B^2-4AC}}{2A}$ $Esin \omega t +F cos \omega t$ $Dsin \omega t, \; \omega = \frac{-B \pm \sqrt{B^2-4AC}}{2A}$ $e^{\omega t}(Esin \omega t +F cos \omega t )$ $Dcos \omega t, \; \omega = \frac{-B \pm \sqrt{B^2-4AC}}{2A}$ $e^{\omega t}(Esin \omega t +F cos \omega t )$