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

Part 1

Problem Statement

Given the two roots and the initial conditions:

${\lambda }_1=-2,{\lambda }_2=5$
$y(0)=1,y^{\prime }(0)=0$

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

 $y^{″}-3y^{\prime }-10=r(x)$

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:

$y(0)=1$

 $1=c_1+c_2$

$y^{\prime }(0)=0$

 $0=-2c_1+5c_2$

Solving these two equations for $c_1$ and $c_2$ yields:

 $c_1=5/4,c_2=-1/4$

Final Solution

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

File:2.1fig1.jpg

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)