# University of Florida/Egm4313/s12.team8.dupre/R5.4

Problem 5.4

## Problem Statement

(1) Show that:

${\displaystyle \displaystyle y_{p}(x)=\sum _{i=0}^{n}y_{p,i}(x)}$ (5.1)

is indeed the overall particular solution of the L2-ODE-VC:

${\displaystyle \displaystyle y''+p(x)y'+q(x)y=r(x)}$ (5.2)

with the excitation:

${\displaystyle \displaystyle r(x)=r_{1}(x)+r_{2}(x)+...+r_{n}(x)=\sum _{i=0}^{n}r_{i}(x)}$ (5.3)

(2) Discuss the choice of ${\displaystyle \displaystyle y_{p}(x)}$ in the above table , e.g., for:

${\displaystyle \displaystyle r(x)=kcos(\omega x)}$

Why would you need to have both ${\displaystyle \displaystyle cos(\omega x),sin(\omega x)}$ in ${\displaystyle \displaystyle y_{p}(x)}$?

### Solution (1)

Using the following equation:

${\displaystyle \displaystyle r_{i}(x)=y_{p,i}''+p(x)y_{p,i}'+q(x)y_{p,i}}$ (5.4)

for different r and y values gives us the following:

${\displaystyle \displaystyle r_{1}(x)=y_{p,1}''+p(x)y_{p,1}'+q(x)y_{p,1}}$ (5.5)

${\displaystyle \displaystyle r_{2}(x)=y_{p,2}''+p(x)y_{p,2}'+q(x)y_{p,2}}$ (5.6)

${\displaystyle \displaystyle r_{3}(x)=y_{p,3}''+p(x)y_{p,3}'+q(x)y_{p,3}}$ (5.7)

Now, adding (5.4),(5.5), and (5.6), gives us:

${\displaystyle \displaystyle r_{1}(x)+r_{2}(x)+r_{3}(x)=(y_{p,1}+y_{p,2}+y_{p,3})''+p(x)(y_{p,1}+y_{p,2}+y_{p,3})'+q(x)(y_{p,1}+y_{p,2}+y_{p,3})}$ (5.8)

Equation (5.8) shows us that the overall particular solution of (5.2) with excitation (5.3), is in fact, equation (5.1).

### Solution (2)

We know that the given example for an excitation is the periodic excitation:

${\displaystyle \displaystyle r(x)=kcos(\omega x)}$

When we decompose a periodic excitation into a Fourier trigonometric series, we find:

${\displaystyle \displaystyle r(x)=a_{0}+\sum _{n=0}^{\infty }[a_{n}cos(n\omega x)+b_{n}sin(\omega x)]}$

Since we know that the particular solution should depend on the excitation, we know that for a periodic excitation ${\displaystyle \displaystyle r(x)}$, we would need both ${\displaystyle \displaystyle cos(\omega x),sin(\omega x)}$ in ${\displaystyle \displaystyle y_{p}(x)}$ to obtain the correct particular solution.