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

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

$$y(x)=c_1e^{\lambda x}+c_2xe^{\lambda x}\!$$

Assuming object starts from rest

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

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

$$y(0)=c_1e^{-3*0}+c_2*0*e^{-3*0}=1\!$$

$$c_1=1\!$$

Taking the derivative and applying our second condition

$$y'(x)=\frac{d}{dx}y(x)=\frac{d}{dx}e^{-3x}+c_2xe^{-3x}\!$$

$$y'(x)=-3e^{-3x}+c_2e^{-3x}-3c_2xe^{-3x}\!$$

$$y'(0)=-3e^{-3*0}+c_2e^{-3*0}-3c_2*0*e^{-3*0}=0\!$$

$$c_2=3\!$$

Giving us the final solution $$y(x)=e^{-3x}+3xe^{-3x}\!$$

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)