University of Florida/Egm4313/s12.team5.R2

R2.1

Question

Given the two roots and the initial conditions:

${\lambda }_1=-2,{\lambda }_2=+5$

(1.0)

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

(1.1)

Part 1.

Find the non-homogeneous L2-ODE-CC in standard form. Find the solution in terms of the initial conditions and the general excitation $r(x).$
Now with no excitation and plot the solution:

$r(x)=0$

(1.2)

Part 2.

Generate 3 non-standard (and non-homogeneous) L2-ODE-CC that admit the two values ${\lambda }_1=-2,{\lambda }_2=+5$ as the two roots of the corresponding characteristic equation.

Solution

Part 1.

Characteristic Equation:

$(\lambda -{\lambda }_1)(\lambda -{\lambda }_2)=0$

(1.3)

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

(1.4)

${\lambda }^2-3\lambda -10=0$

(1.5)

Non-Homogeneous L2-ODE-CC:

$y^{″}-3y^{\prime }-10y=r(x)\to standardform$

(1.6)

Homogenous Solution:

$y_h(x)=c_1{e}^{-2x}+c_2{e}^{5x}$

(1.7)

Overall Solution:

$y(x)=c_1{e}^{-2x}+c_2{e}^{5x}+y_p(x)$

(1.8)

$y^{\prime }(x)=-2c_1{e}^{-2x}+5c_2{e}^{5x}+y{\text{'}}_p(x)$

(1.9)

Satisfy Initial Conditions:

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

(1.10)

$y^{\prime }(0)=0=-2c_1+5c_2+y{\text{'}}_p(0)$

(1.11)

No excitation:

$r(x)=0\to y_p(x)=0\to y{\text{'}}_p(x)=0$

(1.12)

From (1.10):

$c_1+c_2=1\to c_1=1-c_2$

(1.13)

From (1.11):

$-2c_1+5c_2=0$

(1.14)

$5c_2=2c_1$

(1.15)

$\frac{5}{2}{c}_2=c_1$

(1.16)

Plug (1.13) into (1.16):

$\frac{5}{2}{c}_2=1-c_2$

(1.17)

Solve for $c_2$:

$c_2=\frac{2}{7}$

(1.18)

Plug (1.18) into (1.13) and solve for $c_1$:

$c_1=\frac{5}{7}$

(1.19)

Therefore, the final solution in terms of the initial conditions and the general excitation $r(x)$ is:

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

(1.20)

Plot of Solution:

Figure 1File:IEA 2.1 plot.png

Part 2.

Three non-standard and non-homogeneous solutions using the same roots given above:

1.

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

(1.21)

$2{\lambda }^2-6\lambda -20=0$

(1.22)

2.

$3(\lambda -(-2))(\lambda -5)=0$

(1.23)

$3{\lambda }^2-9\lambda -30=0$

(1.24)

3.

$4(\lambda -(-2))(\lambda -5)=0$

(1.25)

$4{\lambda }^2-12\lambda -40=0$

(1.26)

Author

This problem was solved and uploaded by [Derik Bell]

This problem was proofread by: David Herrick

R2.2

Question

From Section 5 in the notes, pg. 5-6

Solve the L2-ODE-CC from pg. 5-5, equation (4)

$y^{″}-10y^{\prime }+25y=r(x)$

For initial conditions:

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

Where there is no excitation, i.e. $r(x)=0$

Solution

${\lambda }^2-10\lambda +25=0$

Factoring:

$(\lambda -5{)}^2=0$

Thus, there is a double root, where $\lambda =5$

Because there is no excitation, the homogenous solution will be the same as the final solution. Thus, the solution will be:

$y_h(x)=y(x)=C_1{e}^{5x}+C_{2}x{e}^{5x}$

Plugging in the initial condition,

$y(0)=1=C_1{e}^0+C_2(0)e^0$

$C_1=1$

In order to solve for the second constant, we need to take a derivative.

$y{\text{'}}_h(x)=y^{\prime }(x)=5C_1{e}^{5x}+C_2{e}^{5x}+5C_{2}x{e}^{5x}$

Plugging in C1 and the other initial condition:

$y{\text{'}}_h(0)=y^{\prime }(0)=0=5(1)e^0+C_2{e}^0+5C_2(0)e^0$

$5+C_2=0\Rightarrow C_2=-5$

So, the final solution is:

$y(x)=e^{5x}-5x{e}^{5x}$

Matlab Code:

x = 0:0.001:1;

f = exp(5*x) - 5*x.*exp(5*x);

plot(x,f)

xlabel ('x')

ylabel ('y(x)')

File:Report2Matlab'.jpg

Author

This problem was solved and uploaded by [David Herrick]
This problem was checked by [William Knapper]

R2.3

Question

Complete problems 3 and 4 from p.59 of K 2011.

Solution

The two problems have the same instructions: "Find a general solution. Check your answer by substitution."

Problem 3

$3.y^{″}+6y^{\prime }+8.96y=0$

We start by using the characteristic equation of this ODE in order to find the roots. We use:

${\lambda }^2+a\lambda +b=0$

where a = 6 and b = 8.96. Taking the discriminant, we find that

$6^2-4(8.96)=0.16>0$

A positive discriminant implies that there exists two real roots, with a general solution of the form:

$y=c_1{e}^{{\lambda }_{1}x}+c_2{e}^{{\lambda }_{2}x}$

To find the unknowns, we find the roots of the above equation:

${\lambda }^2+6\lambda +8.96=0$

${\lambda }_1=-3.2,{\lambda }_2=-2.8$

Therefore, the correct general solution is:

$y=c_1{e}^{-3.2x}+c_2{e}^{-2.8x}$ 

To check that this solution is correct, we take the first and second derivatives:

$y\text{'}=-3.2c_1{e}^{-3.2x}-2.8c_2{e}^{-2.8x}$

$y{\text{'}}^{\prime }=10.24c_1{e}^{-3.2x}+7.84c_2{e}^{-2.8x}$

Now, we plug these values into the original equation:

$10.24c_1{e}^{-3.2x}+7.84c_2{e}^{-2.8x}-19.2c_1{e}^{-3.2x}-16.8c_2{e}^{-2.8x}+8.96c_1{e}^{-3.2x}+8.96c_2{e}^{-2.8x}=0$

Upon further inspection, all the terms on the left side of the equation cancel out and equal zero.

Problem 4

$4.y^{″}+4y^{\prime }+({\pi }^2+4)y=0$

We start by using the characteristic equation of this ODE in order to find the roots. We use:

${\lambda }^2+a\lambda +b=0$

where a = 4 and b = (π^2 + 4). Taking the discriminant, we find that

$4^2-4({\pi }^2+4)=-4{\pi }^2<0$

A negative discriminant implies that there exists complex conjugate roots to this equation. The general solution for this type of equation is

$y=e^{-2x}(Acos(\omega x)+Bsin(\omega x))$

Where

$\omega =\sqrt{b-\frac{1}{4}{a}^2}=\sqrt{({\pi }^2+4)-(\frac{1}{4})4^2}=\pi$

This yields

$y=e^{-2x}(Acos(\pi x)+Bsin(\pi x))$ 

To verify that this is the correct solution, we first find the first and second derivatives of the solution:

$y=e^{-2x}(Acos(\pi x)+Bsin(\pi x))$
$y\text{'}=e^{-2x}(-A\pi sin(\pi x)+B\pi cos(\pi x))-2e^{-2x}(Acos(\pi x)+Bsin(\pi x))$
$y{\text{'}}^{\prime }=e^{-2x}[-A{\pi }^{2}cos(\pi x)-B{\pi }^{2}sin(\pi x)+2A\pi sin(\pi x)-2B\pi sin(\pi x)]-2e^{-2x}[-A\pi sin(\pi x)+B\pi cos(\pi x)-2Acos(\pi x)-2Bsin(\pi x)]$

We then plug these values into the original problem to get this somewhat lengthy equation:

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

Upon further inspection, all the terms on the left side of the equation cancel out to 0.

Author

This problem was solved and uploaded by: (Will Knapper)

This problem was proofread by: David Herrick

R2.4

Question

K 2011 p.59 pbs.5,6. Find a general solution. Check your answer by substitution.

Problem 5

$y^{″}+2\pi y^{\prime }+{\pi }^{2}y=0$

Solution

Determine the characteristic equation and the nature of the determinate:

$y=e^{-\lambda x}$

$y^{\prime }=-\lambda e^{-\lambda x}$

$y^{″}={\lambda }^2{e}^{-\lambda x}$

Characteristic Equation:

${\lambda }^2-2\pi \lambda +{\pi }^2\lambda =0$

$a^2-4b=(2\pi {)}^2-4({\pi }^2)=4{\pi }^2-4{\pi }^2=0$

0 indicates a real double root, therefore:

$y_g=e^{-\frac{a}{2}x}(C_1+C_{2}x)$

$a=2\pi$ from the original equation. Therefore:

$y_g=e^{-\pi x}(C_1+C_{2}x)$

$y{\text{'}}_g=e^{-\pi x}(C_2-\pi C_1-\pi C_{2}x)$

$y^{\prime }{\text{'}}_g=e^{-\pi x}(C_1{\pi }^2+C_2{\pi }^{2}x-2C_2\pi )$

Substituting into the original equation gives:

$e^{-\pi x}[(C_1{\pi }^2+C_2{\pi }^{2}x-2C_2\pi )+2\pi (C_2-\pi C_1-\pi C_{2}x)+{\pi }^2(C_1+C_{2}x)]=0$

$=e^{-\pi x}[C_1{\pi }^2+C_2{\pi }^{2}x-2C_2\pi +2C_2\pi -2C_1{\pi }^2-2C_2{\pi }^{2}x+C_1{\pi }^2+C_2{\pi }^{2}x]$

Canceling gives 0 = 0, confirming the general solution found above.

Author

This problem was solved and uploaded by: (John North)

This problem was proofread by: David Herrick

Question 2

Problem 5

$10y^{″}-32y^{\prime }+25.6y=0$

Solution

Reduce original equation so that the first coefficient is 1:

$y^{″}-3.2y^{\prime }+2.56y=0$

Determine the characteristic equation and the nature of the determinate:

$y=e^{-\lambda x}$

$y^{\prime }=-\lambda e^{-\lambda x}$

$y^{″}={\lambda }^2{e}^{-\lambda x}$

Characteristic Equation:

${\lambda }^2-3.2\lambda +2.56=0$

$a^2-4b=3.2^2-4(2.56)=10.24-10.24=0$

0 indicates a real double root, therefore:

$y_g=e^{-\frac{a}{2}x}(C_1+C_{2}x)$

$a=\frac{-(-3.2)}{2}=1.6$ from the original equation. Therefore:

$y_g=e^{1.6x}(C_1+C_{2}x)$

$y{\text{'}}_g=e^{1.6x}(1.6C_1+1.6C_{2}x+C_2)$

$y^{\prime }{\text{'}}_g=e^{1.6x}(2.56C_1+2.56C_{2}x+3.2C_2)$

Substituting into the original equation gives:

$e^{1.6x}[(2.56C_1+2.56C_{2}x+3.2C_2)-3.2(1.6C_1+1.6C_{2}x+C_2)+2.56(C_1+C_{2}x)]=0$

$=e^{1.6x}[(2.56C_1+2.56C_{2}x+3.2C_2-5.12C_1-5.12C_{2}x-3.2C_2+2.56C_1+2.56C_{2}x]=0$

Canceling gives 0 = 0, confirming that general solution found above.

Author

This problem was solved and uploaded by: (John North)

This problem was proofread by: David Herrick

R2.5

Question

Problems 16 and 17 from pg. 59 of the textbook. The two questions have the same instructions: Find an ODE of the form
$y{\text{'}}^{\prime }+a{y}^{\prime }+by=0$
for the given basis.

Solution

Problem 16

$16.e^{2.6x},e^{-4.3x}$
This basis implies that there are two distinct, real roots: 2.6 and -4.3. We put these roots into the characteristic equation to get: $(\lambda -2.6)(\lambda +4.3)=0\to {\lambda }^2+1.7\lambda -11.18=0$
Then, you convert the characteristic equation to a regular ODE:

$y{\text{'}}^{\prime }+1.7y^{\prime }-11.18y=0$ 

Problem 17

$17.e^{-\sqrt{5}x},x{e}^{-\sqrt{5}x}$
A basis of this form implies that the solution is of the form:
$y=(c_1+c_{2}x)e^{-ax/2}$
The real double root can be found by this equation:
$\lambda =-a/2=-\sqrt{5}$
That means that the characteristic equation can be found by solving:
$(\lambda +\sqrt{5}{)}^2={\lambda }^2+2\sqrt{5}\lambda +5$
Converting the characteristic equation to the ODE, we get:

$y{\text{'}}^{\prime }+2\sqrt{5}{y}^{\prime }+5y=0$ 

Author

This problem was solved and uploaded by: (Will Knapper)
This problem was proofread by (Josh House)

R2.6

Question

Realize spring-dashpot-mass system in series as shown in Fig. p.1-4 with the similar characteristic as in (3)p.5-5, but with the double real root $\lambda =-3$, i.e., find values for parameters k,c,m.

Solution

First we put the double root into the characteristic equation:
$(\lambda +3{)}^2=0$
${\lambda }^2+6\lambda +9=0$

The corresponding L2-ODE is:
$y{\text{'}}^{\prime }+6y^{\prime }+9y=0$

The equation of motion of the spring-dashpot-mass system is:
$m({y_k}^{″}+\frac{k}{c}{y_k}^{\prime })+k{y}_k=0$
or:
$m{y_k}^{″}+m\frac{k}{c}{y_k}^{\prime }+k{y}_k=0$

Solving for the parameters k,c,m:
$m=1$

$m\frac{k}{c}=6$

$k=9$

We get:

$m=1$ 


$k=9$ 


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

Author

This problem was solved and uploaded by: (Radina Dikova)

This problem was proofread by: Mike Wallace

R2.7

Question

Develop the McLauren series (Taylor series at t=0) for e^t, cos(t), and sin(t).

Solution

The general form of the Taylor series expansion at point a is

$f(a)+\frac{f^{\prime }(a)}{1!}(x-a)+\frac{f^{″}(a)}{2!}(x-a{)}^2+\frac{f^{‴}(a)}{3!}(x-a{)}^3+...$

for $e^t$ at t=0:

$e^0+\frac{e^0}{1!}(x)+\frac{e^0}{2!}(x{)}^2+\frac{e^0}{3!}(x{)}^3+...$

$=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...$

for $\cos (t)$ at t=0:

$\cos (0)+\frac{-\sin (0)}{1!}(x)+\frac{-\cos (0)}{2!}(x{)}^2+\frac{sin(0)}{3!}(x{)}^3+\frac{cos(0)}{4!}(x{)}^4+...$

$=1-\frac{x^2}{2!}+\frac{x^4}{4!}+...$

for $\sin (t)$ at t=0:

$\sin (0)+\frac{cos(0)}{1!}(x)+\frac{-\sin (0)}{2!}(x{)}^2+\frac{-cos(0)}{3!}(x{)}^3+\frac{\sin (x)}{4!}(x{)}^4+\frac{\cos (0)}{5!}(x{)}^5+...$

$=x-\frac{x^3}{3!}+\frac{x^5}{5!}+...$

Author

This problem was solved and uploaded by: (John North)
This problem was checked by: William Knapper

R2.8

Question

Find a general solution. Check your answer by substitution.

8. $y^{{\text{'}}^{\prime }}+y^{\text{'}}+3.25y=0$

15. $y^{{\text{'}}^{\prime }}+0.54y^{\text{'}}+(0.0729+\pi )y=0$

Solution

8. $y^{{\text{'}}^{\prime }}+y^{\text{'}}+3.25y=0$

Writing the characteristic equation:

${\lambda }^2+\lambda +3.25=0$

Which is now in the form:

${\lambda }^2+a\lambda +b=0$

Solving for the two roots:

${\lambda }_1=\frac{1}{2}(-a+\sqrt{a^2-4b}),{\lambda }_2=\frac{1}{2}(-a-\sqrt{a^2-4b})$

We will find the discriminant to be less than 0, leading to complex conjugate roots:

$a^2-4b=1^2-4(3.25)=-12$

Which leads to a solution of the form:

$y=e^{-ax/2}(Acos(\omega x)+Bsin(\omega x))$

Where ${\omega }^2=b-\frac{1}{4}{a}^2$, therefore:

${\omega }^2=3.25-\frac{1}{4}(1{)}^2=3\to \omega =\sqrt{3}$

Giving us a solution of:

$y=e^{-x/2}(Acos(\sqrt{3}x)+Bsin(\sqrt{3}x))$

Checking our solution:

$y=e^{-x/2}(Acos(\sqrt{3}x)+Bsin(\sqrt{3}x))$

$y^{\text{'}}=-\frac{1}{2}{e}^{-x/2}(Acos(\sqrt{3}x)+Bsin(\sqrt{3}x))+e^{-x/2}(-\sqrt{3}sin(\sqrt{3}x)+\sqrt{3}Bcos(\sqrt{3}x))$

$y^{{\text{'}}^{\prime }}=-\frac{1}{4}{e}^{-x/2}(Acos(\sqrt{3}x)+Bsin(\sqrt{3}x))-\frac{1}{2}{e}^{-x/2}(-\sqrt{3}Asin(\sqrt{3}x)+\sqrt{3}Bcos(\sqrt{3}x))-\frac{1}{2}{e}^{-x/2}(-\sqrt{3}Asin(\sqrt{3}x)+\sqrt{3}Bcos(\sqrt{3}x))+e^{-x/2}(-3Acos(\sqrt{3}x)-3Bsin(\sqrt{3}x))$

Plugging all of this into the original differential equation and combining like terms gives us:

$[\frac{1}{4}-3-\frac{1}{2}+3.25]e^{-x/2}Acos(\sqrt{3}x)$

$[\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}-\sqrt{3}]e^{-x/2}Asin(\sqrt{3}x)$

$[-\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2}+\sqrt{3}]e^{-x/2}Bcos(\sqrt{3}x)$

$[\frac{1}{4}-3-\frac{1}{2}+3.25]e^{-x/2}Bsin(\sqrt{3}x)$

The terms in the brackets all add up to zero, thus verifying we have a correct solution to our problem.

15. $y^{{\text{'}}^{\prime }}+0.54y\text{'}+(0.0729+\pi )y=0$

Writing the characteristic equation:

${\lambda }^2+0.54\lambda +(0.0729+\pi )=0$

Which is now in the form:

${\lambda }^2+a\lambda +b=0$

Solving for the two roots:

${\lambda }_1=\frac{1}{2}(-a+\sqrt{a^2-4b}),{\lambda }_2=\frac{1}{2}(-a-\sqrt{a^2-4b})$

We will find the discriminant to be less than 0, leading to complex conjugate roots:

$a^2-4b=0.54^2-4(0.0729+\pi )=-4\pi$

Which leads to a solution of the form:

$y=e^{-ax/2}(Acos(\omega x)+Bsin(\omega x))$

Where ${\omega }^2=b-\frac{1}{4}{a}^2$, therefore:

${\omega }^2=(0.0729-\pi )-\frac{1}{4}(0.54{)}^2=\pi \to \omega =\sqrt{\pi }$

Giving us a solution of:

$y=e^{-0.27x}(Acos(\sqrt{\pi }x)+Bsin(\sqrt{\pi }x))$

Checking our solution:

$y=e^{-0.27x}(Acos(\sqrt{\pi }x)+Bsin(\sqrt{\pi }x))$

$y^{\text{'}}=-0.27e^{-0.27x}(Acos(\sqrt{\pi }x)+Bsin(\sqrt{\pi }x))+e^{-0.27x}(-\sqrt{\pi }Asin(\sqrt{\pi }x)+\sqrt{\pi }Bcos(\sqrt{\pi }x))$

$y^{{\text{'}}^{\prime }}=0.0729e^{-0.27x}(Acos(\sqrt{\pi }x)+Bsin(\sqrt{\pi }x))-0.27e^{-0.27x}(-\sqrt{\pi }Asin(\sqrt{\pi }x)+\sqrt{\pi }Bcos(\sqrt{\pi }x))-0.27e^{-0.27x}(-\sqrt{\pi }Asin(\sqrt{\pi }x)+\sqrt{\pi }Bcos(\sqrt{\pi }x))+e^{-0.27x}(-\pi Acos(\sqrt{\pi }x)-\pi Bsin(\sqrt{\pi }x))$

Plugging all of this into the original differential equation and combining like terms gives us:

$[0.0729-\pi -0.1458+0.0729+\pi ]e^{-0.27x}Acos\sqrt{\pi }x$

$[0.27\sqrt{\pi }+0.27\sqrt{\pi }-0.54\sqrt{\pi }]e^{-0.27x}Asin\sqrt{\pi }x$

$[-0.27\sqrt{\pi }-0.27\sqrt{\pi }+0.54\sqrt{\pi }]e^{-0.27x}Bcos\sqrt{\pi }x$

$[0.0729-\pi -0.1458+0.0729+\pi ]e^{-0.27x}Bsin\sqrt{\pi }x$

The terms in the brackets all add up to zero, thus verifying we have a correct solution to our problem.

Author

This problem was solved and uploaded by: (Josh House)
This problem was proofread by: (Radina Dikova)

R2.9

Question

For the given ODE, find and plot the solution for the L2-ODE-CC corresponding to

${\lambda }^2+4\lambda +13=0$

In another figure superpose this figure, Fig. from R2.6 p.5-6, and the Fig. from R2-1 p.3-7.

This corresponds to the L2-ODE-CC in standard form

$y^{″}+4y^{\prime }+13y=r(x)$

Given

Initial conditions

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

Where there is no excitation

$r(x)=0$

Solution

Solving the quadratic equation for $\lambda$ determines the roots of the ODE

${\lambda }_1=\frac{-b+\sqrt{b^2-(4ac)}}{2a}=\frac{-4+\sqrt{(4{)}^2-4(1)(13)}}{2(1)}$

${\lambda }_2=\frac{-b-\sqrt{b^2-(4ac)}}{2a}=\frac{-4-\sqrt{(4{)}^2-4(1)(13)}}{2(1)}$

The roots of the quadratic equation come out to be:

$\lambda =(-2\pm 3i)$

The general homogeneous solution of the ODE will resemble the following form.

$y=C_1{e}^{a_{1}x}cos{b}_{1}x+C_2{e}^{a_{2}x}sin{b}_{2}x$

Plugging in the roots into the equation we get

$y=C_1{e}^{-2x}cos3x+C_2{e}^{-2x}sin3x$

Implementing initial conditions where: $y(0)=1,y^{\prime }(0)=0$

$y(0)=C_1{e}^{-2(0)}cos3(0)+C_2{e}^{-2(0)}sin3(0)=1$

$y^{\prime }(0)=-2C_1{e}^{-2x}cos3x-3C_1{e}^{-2x}sin3x-2C_2{e}^{-2x}sin3x+3C_2{e}^{-2x}cos3x=0$

$y^{\prime }(0)=-2(1)e^{-2(0)}cos3(0)+3C_2{e}^{-2(0)}cos3(0)=-2+3C_2=0;$

So from our initial conditions the constants are determined to be

$C_1=1,C_2=2/3$

The final solution becomes

$y(x)=e^{-2x}cos3x+\frac{2}{3}{e}^{-2x}sin3x$

Figure 1 for R2.9 Superposed Figs. for R2.9

Author

This problem was solved and uploaded by: Mike Wallace
This problem was proofread by John North

Contribution Summary

Problem 2 was solved and Problems 1, 3, and 4 were proofread by David Herrick 15:56, 6 February 2012 (UTC)

Problem 8 was solved and Problem 5 was proofread by Josh House 23:41, 7 February 2012 (UTC)

Problem 6 was solved and Problem 8 was proofread by Radina Dikova 16:26, 7 February 2012 (UTC)

Problem 9 was solved and Problem 6 was proofread by Mike Wallace 23:35, 7 February 2012 (UTC)

Problems 3 and 5 were solved and Problems 2 and 7 were proofread by William Knapper 05:04, 8 February 2012 (UTC)

Problem 1 was solved by Derik Bell 15:44, 8 February 2012 (UTC)

Problems 4 and 7 were solved and Problem 9 was proofread by John North 15:47, 8 February 2012 (UTC)