# 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

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

with ${\displaystyle r(x)=0\!}$

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

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

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

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

${\displaystyle \lambda =-2\pm 3i\!}$

### Homogeneous Solution

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

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

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

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

### Solving for A and B

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

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

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

${\displaystyle A=1\!}$

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

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

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

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

${\displaystyle 0=3B-2\!}$

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

so the solution to our L2-ODE-CC is

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


### Solution to R2.6

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

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

#### Characteristic Equation and Roots

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

${\displaystyle (\lambda +3)(\lambda +3)=0\!}$

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

#### Homogeneous Solution

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

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

Assuming object starts from rest

${\displaystyle y(0)=1\!}$, ${\displaystyle y'(0)=0\!}$

Plugging in ${\displaystyle \lambda }$ and applying our first initial condition

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

${\displaystyle c_{1}=1\!}$

Taking the derivative and applying our second condition

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

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

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

${\displaystyle c_{2}=3\!}$

Giving us the final solution

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


### Plots

#### Solution to this Equation

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

#### Superimposed Graph

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

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

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

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