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 $y(0)=1$, $y^{\prime }(0)=0$

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

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

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

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

$0=3B-2$

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

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

$y^{″}(x)+6y^{\prime }(x)+9y=0$

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

$y(x)=c_1{e}^{\lambda x}+c_{2}x{e}^{\lambda x}$

Assuming object starts from rest

$y(0)=1$, $y^{\prime }(0)=0$

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

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

$y^{\prime }(x)=-3e^{-3x}+c_2{e}^{-3x}-3c_{2}x{e}^{-3x}$

$y^{\prime }(0)=-3e^{-3*0}+c_2{e}^{-3*0}-3c_2*0*e^{-3*0}=0$

$c_2=3$

Giving us the final solution

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

Plots

Solution to this Equation

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

File:Plotr2 9.jpg

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:$y(x)=e^{-3x}+3x{e}^{-3x}$ shown in green

File:R2superposed.jpg

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