User:Egm4313.s12.team9/R2

= R2.1: Non-Homogeneous L2-ODE-CC =

This problem is presented in three parts and will thus be solved in three parts: Part 1 can be found on [[media:Egm4313.s12.team9.3-7.jpg|lecture slide 3-7]] Part 2 can be found on [[media:Egm4313.s12.team9.3-7.jpg|lecture slide 3-7]] Part 3 can be found on [[media:Egm4313.s12.team9.3-9.jpg|lecture slide 3-9]]

Given
Two roots:

$$\lambda_1=-2, \ \lambda_2=-2$$

Initial Conditions:


 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle y(0)=1 $$
 * $$\displaystyle(Eq.1)$$
 * }
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle y'(0)=0 $$
 * $$\displaystyle (Eq.2)$$
 * }

Find the non-homogeneous L2-ODE-CC in standard form and the solution in terms of the initial conditions and the general excitation $$r(x) \ $$.

Methods
From the characteristic equation:

$$(\lambda-\lambda_1)(\lambda-\lambda_2)=0 \ $$

We Have:

$$(\lambda+2)(\lambda-5)=0 \ $$

Expanding to the standard form:

$$\lambda^{2}-3\lambda-10 \ $$

Thus, the non-homogeneous L2-ODE-CC is:

$$y''-3y'-10y=r(x) \ $$

Its corresponding homogeneous solution is:

$$y(x)=c_1e^{5x}+c_2e^{-2x} \ $$

The overall solution $$y(x) \ $$is:

$$y(x)=c_1e^{5x}+c_2e^{-2x}+y_p(x) \ $$

and its derivative $$y'(x) \ $$ is:

$$y'(x)=5c_1e^{5x}-2c_2e^{-2x}+y'_p(x) \ $$

Satisfying the initial conditions (Eq.1 and Eq.2):

$$y(0)=c_1+c_2+y_p(0)=1 \ $$

$$y'(0)=5c_1-2c_2+y'p(0) \ $$

Multiplying (Eq.1) by 2 and adding it to (Eq.2) yields:

$$2=7c_1+0c_2+2y_p(0)+y'_p(0) \ $$

Solving for $$c_1 \ $$:

$$7c_1=2-2y_p(0)-y'_p(0) \ $$

$$c_1=\frac{1}{7}(2-2y_p(0)-y'_p(0)) \ $$

Substituting $$c_1 \ $$ into (Eq.1) yields:

$$1=\frac{1}{7}(2-2y_p(0)-y'_p(0))+c_2+y_p(0) \ $$

Rearranging to solve for $$c_2 \ $$:

$$7=2-2y_p(0)-y'_p(0)+7c_2+7y_p(0) \ $$

$$7c_2=7-2+2y_p(0)-7y_p(0)+y'_p(0) \ $$

$$c_2=\frac{1}{7}(5-5y_p(0)+y'_p(0)) \ $$

Solution
Plugging $$c_1,c_2 \ $$ into $$y(x) \ $$:

$$y(x)=\frac{1}{7}(2-2y_p(0)-y'_p(0))e^{5x}+\frac{1}{7}(5-5y_p(0)+y'_p(0))e^{-2x}+y_p(x) \ $$

Given
Consider the particular case of the non-homogeneous L2-ODE-CC solved above with no excitation:

$$r(x)=0 \ $$

and plot the solution.

Methods
Since $$r(x)=0 \ $$, thus $$y_p(x)=0 \ $$ as well as its corresponding 1st and 2nd derivatives.

We can thus solve for the real values of $$c_1, c_2 \ $$:

$$c_1=\frac{2}{7}, \ c_2=\frac{5}{7}$$

Using MATLAB to plot $$y(x) \ $$ we use the following code:

Solution
Plugging in $$c_1, c_2 \ $$ to $$y(x) \ $$ we get:

$$y(x)=\frac{2}{7}e^{5x}+\frac{5}{7}e^{-2x}$$

The corresponding plot for $$y(x) \ $$ is:



Given
Part 3 asks us to generate 3 non-standard & non-homogeneous L2-ODE-CC that still admit the two initial values:

$$\lambda_1=-2, \ \lambda_2=5$$

found in figure 3a on [[media:Lecture_Slide_3-7.png|Lecture Slide 3-7]] as the two roots in the corresponding characteristic equation.

Methods
The standard form can be found from Eq.1 on [[media:Lecture_Slide_3-8.png|Lecture Slide 3-8]] when $$M=1 \ $$. Non-standard forms can be found using any integer $$M\neq0,1 \ $$.

Thus, three non-standard corresponding equations can be found by using the values:

$$M=2,3,5 \ $$

and the characteristic equation:

$$M(\lambda-\lambda_1)(\lambda-\lambda_2)=0 \ $$.

Plugging in these values of $$M \ $$ we get:

$$M=2 \rightarrow 2\lambda^{2}-6\lambda-20 \ $$

$$M=3 \rightarrow 3\lambda^{2}-9\lambda-30 \ $$

$$M=5 \rightarrow 5\lambda^{2}-15\lambda-50 \ $$

Solution
The corresponding non-standard and non-homogeneous L2-ODE-CC equations with matching roots are:

$$M=2 \rightarrow r(x)=2y''-6y'-20y \ $$

$$M=3 \rightarrow r(x)=3y''-9y'-30y \ $$

$$M=5 \rightarrow r(x)=5y''-15y'-50y \ $$

Author
Solved and typed by Egm4313.s10.team9.delp 04:24, 7 February 2012 (UTC)

Reviewed by Egm4313.s10.team9.lebo 13:25, 8 February 2012 (UTC)

Edited by Egm4313.s10.team9.lebo 13:25, 8 February 2012 (UTC)

= R2.2: Solution for Linear 2nd Order ODE (CC) with Double Root = This problem can be found on [[media:Egm4313.s12.team9.5-6.jpg|lecture slide 5-6]].

Given
Given the following linear, 2nd-order ODE with constant coefficients,


 * {| style="width:100%" border="0" align="left"

$$ The initial conditions:
 * $$\displaystyle {y}''-10{y}'+25y=r(x) \! $$
 * $$\displaystyle (Eq.1)
 * $$\displaystyle (Eq.1)
 * }
 * }


 * {| style="width:100%" border="0" align="left"

$$ And no excitation:
 * $$\displaystyle y(0)=1\text{, }{y}'(0)=0 \! $$
 * $$\displaystyle (Eq.1b)
 * $$\displaystyle (Eq.1b)
 * }
 * }


 * {| style="width:100%" border="0" align="left"

$$ Find and plot the solution.
 * $$\displaystyle r(x)=0 \! $$
 * $$\displaystyle (Eq.1c)
 * $$\displaystyle (Eq.1c)
 * }
 * }

Methods
Let $$\lambda =\frac{dy}{dx}$$ in Equation (1).

The characteristic equation of Equation (1) is


 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle {{\lambda }^{2}}-10\lambda +25=0 \! $$
 * $$\displaystyle (Eq.2)
 * $$\displaystyle (Eq.2)
 * }
 * }

The case of the equation can be determined by $${{a}^{2}}-4b \! $$ where $$a=-10 \! $$ and $$b=25 \! $$ as shown below:


 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle {{a}^{2}}-4b={{(-10)}^{2}}-4(25)=100-100=0 \! $$
 * }
 * }
 * }

Since this is equal to 0, the equation is a Case II equation with a real double root.

Solution
Equation (2) can be factored as shown below to determine the root:


 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle {{(\lambda -5)}^{2}}=0 \! $$
 * }
 * {| style="width:100%" border="0" align="left"
 * }
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle \text{double root: }\lambda =5 \! $$
 * }
 * }
 * }

From [[media:Egm4313.s12.team9.5-4.jpg|lecture slide 5-4]] and from Kreyszig 2011 pg.55, the general homogenous solution for the equation is:


 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle {{y}_{h}}={{c}_{1}}{{e}^{\lambda x}}+{{c}_{2}}x{{e}^{\lambda x}} \! $$
 * }
 * }
 * }

Therefore, the general solution for this example is:


 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle {{y}_{h}}(x)={{c}_{1}}{{e}^{5x}}+{{c}_{2}}x{{e}^{5x}} \! $$
 * $$\displaystyle (Eq.3)
 * $$\displaystyle (Eq.3)
 * }
 * }

The initial conditions are used to solve for $${{c}_{1}}\text{ and }{{c}_{2}} \! $$, as shown below.

From Equations (1b) and (3),


 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle y(0)={{c}_{1}}{{e}^{0}}+(0){{c}_{2}}{{e}^{0}}={{c}_{1}}=1 \! $$
 * }
 * }
 * }
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle {y}'(0)=5{{c}_{1}}{{e}^{5x}}+{{c}_{2}}{{e}^{5x}}+x(5{{c}_{2}}{{e}^{5x}})={{e}^{5x}}(5{{c}_{1}}+{{c}_{2}}+5x{{c}_{2}})=5{{c}_{1}}+{{c}_{2}}+5(0){{c}_{2}}=5(1)+{{c}_{2}}=0 \! $$
 * }
 * }
 * }
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle \therefore {{c}_{2}}=-5 \! $$
 * }
 * }
 * }

Substituting the solutions for $${{c}_{1}}\text{ and }{{c}_{2}} \! $$ into Equation (3), the final solution is:


 * {| style="width:100%" border="0" align="left"

$$\displaystyle y(x)={{e}^{5x}}-5x{{e}^{5x}} \!$$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * $$\displaystyle $$
 * }

Using MATLAB to plot $$y(x) \ $$ we use the following code:

The corresponding plot for $$y(x) \ $$ is:



Author
Solved and typed by --Egm4313.s12.team9.wright 05:53, 8 February 2012 (UTC)

Reviewed by --Egm4313.s12.team9.ehlers.km 15:08, 8 February 2012 (UTC)

Edited by

=R2.3: Find a general solution for problems 3 and 4 (K2011 pg. 59)= This problem can be found on [[media:Egm4313.s12.team9.5-6.jpg|lecture slide 5-6]].

Given
$$ y'' + 6y' + 8.96y = 0 \! $$ $$ \lambda^{2} + 6\lambda + 8.96 = 0 \! $$

Methods
Using the Quadratic Equation: $$ a=1, b=6, c=8.96 \! $$  $$ \frac{-b\frac{+}{-}\sqrt{b^{2} - 4ac}}{2a} \! $$ $$ \frac{-6\frac{+}{-}\sqrt{6^{2} - 35.84}}{2} \! $$ $$ \frac{-6 + 0.4}{2} \! $$ $$ \frac{-6 - 0.4}{2} \! $$ Root 1: $$ -2.8 \! $$ Root 2: $$ -3.2 \! $$

Solution
General Solution:

Given
From Kreyszig 2011 pg. 59 problem 4,


 * {| style="width:100%" border="0" align="left"

{y}''+4{y}'+({{\pi }^{2}}+4)y=0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq.1)
 * }
 * }

Find a general solution for Equation (1) and check the answer by substitution.

Methods
Let $$\lambda =\frac{dy}{dx}$$ in Equations (1). The characteristic equation of Equation (1) is


 * {| style="width:100%" border="0" align="left"

{{\lambda }^{2}}+4\lambda +({{\pi }^{2}}+4)=0 $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

The case of the equation can be determined by $$\displaystyle {{a}^{2}}-4b$$, where $$\displaystyle a=4$$ and $$\displaystyle b={{\pi }^{2}}+4$$ , as shown below:


 * {| style="width:100%" border="0" align="left"

{{a}^{2}}-4b={{4}^{2}}-4({{\pi }^{2}}+4)=16-4{{\pi }^{2}}-16=-4{{\pi }^{2}} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Since this is below 0, the equation is a Case III equation with complex conjugate roots.

Solution
The roots of the characteristic equation is given by:


 * {| style="width:100%" border="0" align="left"

& \lambda =\frac{1}{2}\left( -a\pm \sqrt{{{a}^{2}}-4b} \right) \\ & \lambda =\frac{1}{2}\left( -4\pm \sqrt{-4{{\pi }^{2}}} \right) \\ & \lambda =-2\pm \pi i \\ \end{align}$$
 * $$\displaystyle\begin{align}
 * $$\displaystyle\begin{align}
 * }
 * }

Therefore, in this problem,
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle \omega =\pi$$
 * }
 * }
 * }

From page 57, a real general solution in Case III is


 * {| style="width:100%" border="0" align="left"

y={{e}^{-ax/2}}(A\cos \omega x+B\sin \omega x)$$ where A and B are arbitrary constant coefficients. In this example, the general solution is
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }
 * {| style="width:0%" border="0" align="left"

$$\displaystyle y={{e}^{-2x}}(A\cos \pi x+B\sin \pi x) $$
 * style="width:100%; padding:10px; border:2px solid #8888aa" |
 * style="width:100%; padding:10px; border:2px solid #8888aa" |
 * style= |
 * }
 * }

To check by substitution, the first and second derivatives of the general solution are taken and substituted into Equation (1).


 * {| style="width:100%" border="0" align="left"

& {y}'=-2{{e}^{-2x}}(A\cos \pi x+B\sin \pi x)+{{e}^{-2x}}\left( -A\pi \sin \pi x+B\pi \cos \pi x \right) \\ & {y}''=4{{e}^{-2x}}(A\cos \pi x+B\sin \pi x)-2{{e}^{-2x}}(-A\pi \sin \pi x+B\pi \cos \pi x)-2{{e}^{-2x}}(-A\pi \sin \pi x+B\pi \cos \pi x)+{{e}^{-2x}}(-A{{\pi }^{2}}\cos \pi x-B{{\pi }^{2}}\sin \pi x) \\ & {y}''=4{{e}^{-2x}}(A\cos \pi x+B\sin \pi x)-4{{e}^{-2x}}(-A\pi \sin \pi x+B\pi \cos \pi x)+{{e}^{-2x}}(-A{{\pi }^{2}}\cos \pi x-B{{\pi }^{2}}\sin \pi x) \\ \end{align}$$
 * $$\displaystyle\begin{align}
 * $$\displaystyle\begin{align}
 * }
 * }

Substituting into Equation (1),


 * {| style="width:100%" border="0" align="left"

4{{e}^{-2x}}(A\cos \pi x+B\sin \pi x)-4{{e}^{-2x}}(-A\pi \sin \pi x+B\pi \cos \pi x)+{{e}^{-2x}}(-A{{\pi }^{2}}\cos \pi x-B{{\pi }^{2}}\sin \pi x)+4\left( -2{{e}^{-2x}}(A\cos \pi x+B\sin \pi x)+{{e}^{-2x}}(-A\pi \sin \pi x+B\pi \cos \pi x) \right)+({{\pi }^{2}}+4)\left( {{e}^{-2x}}(A\cos \pi x+B\sin \pi x) \right)=0$$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Multiplying and rearranging terms,
 * {| style="width:100%" border="0" align="left"

4{{e}^{-2x}}(A\cos \pi x+B\sin \pi x)-8{{e}^{-2x}}(A\cos \pi x+B\sin \pi x)+4{{e}^{-2x}}(A\cos \pi x+B\sin \pi x)-4{{e}^{-2x}}(-A\pi \sin \pi x+B\pi \cos \pi x)+4{{e}^{-2x}}(-A\pi \sin \pi x+B\pi \cos \pi x)+{{e}^{-2x}}(-A{{\pi }^{2}}\cos \pi x-B{{\pi }^{2}}\sin \pi x)+{{\pi }^{2}}{{e}^{-2x}}(A\cos \pi x+B\sin \pi x)=0$$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }


 * {| style="width:100%" border="0" align="left"

0=0$$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Since all terms cancel, the solution is
 * {| style="width:100%" border="0" align="left"

y={{e}^{-2x}}(A\cos \pi x+B\sin \pi x)$$.
 * $$\displaystyle
 * }
 * }

Author
Solved and typed by --Egm4313.s12.team9.tjl 18:17, 8 February 2012 (UTC)

Reviewed by --Egm4313.s12.team9.wright 05:53, 8 February 2012 (UTC)

Edited by --Egm4313.s12.team9.wright 05:53, 8 February 2012 (UTC)

= R2.4: Find a general solution for problems 5 and 6 (K2011 pg. 59) =

Problem 5
This problem can be found on [[media:Egm4313.s12.team9.5-6.jpg|lecture slide 5-6]] and from Kreyszig 2011 pg. 59 problem 5.

Given
Problem Statement: Find a general solution and check by substitution.

Methods
Becomes:

Let $$\lambda =\frac{dy}{dx}$$ in Equation (1).

Equation (1) becomes:

Solving for $$\lambda \! $$ through factoring,

Root: $$\lambda =-\pi \! $$

Solution
General Solution:

So the general solution becomes:

$$ \; y = c_1 e^{-\pi x} + c_2 x e^{-\pi x} $$

Check by Substitution:

To check by substitution, you need to find $$ \; y, y', y'' $$ and plug them into the original ODE.

$$ \; y= e ^ {-\pi x} (c_1 + c_2 x) $$

$$ \; y'= -\pi c_1e ^ {-\pi x} + c_2 (e ^ {-\pi x} -\pi x e ^ {-\pi x})= - \pi e ^{- \pi x} (c_1 + c_2 x) + c_2 e^ {- \pi x} $$

$$ \; y''= \pi ^{2} c_1e ^ {- \pi x} - \pi c_2 e ^ {-\pi x} -\pi c_2(e ^ {- \pi x} - \pi x e^ {- \pi x} )= \pi ^ {2} e^ {- \pi x} (c_1 + c_2 x) -2 \pi c_2 e^ {- \pi x} $$

Substitute the values of $$ \; y, y', y $$ into $$ \; {y}+2\pi {y}'+\pi ^{2}y=0 $$ :

$$ \; [ \pi ^ {2} e^ {- \pi x} (c_1 + c_2 x) - 2 \pi c_2e^ {- \pi x}] + 2 \pi [ - \pi e^ {- \pi x} (c_1 + c_2 x) + c_2e^ {- \pi x} ] + \pi ^ {2} [ e^ {- \pi x} (c_1 + c_2 x)]=0 $$

After substitution, it proves the general solution because the substitution equals zero.

Problem 6
This problem can be found on [[media:Egm4313.s12.team9.5-6.jpg|lecture slide 5-6]] and from Kreyszig 2011 pg. 59 problem 6.

Given
Problem Statement: Find a general solution and check by substitution.

Methods
Equation (1) can be rewritten as:

Let $$\lambda =\frac{dy}{dx}$$ in Equation (2).

The characteristic equation can then be written as:

Solve for $$ \; \lambda $$ by using the quadratic equation.

$$ \; \lambda=\frac{-b\pm\sqrt{b^2-4ac}}{2a} $$

$$ \; \lambda = \frac{8}{5} $$

Solution
General solution:

The general solution then becomes:

Check by Substitution:

To check by substitution, you need to find $$ \; y, y', y'' $$ and plug them into the original ODE.

$$ \; y = e^{\frac{8}{5}x}({c_1 + c_2 x}) $$

$$ \; y' = \frac{8}{5} e^{\frac{8}{5}x}(c_1 + c_2 x) + c_2 e^{\frac{8}{5}x} $$

$$ \; y'' = \frac{64}{25} e^{\frac{8}{5}x}(c_1 + c_2 x) + \frac{16}{5}c_2 e^{\frac{8}{5}x} $$

Substitute the values of $$ \; y, y', y $$ into $$ \; 10 y - 32 y' + 25.6 y = 0 $$ :

$$ \; \frac{64}{25} e^{\frac{8}{5}x}(c_1 + c_2 x) - \frac{64}{25} e^{\frac{8}{5}x}(c_1 + c_2 x) + 32 c_2 e^{\frac{8}{5}x} - 32 c_2 e^{\frac{8}{5}x} = 0 $$

After substitution, it proves the general solution because the substitution equals zero.

Author
Solved and typed by --Egm4313.s12.team9.ehlers.km 15:03, 8 February 2012 (UTC)

Reviewed by --Egm4313.s12.team9.wright 05:53, 8 February 2012 (UTC)

Edited by --Egm4313.s12.team9.wright 05:53, 8 February 2012 (UTC)

= R2.5: Find an ODE for problems 16 and 17 (K2011 pg. 59) = This problem can be found on [[media:Egm4313.s12.team9.5-6.jpg|lecture slide 5-6]].

Given
The DE given is:


 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle {y}''+a{y}'+by = 0 $$
 * $$\displaystyle (Eq.1)
 * $$\displaystyle (Eq.1)
 * }
 * }

We are being asked to find an ODE for the given basis solutions shown below.

For problem 16:


 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle e^{2.6x}\,,\,e^{-4.3x} $$
 * }
 * }
 * }

Methods
From Kreyszig 2011 pg. 58:

In the case of two real distinct roots, the following general solution should be used:


 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle y = c_{1}e^{\lambda x} + c_{2}e^{\lambda x} $$
 * <p style="text-align:right;">$$\displaystyle (Eq.2)
 * <p style="text-align:right;">$$\displaystyle (Eq.2)
 * }
 * }

Solution
Let


 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle y_{1} = e^{2.6x}\,,\,y_{2} = e^{-4.3x} $$
 * }
 * }
 * }

Since $$ y_{1} \!$$ and $$ y_{2} \!$$ are defined for all real values, the general solution of the ODE would look like this


 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle y = c_{1}e^{2.6x} + c_{2}e^{-4.3x} $$
 * <p style="text-align:right;">$$\displaystyle (Eq.3)
 * <p style="text-align:right;">$$\displaystyle (Eq.3)
 * }
 * }

Comparing the general solution (3) to the "model" general solution (2), we find that


 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle \lambda_{1}=2.6\,,\,\lambda_{2}=-4.3 $$
 * }
 * }
 * }

Which are distinct real values; therefore, the characteristic equation can be obtained by


 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle (\lambda-2.6)\,[\lambda-(-4.3)]=0 $$
 * }
 * }
 * }

Distributing,


 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle \lambda^{2}-2.6\lambda+4.3\lambda-11.18=0 $$
 * }
 * }
 * }

So, the ODE with the given basis solution that we are looking for is:


 * {| style="width:0%" border="0" align="left"

$$\displaystyle {y}''+1.7{y}'-11.18y = 0 $$
 * style="width:100%; padding:10px; border:2px solid #8888aa" |
 * style="width:100%; padding:10px; border:2px solid #8888aa" |
 * style= |
 * }
 * }

Given
The DE given is:


 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle {y}''+a{y}'+by = 0 $$
 * }
 * }
 * }

We are being asked to find an ODE for the given basis solutions shown below.

For problem 17:


 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle e^{-\sqrt{5}x}\,,\,xe^{-\sqrt{5}x} $$
 * }
 * }
 * }

Methods
From Kreyszig 2011 pg. 58:

In the case of a real double root, the following general solution should be used:


 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle y = (c_{1}+xc_{2})e^{\lambda x} $$
 * <p style="text-align:right;">$$\displaystyle (Eq.4)
 * <p style="text-align:right;">$$\displaystyle (Eq.4)
 * }
 * }

Solution
Let


 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle y_{1} = e^{-\sqrt{5}x}\,,\,y_{2} = xe^{-\sqrt{5}x} $$
 * }
 * }
 * }

Since $$ y_{1} \!$$ and $$ y_{2} \! $$ are defined for all real values, the general solution of the ODE is written as


 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle y = (c_{1}+c_{2}x)e^{-\sqrt{5}x} $$
 * <p style="text-align:right;">$$\displaystyle (Eq.5)
 * <p style="text-align:right;">$$\displaystyle (Eq.5)
 * }
 * }

Comparing the general solution (5) to the "model" general solution (4), we obtain


 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle \lambda=-\sqrt{5} $$
 * }
 * }
 * }

Which is a real repeated root; therefore, the characteristic equation can be obtained by


 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle [\lambda-(-\sqrt{5})]\,[\lambda-(-\sqrt{5})]=0 $$
 * }
 * }
 * }

Distributing,


 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle \lambda^{2}+2\sqrt{5}\lambda+5=0 $$
 * }
 * }
 * }

So, the ODE with the given basis solution that we are looking for is:


 * {| style="width:0%" border="0" align="left"

$$\displaystyle {y}''+2\sqrt{5}{y}'+5y = 0 $$
 * style="width:100%; padding:10px; border:2px solid #8888aa" |
 * style="width:100%; padding:10px; border:2px solid #8888aa" |
 * style= |
 * }
 * }

Author
Solved and typed by E.Beuses 16:15, 8 February 2012 (UTC)

Reviewed by --Egm4313.s12.team9.wright 08:58, 8 February 2012 (UTC)

Edited by

= R2.6 = This problem can be found on [[media:Egm4313.s12.team9.5-6.jpg|lecture slide 5-6]].

Given
We are given an equation from [[media:Lecture_Slide_5-5.png|lecture slide 5-5]]:


 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle (\lambda - 5)^2 = (\lambda)^2 - 10(\lambda) + 25 $$
 * <p style="text-align:right;">$$\displaystyle (Eq.1)
 * <p style="text-align:right;">$$\displaystyle (Eq.1)
 * }
 * }

and told to use:


 * {| style="width:100%" border="0" align="left"

$$ as our double real root.
 * $$\displaystyle \lambda = -3 $$
 * <p style="text-align:right;">$$\displaystyle (Eq.2)
 * <p style="text-align:right;">$$\displaystyle (Eq.2)
 * }
 * }

We were asked to find the variables k, c, and m, which correlate to the spring-dashpot-mass system in series shown in [[media:Lecture_Slide_1-4.png|lecture slide 1-4]]:



Methods
So we sub Eq.2 into Eq.1 to get:
 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle (\lambda + 3)^2 = 0 $$
 * <p style="text-align:right;">$$\displaystyle (Eq.3)
 * <p style="text-align:right;">$$\displaystyle (Eq.3)
 * }
 * }

Now, we expand Eq.3:
 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle {y}'' + 6{y}' + 9 = 0 $$
 * <p style="text-align:right;">$$\displaystyle (Eq.4)
 * <p style="text-align:right;">$$\displaystyle (Eq.4)
 * }
 * }

Solution
With the expansion of the equation, we see (from [[media:Egm4313.s12.team9.1-5.jpg|lecture slide 1-5]], equation 3) how the coefficients relate to the variables we want to find. So the coefficient in front of $$ y'' \ $$ relates to m, the coefficient in front of $$ y' \ $$ relates to $$\frac{mk}{c}$$, and the coefficient in front of $$ y \ $$ relates to k:


 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle m=1 $$
 * }
 * }
 * }


 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle c=\frac{3}{2} $$
 * }
 * }
 * }


 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle k=9 $$
 * }
 * }
 * }

Author
Solved and typed by --Egm4313.s12.team9.lebo 18:23, 8 February 2012 (UTC)

Reviewed by --Egm4313.s10.team9.delp 19:56, 8 February 2012 (UTC)

Edited by

=R2.7: Develop the MacLaurin Series for $$ e^{t}, cos(t), sin(t) \! $$= This problem can be found on [[media:Egm4313.s12.team9.6-6.jpg|lecture slide 6-6]].

$$ P_n(t) = f(a) + f'(a)(t-a) + \frac{f''(a)}{2!}(t-a)^{2} +...+\frac{f^{n}(a)}{n!}(t-a)^{n} \! $$

MacLaurin Series(Taylor Series at t=0) for $$ e^{t} \! $$
$$ f(t) = e^{t}.....f(0)=1 \! $$ $$ f'(t) = e^{t}.....f'(0)=1 \! $$ $$ f(t) = e^{t}.....f(0)=1 \! $$ $$ f(t) = e^{t}.....f(0)=1 \! $$ $$ f'(t) = e^{t}.....f'(0)=1 \! $$ $$ P_5(t) = 1 + t + \frac{t^{2}}{2!} + \frac{t^{3}}{3!} + \frac{t^{4}}{4!} \! $$

MacLaurin Series(Taylor Series at t=0) for $$ cos(t) \! $$
$$ f(t) = \cos{t}.....f(0)=1 \! $$ $$ f'(t) = -\sin{t}.....f'(0)=0 \! $$ $$ f(t) = -\cos{t}.....f(0)=-1 \! $$ $$ f(t) = \sin{t}.....f(0)=0 \! $$ $$ f'(t) = \cos{t}.....f'(0)=1 \! $$ $$ P_5(t) = 1 - \frac{t^{2}}{2!} + \frac{t^{4}}{4!} \! $$

MacLaurin Series(Taylor Series at t=0) for $$ sin(t) \! $$
$$ f(t) = \sin{t}.....f(0)=0 \! $$ $$ f'(t) = \cos{t}.....f'(0)=1 \! $$ $$ f(t) = -\sin{t}.....f(0)=0 \! $$ $$ f(t) = -\cos{t}.....f(0)=-1 \! $$ $$ f'(t) = \sin{t}.....f'(0)=0 \! $$ $$ P_5(t) = t - \frac{t^{3}}{3!} \! $$

Author
Solved and typed by --Egm4313.s12.team9.tjl 18:17, 8 February 2012 (UTC) Reviewed by --Egm4313.s12.team9.murray 09:50, 8 February 2012 (UTC)

Edited by

= R2.8: General Solution for Linear 2nd Order ODEs (CC) with Complex Conjugate Roots = This problem can be found on [[media:Egm4313.s12.team9.6-6.jpg|lecture slide 6-6]].

Given
From Kreyszig pg. 59 problem 8,
 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle {y}''+{y}'+3.25y=0 \ $$
 * <p style="text-align:right;">$$\displaystyle (Eq.1)
 * <p style="text-align:right;">$$\displaystyle (Eq.1)
 * }
 * }

Find a general solution for Equation (1) and check the answer by substitution.

Methods
Let $$\lambda =\frac{dy}{dx}$$ in Equation (1).

The characteristic equation of Equation (1) is
 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle {{\lambda }^{2}}+\lambda +3.25=0 \ $$
 * <p style="text-align:right;">$$\displaystyle (Eq.2)
 * <p style="text-align:right;">$$\displaystyle (Eq.2)
 * }
 * }

The case of the equation can be determined by $${{a}^{2}}-4b \!$$ where $$a=1 \!$$ and $$b=3.25 \!$$, as shown below:
 * {| style="width:100%" border="0" align="left"

Since this is below 0, the equation is a Case III equation with complex conjugate roots.
 * $$\displaystyle {{a}^{2}}-4b={{(1)}^{2}}-4(3.25)=1-13=-12 \ $$
 * }
 * }
 * }

Solution
The roots of the characteristic equation are given by:


 * {| style="width:100%" border="0" align="left"

& \lambda =\frac{1}{2}\left( -a\pm \sqrt{{{a}^{2}}-4b} \right) \\ & \lambda =\frac{1}{2}\left( -1\pm \sqrt{-12} \right)=\frac{1}{2}\left( -1\pm \left( 2\sqrt{3} \right)i \right) \\ & \lambda =-\frac{1}{2}\pm \left( \sqrt{3} \right)i \\ \end{align}$$
 * $$\displaystyle \begin{align}
 * $$\displaystyle \begin{align}
 * }
 * }

Therefore, in this problem,
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle \omega =\sqrt{3} $$
 * }
 * }
 * }

From [[media:Egm4313.s12.team9.6-5.jpg|lecture slide 6-5]] and from Kreyszig page 57, a general homogeneous solution for a Case III equation is
 * {| style="width:100%" border="0" align="left"

{{y}_{h}}={{e}^{-ax/2}}(A\cos \omega x+B\sin \omega x) \ $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

where A and B are arbitrary constant coefficients. In this example, the general solution is
 * {| style="width:100%" border="0" align="left"

$$\displaystyle y={{e}^{-0.5x}}\left( A\cos (\sqrt{3})x+B\sin (\sqrt{3})x \right) \!$$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right;">$$\displaystyle $$
 * }

To check by substitution, the first and second derivatives of the general solution are taken and substituted into Equation (1).
 * {| style="width:100%" border="0" align="left"

{y}'=-0.5{{e}^{-0.5x}}\left( A\cos (\sqrt{3})x+B\sin (\sqrt{3})x \right)+{{e}^{-0.5x}}\left( -A(\sqrt{3})\sin (\sqrt{3})x+B(\sqrt{3})\cos (\sqrt{3})x \right)$$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }


 * {| style="width:100%" border="0" align="left"

{y}''=0.25{{e}^{-0.5x}}\left( A\cos (\sqrt{3})x+B\sin (\sqrt{3})x \right)-0.5{{e}^{-0.5x}}\left( -A(\sqrt{3})\sin (\sqrt{3})x+B(\sqrt{3})\cos (\sqrt{3})x \right)-0.5{{e}^{-0.5x}}\left( -A(\sqrt{3})\sin (\sqrt{3})x+B(\sqrt{3})\cos (\sqrt{3})x \right)+{{e}^{-0.5x}}\left( -3A\cos (\sqrt{3})x-3B\sin (\sqrt{3})x \right)$$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }


 * {| style="width:100%" border="0" align="left"

{y}''=0.25{{e}^{-0.5x}}\left( A\cos (\sqrt{3})x+B\sin (\sqrt{3})x \right)-{{e}^{-0.5x}}\left( -A(\sqrt{3})\sin (\sqrt{3})x+B(\sqrt{3})\cos (\sqrt{3})x \right)+{{e}^{-0.5x}}\left( -3A\cos (\sqrt{3})x-3B\sin (\sqrt{3})x \right)$$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Substituting into Equation (1),
 * {| style="width:100%" border="0" align="left"

0.25{{e}^{-0.5x}}\left( A\cos (\sqrt{3})x+B\sin (\sqrt{3})x \right)-{{e}^{-0.5x}}\left( -A(\sqrt{3})\sin (\sqrt{3})x+B(\sqrt{3})\cos (\sqrt{3})x \right)+{{e}^{-0.5x}}\left( -3A\cos (\sqrt{3})x-3B\sin (\sqrt{3})x \right)-0.5{{e}^{-0.5x}}\left( A\cos (\sqrt{3})x+B\sin (\sqrt{3})x \right)+{{e}^{-0.5x}}\left( -A(\sqrt{3})\sin (\sqrt{3})x+B(\sqrt{3})\cos (\sqrt{3})x \right)+3.25{{e}^{-0.5x}}\left( A\cos (\sqrt{3})x+B\sin (\sqrt{3})x \right)=0$$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Multiplying and rearranging terms,


 * {| style="width:100%" border="0" align="left"

0.25{{e}^{-0.5x}}\left( A\cos (\sqrt{3})x+B\sin (\sqrt{3})x \right)-0.5{{e}^{-0.5x}}\left( A\cos (\sqrt{3})x+B\sin (\sqrt{3})x \right)-3{{e}^{-0.5x}}\left( A\cos (\sqrt{3})x+B\sin (\sqrt{3})x \right)+3.25{{e}^{-0.5x}}\left( A\cos (\sqrt{3})x+B\sin (\sqrt{3})x \right)-{{e}^{-0.5x}}\left( -A(\sqrt{3})\sin (\sqrt{3})x+B(\sqrt{3})\cos (\sqrt{3})x \right)+{{e}^{-0.5x}}\left( -A(\sqrt{3})\sin (\sqrt{3})x+B(\sqrt{3})\cos (\sqrt{3})x \right)=0$$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }


 * {| style="width:100%" border="0" align="left"


 * $$0=0 \!$$
 * }
 * }
 * }

Since all terms cancel, the solution is
 * {| style="width:100%" border="0" align="left"

y={{e}^{-0.5x}}\left( A\cos (\sqrt{3})x+B\sin (\sqrt{3})x \right)$$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Given
From Kreyszig pg. 59 problem 15,
 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle {y}''+0.54{y}'+(0.0729+\pi )y=0 \! $$
 * <p style="text-align:right;">$$\displaystyle (Eq.3)
 * <p style="text-align:right;">$$\displaystyle (Eq.3)
 * }
 * }

Find a general solution for Equation (3) and check the answer by substitution.

Methods
Let $$\lambda =\frac{dy}{dx}$$ in Equation (3).

Using the same method, the characteristic equation of Equation (3) is
 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle {{\lambda }^{2}}+0.54\lambda +(0.0729+\pi )=0 \ $$
 * <p style="text-align:right;">$$\displaystyle (Eq.4)
 * <p style="text-align:right;">$$\displaystyle (Eq.4)
 * }
 * }

The case of the equation can be determined by $${{a}^{2}}-4b \! $$ where $$a=0.54 \! $$ and $$b=0.0729+\pi \! $$, as shown below:
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle {{a}^{2}}-4b={{(0.54)}^{2}}-4(0.0729+\pi )=0.2916-0.2916-4\pi =-4\pi \ $$
 * }
 * }
 * }

Since this is below 0, the equation is also a Case III equation with complex conjugate roots.

Solution
The roots of the characteristic equation are given by:


 * {| style="width:100%" border="0" align="left"

& \lambda =\frac{1}{2}\left( -a\pm \sqrt{{{a}^{2}}-4b} \right) \\ & \lambda =\frac{1}{2}\left( -0.54\pm \sqrt{-4\pi } \right)=\frac{1}{2}\left( -0.54\pm \left( 2\sqrt{\pi } \right)i \right) \\ & \lambda =-0.27\pm \left( \sqrt{\pi } \right)i \\ \end{align}$$
 * $$\displaystyle \begin{align}
 * $$\displaystyle \begin{align}
 * }
 * }

Therefore, in this problem,


 * {| style="width:100%" border="0" align="left"


 * $$\omega =\sqrt{\pi }$$
 * }
 * }
 * }

From [[media:Egm4313.s12.team9.6-5.jpg|lecture slide 6-5]] and from Kreyszig page 57, a general homogeneous solution for a Case III equation is
 * {| style="width:100%" border="0" align="left"


 * $${{y}_{h}}={{e}^{-ax/2}}(A\cos \omega x+B\sin \omega x) \! $$
 * }
 * }
 * }

where A and B are arbitrary constant coefficients. In this example, the general solution is
 * {| style="width:100%" border="0" align="left"

$$\displaystyle y={{e}^{-0.27x}}\left( A\cos (\sqrt{\pi }x)+B\sin (\sqrt{\pi }x) \right) \! $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right;">$$\displaystyle $$
 * }

To check by substitution, the first and second derivatives of the general solution are taken and substituted into Equation (2).
 * {| style="width:100%" border="0" align="left"


 * $${y}'=-0.27{{e}^{-0.27x}}\left( A\cos (\sqrt{\pi }x)+B\sin (\sqrt{\pi }x) \right)+{{e}^{-0.27x}}\left( -\sqrt{\pi }A\sin (\sqrt{\pi }x)+\sqrt{\pi }B\cos (\sqrt{\pi }x) \right)$$
 * }
 * }
 * }


 * {| style="width:100%" border="0" align="left"


 * $${y}''=0.25{{e}^{-0.5x}}\left( A\cos (\sqrt{3})x+B\sin (\sqrt{3})x \right)-0.5{{e}^{-0.5x}}\left( -A(\sqrt{3})\sin (\sqrt{3})x+B(\sqrt{3})\cos (\sqrt{3})x \right)-0.5{{e}^{-0.5x}}\left( -A(\sqrt{3})\sin (\sqrt{3})x+B(\sqrt{3})\cos (\sqrt{3})x \right)+{{e}^{-0.5x}}\left( -3A\cos (\sqrt{3})x-3B\sin (\sqrt{3})x \right)$$
 * }
 * }
 * }


 * {| style="width:100%" border="0" align="left"


 * $${y}''=0.25{{e}^{-0.5x}}\left( A\cos (\sqrt{3})x+B\sin (\sqrt{3})x \right)-{{e}^{-0.5x}}\left( -A(\sqrt{3})\sin (\sqrt{3})x+B(\sqrt{3})\cos (\sqrt{3})x \right)+{{e}^{-0.5x}}\left( -3A\cos (\sqrt{3})x-3B\sin (\sqrt{3})x \right)$$
 * }
 * }
 * }

Substituting into Equation (2),
 * {| style="width:100%" border="0" align="left"


 * $$0.0729{{e}^{-0.27x}}\left( A\cos (\sqrt{\pi }x)+B\sin (\sqrt{\pi }x) \right)-0.54{{e}^{-0.27x}}\left( -\sqrt{\pi }A\sin (\sqrt{\pi }x)+\sqrt{\pi }B\cos (\sqrt{\pi }x) \right)+{{e}^{-0.27x}}\left( -\pi A\cos (\sqrt{\pi }x)-\pi B\sin (\sqrt{\pi }x) \right)+0.54\left[ -0.27{{e}^{-0.27x}}\left( A\cos (\sqrt{\pi }x)+B\sin (\sqrt{\pi }x) \right)+{{e}^{-0.27x}}\left( -\sqrt{\pi }A\sin (\sqrt{\pi }x)+\sqrt{\pi }B\cos (\sqrt{\pi }x) \right) \right]+\left( 0.0729+\pi \right)\left[ {{e}^{-0.27x}}\left( A\cos (\sqrt{\pi }x)+B\sin (\sqrt{\pi }x) \right) \right]=0$$
 * }
 * }
 * }

Multiplying and rearranging terms,
 * {| style="width:100%" border="0" align="left"


 * $$0.0729{{e}^{-0.27x}}\left( A\cos (\sqrt{\pi }x)+B\sin (\sqrt{\pi }x) \right)+0.0729{{e}^{-0.27x}}\left( A\cos (\sqrt{\pi }x)+B\sin (\sqrt{\pi }x) \right)-0.1458{{e}^{-0.27x}}\left( A\cos (\sqrt{\pi }x)+B\sin (\sqrt{\pi }x) \right)+\pi {{e}^{-0.27x}}\left( A\cos (\sqrt{\pi }x)+B\sin (\sqrt{\pi }x) \right)-\pi {{e}^{-0.27x}}\left( A\cos (\sqrt{\pi }x)+B\sin (\sqrt{\pi }x) \right)-0.54{{e}^{-0.27x}}\left( -\sqrt{\pi }A\sin (\sqrt{\pi }x)+\sqrt{\pi }B\cos (\sqrt{\pi }x) \right)+0.54{{e}^{-0.27x}}\left( -\sqrt{\pi }A\sin (\sqrt{\pi }x)+\sqrt{\pi }B\cos (\sqrt{\pi }x) \right)=0$$
 * }
 * }
 * }


 * {| style="width:100%" border="0" align="left"


 * $$0=0 \!$$
 * }
 * }
 * }

Since all terms cancel, the solution is
 * {| style="width:100%" border="0" align="left"


 * $$y={{e}^{-0.27x}}\left( A\cos (\sqrt{\pi }x)+B\sin (\sqrt{\pi }x) \right)$$
 * }
 * }
 * }

Author
Solved and typed by --Egm4313.s12.team9.wright 08:18, 8 February 2012 (UTC)

Reviewed by Egm4313.s10.team9.delp 16:16, 8 February 2012 (UTC)

Edited by

= R2.9 =

Given:
$$\lambda^2 + 4\lambda +13 = 0 \! $$ Characteristic equation from the ODE: $$y''+4y'+13y=r(x) \! $$ With the following initial conditions: $$ y(0)=1,\, y'(0)=0,\,  r(x)=0 \! $$

Solution I:
First, we solve the Characteristic Equation for $$\lambda$$ using the Quadratic Formula: $$\lambda = \frac{-b \pm \sqrt{b^2-4(a)(c)}}{2(a)} \! $$ $$\lambda = \frac{-4 \pm \sqrt{4^2-4(1)(13)}}{2(1)} \! $$ $$\lambda = \frac{-4 \pm 6i}{2} \! $$ $$\lambda = {-2 \pm 3i} \! $$ Where $$ a=-2 \,\! $$ and $$ \,\omega = 3 \! $$ The solution form for a Linear Second Order Differential Equation with Constant Coefficients is: $$ y(x) = e^{-ax}(C_1\cos({\omega x}) + C_2\sin({\omega x})) \! $$ Now we take the derivative of $$ y(x) $$ $$ y(x) = e^{-2x}(C_1\cos({3x}) + C_2\sin({3x})) \! $$ $$ y(x)' = e^{-2x}(-3C_1\sin({3x}) + 3C_2\cos({3x})) - 2e^{-2x}(C_1\cos({3x}) + C_2\sin({3x})) \! $$ Now, we use the initial conditions listed above to determine the coefficients $$C_1$$ and $$C_2$$. $$y(0)= 1 = e^{-2(0)}(C_1\cos({3(0)}) + C_2\sin({3(0)})) \! $$ $$ y(0)' = 0 = e^{-2(0)}(-3C_1\sin({3(0)}) + 3C_2\cos({3(0)})) - 2e^{-2(0)}(C_1\cos({3(0)}) + C_2\sin({3(0)})) \! $$ $$C_1=1 \! $$ $$C_2=\frac{2}{3} \! $$ Therefore, the final solution will be,

Solution II:
We are then asked to create a superposed plot with solutions from R2.1 and R2.6 using the following equations:

Author
Solved and typed by --Egm4313.s12.team9.murray 09:47, 8 February 2012 (UTC)

Reviewed by E.Beuses 17:15, 8 February 2012 (UTC)

Edited by

= References =

=Contributing Team Members=

--Egm4313.s12.team9.wright 09:28, 8 February 2012 (UTC)

--Egm4313.s10.team9.delp 16:20, 8 February 2012 (UTC)

--Egm4313.s12.team9.murray 18:53, 8 February 2012 (UTC)

--Egm4313.s12.team9.ehlers.km 15:14, 8 February 2012 (UTC)

--Egm4313.s12.team9.beuses 17:23, 8 February 2012 (UTC)

--Egm4313.s12.team9.tjl 18:23, 8 February 2012 (UTC)

--Egm4313.s12.team9.lebo 18:53, 8 February 2012 (UTC)

Table code referenced from Team 4- R1