University of Florida/Egm4313/s12.team11.imponenti/R2.9

Report 2, Problem 9

Problem Statement

Find and plot the solution for the L2-ODE-CC corresponding to

$\lambda ^{2}+4\lambda +13\!$

with $r(x)=0\!$

and initial conditions $y(0)=1\!$, $y'(0)=0\!$

In another figure, superimpose 3 figs.:(a)this fig. (b) the fig. in R2.6 p.5-6, and (c) the fig. in R2.1 p.3-7

Quadratic Equation

$\lambda ={\frac {-b\pm {\sqrt {(}}b^{2}-4ac)}{2a}}\!$ with $a=1,b=4,c=13\!$

$\lambda ={\frac {-4\pm {\sqrt {(}}4^{2}-4*1*13)}{2*1}}=-2\pm 3i\!$

$\lambda =-2\pm 3i\!$

Homogeneous Solution

The solution to a L2-ODE-CC with two complex roots is given by

$y(x)=e^{-{\frac {a}{2}}x}[Acos(\omega x)+Bsin(\omega x)]\!$

where $\lambda =-{\frac {a}{2}}\pm \omega i=-2\pm 3i\!$

$y(x)=e^{-2x}[Acos(3x)+Bsin(3x)]\!$

Solving for A and B

first initial condition $y(0)=1\!$

$y(x)=e^{-2x}[Acos(3x)+Bsin(3x)]\!$

$y(0)=e^{-2*0}[Acos(3*0)+Bsin(3*0)]=1\!$

$A=1\!$

second initial condition $y'(0)=0\!$

$y'(x)={\frac {d}{dx}}y(x)={\frac {d}{dx}}e^{-2x}[cos(3x)+Bsin(3x)]\!$

$y'(x)=e^{-2x}[(-2B-3)sin(3x)+(3B-2)cos(3x)]\!$

$y'(0)=e^{-2*0}[(-2B-3)sin(3*0)+(3B-2)cos(3*0)]\!$

$0=3B-2\!$

$B={\frac {2}{3}}\!$

so the solution to our L2-ODE-CC is

                      $y(x)=e^{-2x}[cos(3x)+{\frac {2}{3}}sin(3x)]\!$

Solution to R2.6

After solving for the constants ${\frac {k}{c}}$ and ${\frac {k}{m}}$ we have the following homogeneous equation

$y''(x)+6y'(x)+9y=0\!$

Characteristic Equation and Roots

$\lambda ^{2}+6\lambda +9=0\!$

$(\lambda +3)(\lambda +3)=0\!$

We have a real double root $\lambda =-3\!$

Homogeneous Solution

We know the homogeneous solution to a L2-ODE-CC with a double real root to be

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

Assuming object starts from rest

$y(0)=1\!$, $y'(0)=0\!$

Plugging in $\lambda$ and applying our first initial condition

$y(0)=c_{1}e^{-3*0}+c_{2}*0*e^{-3*0}=1\!$

$c_{1}=1\!$

Taking the derivative and applying our second condition

$y'(x)={\frac {d}{dx}}y(x)={\frac {d}{dx}}e^{-3x}+c_{2}xe^{-3x}\!$

$y'(x)=-3e^{-3x}+c_{2}e^{-3x}-3c_{2}xe^{-3x}\!$

$y'(0)=-3e^{-3*0}+c_{2}e^{-3*0}-3c_{2}*0*e^{-3*0}=0\!$

$c_{2}=3\!$

Giving us the final solution

                 $y(x)=e^{-3x}+3xe^{-3x}\!$

Plots

Solution to this Equation

$y(x)=e^{-2x}[cos(3x)+{\frac {2}{3}}sin(3x)]\!$

Superimposed Graph

Our solution: $y(x)=e^{-2x}[cos(3x)+{\frac {2}{3}}sin(3x)]\!$ shown in blue

Equation for fig. in R2.1 p.3-7: $y(x)={\frac {5}{4}}e^{-2x}-{\frac {1}{4}}e^{5x}\!$ shown in red

Equation for fig. in R2.6 p.5-6:$y(x)=e^{-3x}+3xe^{-3x}\!$ shown in green

Egm4313.s12.team11.imponenti 03:38, 8 February 2012 (UTC)