University of Florida/Egm4313/s12.team11.gooding/R2/2.1

Problem R2.1

Part 1

Problem Statement

Given the two roots and the initial conditions:

$\lambda _{1}=-2,\lambda _{2}=5\!$
$y(0)=1,y'(0)=0\!$

Find the non-homogeneous L2-ODE-CC in standard form and the solution in terms of the initial conditions and the general excitation $r(x)\!$ .
Consider no excitation:
$r(x)=0\!$
Plot the solution

Solution

Characteristic Equation:

$(\lambda -\lambda _{1})(\lambda -\lambda _{2})=0\!$
$(\lambda +2)(\lambda -5)=\lambda ^{2}+2\lambda -5\lambda -10=0\!$

  $\lambda ^{2}-3\lambda -10=0\!$

Non-Homogeneous L2-ODE-CC

  $y''-3y'-10=r(x)\!$

Homogeneous Solution:

$y_{h}(x)=c_{1}e^{-2x}+c_{2}e^{5x}\!$
$y(x)=c_{1}e^{-2x}+c_{2}e^{5x}+y_{p}(x)\!$
Since there is no excitation,
$y_{p}(x)=0\!$

  $y(x)=c_{1}e^{-2x}+c_{2}e^{5x}\!$

Substituting the given initial conditions:

$y(0)=1\!$

  $1=c_{1}+c_{2}\!$

$y'(0)=0\!$

  $0=-2c_{1}+5c_{2}\!$

Solving these two equations for $c_{1}\!$ and $c_{2}\!$ yields:

  $c_{1}=5/4,c_{2}=-1/4\!$

Final Solution

  $y(x)=(5/4)e^{-2x}-(1/4)e^{5x}\!$

Part 2

Problem Statement

Generate 3 non-standard (and non-homogeneous) L2-ODE-CC that admit the 2 values in (3a) p.3-7 as the 2 roots of the corresponding characteristic equation.

Solutions

$2(\lambda +2)(\lambda -5)=2\lambda ^{2}+4\lambda -10\lambda -20=0\!$

  $2\lambda ^{2}-6\lambda -20=0\!$

$3(\lambda +2)(\lambda -5)=3\lambda ^{2}+6\lambda -15\lambda -30=0\!$

  $3\lambda ^{2}-9\lambda -30=0\!$

$4(\lambda +2)(\lambda -5)=4\lambda ^{2}+8\lambda -20\lambda -40=0\!$

  $4\lambda ^{2}-12\lambda -40=0\!$

--Egm4313.s12.team11.gooding 02:01, 7 February 2012 (UTC)

This article is issued from Wikiversity. The text is licensed under Creative Commons - Attribution - Sharealike. Additional terms may apply for the media files.