University of Florida/Egm4313/s12.teamboss/R3

Report 3

Problem R3.1: Using the Method of Undetermined Coefficients and the Modification Rule to find a solution

Statement

Given the double root $\lambda =5\!$ and the excitation $r(x)=7e^{5x}-2x^{2}\!$, with the initial conditions $y(0)=4,y'(0)=5\!$ find the solution $y(x)\!$.
Plot this solution and the solution to the same problem except with the excitation $r(x)=7e^{5x}\!$.

Solution

The solution $y(x)\!$ is composed of a general solution and a particular solution so that $y(x)=y_{g}(x)+y_{p}(x)\!$.
Using the given double root, we can find the equation for the characteristic equation which leads to the homogeneous solution:

$\displaystyle {(\lambda -5)^{2}=\lambda ^{2}-10\lambda +25=0}$

(1.0)

Homogeneous solution:

$\displaystyle {y''-10y'+25y=r(x)}$

(1.1)

$\displaystyle {y''-10y'+25y=7e^{5x}-2x^{2}}$

(1.2)

First, by using the Modification Rule, we find that the general equation associated with the given double root is:

$\displaystyle {y_{g}(x)=C_{1}e^{5x}+C_{2}xe^{5x}}$

(1.3)

We need to find the particular solution to the excitation $r(x)=7e^{5x}-2x^{2}\!$. In analyzing the excitation, it is found that the particular solution looks like this:

$\displaystyle {y_{p}(x)=Cx^{2}e^{5x}-[K_{2}x^{2}+K_{1}x+K_{0}]}$

(1.4)

Now, we need to find the values for the constants $C,K_{2},K_{1},K_{0}\!$, by taking the first and second derivatives of the particular solution:

$\displaystyle {y_{p}(x)=Cx^{2}e^{5x}-[K_{2}x^{2}+K_{1}x+K_{0}]}$

(1.4)

$\displaystyle {y_{p}(x)'=2Cxe^{5x}+5Cx^{2}e^{5x}-[2K_{2}x+K_{1}]}$

(1.5)

$\displaystyle {y_{p}(x)''=2Ce^{5x}+20Cxe^{5x}+25Cx^{2}e^{5x}-[2K_{2}]}$

(1.6)

Now, plug these into the homogeneous solution (1.2) and simplify:

$\displaystyle {2Ce^{5x}-2K_{2}+20K_{2}x+10K_{1}-25K_{2}x^{2}-25K_{1}x-25K_{0}=7e^{5x}-2x^{2}}$

(1.7)

Comparing the coefficients for $e^{5x},x^{2}\!$ and $x\!$ allows us to solve for the unknown coefficients:

$\displaystyle {e^{5x}:2C=7\rightarrow C=7/2}$

(1.8)

$\displaystyle {x^{2}:-25K_{2}=-2\rightarrow K_{2}=0.08}$

(1.9)

$\displaystyle {x:20K_{2}-25K_{1}=0\rightarrow K_{1}=0.064}$

(1.10)

$\displaystyle {-2K_{2}+10K_{1}-25K_{0}=0\rightarrow K_{0}=0.0192}$

(1.11)

Plug into $y_{p}(x)\!$ (1.4):

$\displaystyle {y_{p}(x)={\frac {7}{2}}x^{2}e^{5x}-[0.08x^{2}+0.064x+0.0192]}$

(1.12)

Now that we have the particular solution, we can use the initial conditions to solve for the unknown constants in the general solution, $y_{g}(x)\!$, and then we will be able to solve for the final solution $y(x)\!$:

$\displaystyle {y(x)=C_{1}e^{5x}+C_{2}xe^{5x}+{\frac {7}{2}}x^{2}e^{5x}-[0.08x^{2}+0.064x+0.0192]}$

(1.13)

Take derivative of (1.13):

$\displaystyle {y(x)'=5C_{1}e^{5x}+C_{2}e^{5x}+5C_{2}xe^{5x}+7xe^{5x}+{\frac {35}{2}}x^{2}e^{5x}-[0.16x+0.064]}$

(1.14)

Using the given initial conditions and equations (1.13) and (1.14):

$\displaystyle {y(0)=4\rightarrow 4=C_{1}-0.0192\rightarrow C_{1}=4.0192}$

(1.15)

$\displaystyle {y'(0)=5\rightarrow 5=5C_{1}+C_{2}-0.064\rightarrow C_{2}=-15.032}$

(1.16)

Plug these constants back into the solution (1.13) to obtain the final solution:

$\displaystyle y(x)=4.0192e^{5x}-15.032xe^{5x}+{\frac {7}{2}}x^{2}e^{5x}-[0.08x^{2}+0.064x+0.0192]$

(1.17)

The solution to the same problem but with excitation $r(x)=7e^{5x}\!$ is:

$\displaystyle {y(x)=4e^{5x}-25xe^{5x}+{\frac {7}{2}}x^{2}e^{5x}}$

(1.18)

This is the plot for (1.17) and (1.18):

Author

Solved and Typed By - --Egm4313.s12.team1.wyattling 22:00, 20 February 2012 (UTC)
Reviewed By - Egm4313.s12.team1.armanious 22:39, 21 February 2012 (UTC)

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Problem R3.2 Perturbation Method for Double Real Root

Statement

Perturbation method for double real root:
Developing the 2nd homogeneous solution for the case of double real root as a limiting case of distinct roots (see Sec7 p. 7-5 ). Consider two distinct real roots of the form:

$\lambda _{1}=\lambda ,\lambda _{2}=\lambda +\epsilon \!$

1) Find the homogeneous L2-ODE-CC having the above distinct roots.
2) Show that the following is a homogeneous solution:
${\frac {e^{(\lambda +\epsilon )x}-e^{\lambda x}}{\epsilon }}\!$

The fraction in (3) p.7-5, for small $\epsilon \!$, is a finite difference formula that approximates the derivative

${\frac {e^{(\lambda +\epsilon )x}-e^{\lambda x}}{\epsilon }}\approx {\frac {d}{d\lambda }}e^{\lambda x}\!$

In fact,
${\frac {d}{d\lambda }}e^{\lambda x}=\lim _{\epsilon \to 0}{\frac {e^{(\lambda +\epsilon )x}-e^{\lambda x}}{\epsilon }}\!$

3)Find the limit of the homogeneous solution in (3) p.7-5 as [epsilon goes to 0] (think l'Hopital's rule)

4)Take the derivative of $e^{\lambda x}\!$ with respect to $\lambda \!$
5)Compare the results in Parts (3) and (4), and relate to the result by variation of parameters.
6)Numerical experiment: Compute (3) p.7-5 using at $\lambda =5\!$ and with $\epsilon =0.001\!$, and compare to get the value obtained from the exact 2nd homogeneous solution.

Solution

In order to find the corresponding L2-ODE-CC, the characteristic equation corresponding to the given solution must be found:

$\displaystyle (\lambda '-\lambda )(\lambda '-\lambda -\epsilon )=0$

(2.0)

$\displaystyle (\lambda ')^{2}-2\lambda '\lambda +\lambda ^{2}-\lambda '\epsilon +\lambda \epsilon =(\lambda ')^{2}-(2\lambda +\epsilon )\lambda '+(\lambda ^{2}+\lambda \epsilon )=0$

(2.1)

Therefore the corresponding L2-ODE-CC is:

$\displaystyle y''-(2\lambda +\epsilon )y'+(\lambda ^{2}+\lambda \epsilon )y=0$

(2.2)

Note: if ε is equal to zero, the characteristic equation of 2.2 has a double real root at λ.

To show that the following is a homogeneous solution, the first and second derivatives must be taken:

$\displaystyle y(x)={\frac {e^{(\lambda +\epsilon )x}-e^{\lambda x}}{\epsilon }}$

(2.3)

$\displaystyle y'(x)={\frac {(\lambda +\epsilon )e^{(\lambda +\epsilon )x}-\lambda e^{\lambda x}}{\epsilon }}$

(2.4)

$\displaystyle y''(x)={\frac {(\lambda +\epsilon )^{2}e^{(\lambda +\epsilon )x}-\lambda ^{2}e^{\lambda x}}{\epsilon }}$

(2.5)

Using these values in 2.2 yields:

$\displaystyle {\frac {(\lambda +\epsilon )^{2}e^{(\lambda +\epsilon )x}-\lambda ^{2}e^{\lambda x}}{\epsilon }}-{\frac {(2\lambda +\epsilon )[(\lambda +\epsilon )e^{(\lambda +\epsilon )x}-\lambda e^{\lambda x}]}{\epsilon }}+{\frac {(\lambda ^{2}+\lambda \epsilon )[e^{(\lambda +\epsilon )x}-e^{\lambda x}]}{\epsilon }}=0$

(2.6)

Rearranging and simplifying yields:

$\displaystyle {\frac {[\lambda ^{2}+2\lambda \epsilon +\epsilon ^{2}-2\lambda ^{2}-3\lambda \epsilon -\epsilon ^{2}+\lambda ^{2}+\lambda \epsilon ]e^{(\lambda +\epsilon )x}+[-\lambda ^{2}+2\lambda ^{2}+\lambda \epsilon -\lambda ^{2}-\lambda \epsilon ]e^{\lambda x}}{\epsilon }}=0$

(2.7)

All of the terms in the above equation cancel to yield:

$\displaystyle {\frac {0e^{(\lambda +\epsilon )x}+0e^{\lambda x}}{\epsilon }}=0$

(2.8)

Using l'Hopital's rule:

$\displaystyle \lim _{\epsilon \to 0}{\frac {e^{(\lambda +\epsilon )x}-e^{\lambda x}}{\epsilon }}=\lim _{\epsilon \to 0}{\frac {{\frac {d}{d\epsilon }}(e^{(\lambda +\epsilon )x}-e^{\lambda x})}{{\frac {d}{d\epsilon }}(\epsilon )}}=\lim _{\epsilon \to 0}{\frac {{\frac {d}{d\epsilon }}(e^{(\lambda +\epsilon )x})-0}{1}}$

(2.9)

Simplifying further:

$\displaystyle \lim _{\epsilon \to 0}{\frac {d}{d\epsilon }}(e^{(\lambda +\epsilon )x})=\lim _{\epsilon \to 0}xe^{(\lambda +\epsilon )x}=x\lim _{\epsilon \to 0}e^{(\lambda +\epsilon )x}$

(2.10)

This ultimately yields:

$\displaystyle \lim _{\epsilon \to 0}{\frac {e^{(\lambda +\epsilon )x}-e^{\lambda x}}{\epsilon }}=xe^{\lambda x}$

(2.11)

The following should also be taken into consideration:

$\displaystyle {\frac {d}{d\lambda }}(e^{\lambda x})=xe^{\lambda x}$

(2.12)

Clearly, the results of 2.11 and 2.12 are equivalent. This shows that $y(x)=xe^{\lambda x}\!$ is an appropriate solution to a homogeneous L2-ODE-CC having one double root.

To test this, test values will be used to solve for the approximate (2.12) and exact (2.13) solutions of the ODE. For this test, $\lambda =5\!$ and $\epsilon =0.001\!$

$\displaystyle y_{approx}(x)={\frac {e^{(5+0.001)x}-e^{5x}}{0.001}}=1000(e^{0.001x}-1)e^{5x}$

(2.13)

$\displaystyle y_{exact}(x)=xe^{5x}$

(2.14)

If the above derivations are true, then the following approximation must also be true:

$\displaystyle 1000(e^{0.001x}-1)\approx x$

(2.15)

Plotting both sides of 2.15 as a function of x shows that this is a valid approximation for most values of x.

Figure 3.2-1

Author

Solved and Typed By - Egm4313.s12.team1.armanious 05:51, 21 February 2012 (UTC)
Reviewed By - --Egm4313.s12.team1.durrance 02:53, 22 February 2012 (UTC)

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Problem R3.3 Finding Solution of ODE with Polynomial Excitation

Statement

Find the complete solution for $y''-3y'+2y=4x^{2}\!$, with the initial conditions

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

Plot the solution $y(x).\!$

Solution

First we create the characteristic equation in standard form:

$\displaystyle {\lambda ^{2}-3\lambda +2=0}$

(3.0)

Then, by setting it equal to zero, we can find what $\lambda \!$ equals:

$\displaystyle {(\lambda -2)(\lambda -1)=0}$

(3.1)

$\displaystyle {\lambda =2,\lambda =1}$

(3.2)

Given two, distinct, real roots, the general solution looks like this:

$\displaystyle y_{g}(x)=C_{1}e^{2x}+C_{2}e^{x}$

(3.3)

By using the method of undetermined coefficients, the excitation $4x^{2}\!$ is analyzed to yield a particular solution:
In assessing a polynomial with a second power, the form of the particular solution will look like this:

$\displaystyle y_{p}(x)=A_{2}x^{2}+A_{1}x+A_{0}$

(3.5)

It's derivative would look like this:

$\displaystyle y_{p}'(x)=2A_{2}x+A_{1}$

(3.6)

And the second derivative to follow would then become:

$\displaystyle y_{p}''(x)=2A_{2}$

(3.7)

Based on the coefficients, the following system of equations exists:

$\displaystyle 2A_{2}=4$

(3.8)

$\displaystyle -6A_{2}+A_{1}=0$

(3.9)

$\displaystyle 2A_{2}-3A_{1}+A_{0}=0$

(3.10)

The results of this set of equations make the coefficients of A's:

$\displaystyle A_{2}=2$

$\displaystyle A_{1}=12$

$\displaystyle A_{0}=32$

The resulting particular equation looks like this:

$\displaystyle y_{p}(x)=2x^{2}+12x+32$

(3.11)

By adding the particular and general solutions, we get the complete solution:

$\displaystyle 2x^{2}+12x+32+C_{1}e^{2x}+C_{2}e^{x}=y$

(3.12)

We consider the initial conditions by taking the first derivative of the complete solution:

$\displaystyle 4x+12+2C_{1}e^{2x}+C_{2}e^{x}=y'$

(3.13)

By plugging in 0 for x, 1 for y, and 0 for y', we can solve for the constants $C_{1},C_{2}\!$:

$\displaystyle y(0)=1=2*0^{2}+12*0+32+C_{1}e^{2*0}+C_{2}e^{0}=32+C_{1}+C_{2}=1$

(3.14)

$\displaystyle y'(0)=0=4*0+12+2C_{1}e^{2*0}+C_{2}e^{0}=12+2C_{1}+C_{2}=0$

(3.15)

Solving the equations proves that $C_{1}=19,C_{2}=-50\!$:
The resulting complete solution with consideration for initial conditions then becomes:

$\displaystyle 2x^{2}+12x+32+19e^{2x}-50e^{x}=y$

(3.12)

y plotted looks like this:

Author

Solved and Typed By -Egm4313.s12.team1.silvestri 15:56, 19 February 2012 (UTC)
Reviewed By -Egm4313.s12.team1.armanious 03:02, 22 February 2012 (UTC)

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Problem R3.4 Solving for Particular Solution in Summation Form

Statement

From R3.4 in Sec 3 p. 7-11; Use the Basic Rule 1 and the Sum Rule to show that the appropriate particular solution for

$\displaystyle y''-3y'+2y=4x^{2}-6x^{5}$

(4.1)

is of the form

$\displaystyle y_{p}(x)=\sum _{j=0}^{n}c_{j}x_{j}$

(4.2)

i.e.,

$\displaystyle y_{p}(x)=\sum _{j=0}^{5}c_{j}x_{j}$

(4.3)

Basic Rule: Select $y_{p}(x)\!$ from the table and determine the coefficients by substituting $y_{p}(x)\!$ in

$\displaystyle y''+ay'+by=r(x)$

(4.4)

Sum Rule: If $r(x)\!$ is the sum of the terms in the 1st column of table 2.1 then $y_{p}(x)\!$ is the sum of the corresponding terms in the 2nd column of this table.

Table 2.1

Solution

According to the table the Homogeneous equation has two $r(x)\!$ values of the form $kx^{n}(n=0,1,2...)$. Using the Basic Rule this means that there is a particular solution of the form $K_{n}x^{n}+K_{n-1}x^{n-1}+...+K_{1}x+K_{0}\!$ for each $r(x)\!$.

$\displaystyle r_{1}(x)=4x^{2}\rightarrow r_{1}(x)=Kx^{n}(n=2)$

(4.5)

So the particular solution for this $r_{1}(x)\!$ should be:

$\displaystyle y_{p1}=K_{2}x^{2}+K_{1}x^{1}+K_{0}$

(4.6)

Where $K=4\!$. Which simplifies to:

$\displaystyle y_{p1}=\sum _{j=0}^{2}K_{j}x^{j}$

(4.7)

For the second $r_{2}(x)\!$:

$\displaystyle r_{2}(x)=-6x^{5}\rightarrow r_{2}(x)=Cx^{n}(n=5)$

(4.8)

So the particular solution for this $r_{2}(x)\!$ should be:

$\displaystyle y_{p2}=C_{5}x^{5}+C_{4}x^{4}+C_{3}x^{3}+C_{2}x^{2}+C_{1}x+C_{0}$

(4.9)

Where $C=4\!$. Which simplifies to:

$\displaystyle y_{p1}=\sum _{j=0}^{5}C_{j}x^{j}$

(4.10)

Using the sum rule:

$\displaystyle y_{p}=\sum _{j=0}^{2}K_{j}x^{j}+\sum _{j=0}^{5}Cx^{j}$

(4.11)

Now using the Sum Rule which just states if there are two $r(x)\!$ values in any form on the left side of the table, then the particular solution is the sum of the solutions for $r(x)\!$ on the right side of the table.

So the particular solution for $y''-3y'+2y=4x^{2}-6x^{5}\!$ where $c=K+C$

$\displaystyle y_{p}=\sum _{j=0}^{5}c_{j}x^{j}$

(4.12)

Author

Solved and Typed By - User:Egm4313.s12.team1.stewart 22:47, 21 February 2012 (UTC)
Reviewed By - Egm4313.s12.team1.silvestri 04:57, 22 February 2012 (UTC)

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Problem R3.5: Finding Particular Solutions by Combining Linear Coefficient Series Expansions

Statement

Given:

$\displaystyle y''-3y'+2y=4x^{2}-6x^{5}$

(5.0)

$\displaystyle y_{p}(x)=\sum _{j=0}^{5}c_{j}x^{j}=4x^{2}-6x^{5}$

(5.1)

Find:
-Coefficients of particular solution using series expansion and matrix back-substitution
-Solution $\displaystyle y(x)$ using initial conditions $\displaystyle y(0)=1,y'(0)=0$
Plot solution.

Solution

Using Sec7b-1 p.7-13 Eq. (1):

$\displaystyle \sum _{j=0}^{3}c_{j+2}(j+2)(j+1)x^{j}-3\sum _{j=0}^{4}c_{j+1}(j+1)x^{j}+2\sum _{j=0}^{5}c_{j}x^{j}=4x^{2}-6x^{5}$

(5.2)

Which is a combined series expansion for the general coefficient series expansion of Sec7b-1 p.7-12 Eq. (4):

$\displaystyle \sum _{j=2}^{5}c_{j}j(j-1)x^{j-2}-3\sum _{j=1}^{5}c_{j}jx^{j-1}+2\sum _{j=0}^{5}c_{j}x^{j}=4x^{2}-6x^{5}$

(5.3)

We can find a simultaneous linear system of equations to solve for the coefficients of $\displaystyle y_{p}(x)$.

Note: by combining the summations in Eq. (5.2) by setting the upper bounds of the summations to a common value 3, the equation can be simplified to:

$\displaystyle \sum _{j=0}^{3}[c_{j+2}(j+2)(j+1)-3c_{j+1}(j+1)+2c_{j}]x^{j}+(2c_{4}-15c_{5})x^{4}+2c_{5}x^{5}=4x^{2}-6x^{5}$

(5.4)

Therefore providing a system of equations by equating coefficients from the summations to the $\displaystyle y_{p}(x)$:

$\displaystyle j=0:[c_{2}(2)(1)-3c_{1}(1)+2c_{0}]x^{0}=0x^{0}=[2c_{0}-3c_{1}+2c_{2}]x^{0}$

$\displaystyle j=1:[c_{3}(3)(2)-3c_{2}(2)+2c_{1}]x^{1}=0x^{1}=[2c_{1}-6c_{2}+6c_{3}]x^{1}$

$\displaystyle j=2:[c_{4}(4)(3)-3c_{3}(3)+2c_{2}]x^{2}=4x^{2}=[2c_{2}-9c_{3}+12c_{4}]x^{2}$

$\displaystyle j=3:[c_{5}(5)(4)-3c_{4}(4)+2c_{3}]x^{3}=0x^{3}=[2c_{3}-12c_{4}+20c_{5}]x^{3}$

$\displaystyle [2c_{4}-15c_{5}]x^{4}=0x^{4}$

$\displaystyle 2c_{5}x^{5}=-6x^{5}$

(5.5)

Eqs. (5.5) can be verified by solving for the coefficients by Eq. (5.2) to prove that the summations were combined correctly:

$\displaystyle j=0:2c_{0}x^{0}=2c_{0}$

$\displaystyle j=1:2c_{1}x^{1}-3c_{1}(1)x^{0}=2c_{1}x-3c_{1}$

$\displaystyle j=2:2c_{2}x^{2}-3c_{2}(2)x^{1}+c_{2}(2)(1)x^{0}=2c_{2}x^{2}-6c_{2}x+3c_{2}$

$\displaystyle j=3:2c_{3}x^{3}-3c_{3}(3)x^{2}+c_{3}(3)(2)x^{1}=2c_{3}x^{3}-9c_{3}x^{2}+6c_{3}x$

$\displaystyle j=4:2c_{4}x^{4}-3c_{4}(4)x^{3}+c_{4}(4)(3)x^{2}=2c_{4}x^{4}-12c_{4}x^{3}+12c_{4}x^{2}$

$\displaystyle j=5:2c_{5}x^{5}-3c_{5}(5)x^{4}+c_{5}(5)(4)x^{3}=2c_{5}x^{5}-15c_{5}x^{4}+20c_{5}x^{3}$

(5.6)

By summing the terms in (5.6) and grouping like terms, then equating it to the $\displaystyle y_{p}(x)$, we find the coefficient equations are the same as Eqs. (5.5):

$\displaystyle [2c_{0}-3c_{1}+2c_{2}]x^{0}+[2c_{1}-6c_{2}+6c_{3}]x^{1}+[2c_{2}-9c_{3}+12c_{4}]x^{2}+[2c_{3}-12c_{4}+20c_{5}]x^{3}+[2c_{4}-15c_{5}]x^{4}+2c_{5}x^{5}=0x^{0}+0x^{1}+4x^{2}+0x^{3}+0x^{4}-6x^{5}$

(5.7)

The linear system of equations in Eqs. (5.5) can be placed in matrix form:

$\left[{\begin{array}{cccccc}2&-3&2&0&0&0\\0&2&-6&6&0&0\\0&0&2&-9&12&0\\0&0&0&2&-12&20\\0&0&0&0&2&-15\\0&0&0&0&0&2\end{array}}\right]$ $\left({\begin{array}{cccccc}{c}_{0}\\{c}_{1}\\{c}_{2}\\{c}_{3}\\{c}_{4}\\{c}_{5}\end{array}}\right)\ =\left[{\begin{array}{cccccc}0\\0\\4\\0\\0\\-6\end{array}}\right]\$

(5.8)

Solving this system of equations by back-substitution yields the values of the coefficients for the $\displaystyle y_{p}(x)$:

$\displaystyle c=\{-701.75,-691.5,-335.5,-105,-22.5,-3\}$

(5.9)

Substituting these coefficients into Eq. (5.1):

$\displaystyle y_{p}(x)=-701.75x^{0}-691.5x^{1}-335.5x^{2}-105x^{3}-22.5x^{4}-3x^{5}$

(5.10)

The general solution is the sum of the homogeneous (found in R3.3) and particular solutions:

$\displaystyle y(x)=-701.75-691.5x^{1}-335.5x^{2}-105x^{3}-22.5x^{4}-3x^{5}+C_{1}e^{2x}+C_{2}e^{x}$

(5.11)

Considering the initial value condition $\displaystyle y(0)=1$ in the general solution:

$\displaystyle y(0)=1=-701.75+C_{1}e^{0}+C_{2}e^{0}=-701.75+C_{1}+C_{2}$

(5.12)

Taking the first derivative of Eq. (5.11) to consider the initial value condition $\displaystyle y'(0)=0$:

$\displaystyle y'(x)=-691.5-671x-315x^{2}-90x^{3}-15x^{4}+2C_{1}e^{2x}+C_{2}e^{x}$

(5.13)

$\displaystyle y'(0)=0=-691.5+2C_{1}+C_{2}$

(5.14)

Solving Eq. (5.12) and Eq. (5.13) yield the coefficients:

$\displaystyle C_{1}=-11.25,C_{2}=714$

(5.15)

This yields our final solution by substituting the coefficients into Eq. (5.11): Solving Eq. (5.12) and Eq. (5.13) yield the coefficients:

$\displaystyle y(x)=-701.75-691.5x^{1}-335.5x^{2}-105x^{3}-22.5x^{4}-3x^{5}-11.25e^{2x}+714e^{x}$

(5.16)

A plot of the solution:

Figure 3.5-1

Author

Solved and Typed By - --Egm4313.s12.team1.durrance 22:56, 21 February 2012 (UTC)--
Reviewed By - Egm4313.s12.team1.silvestri 03:28, 22 February 2012 (UTC)

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Problem R3.6: Superimposing Particular Solutions

Statement

Solve the L2-ODE-CC (2) p.7-11 with initial conditions (2b) p.3-6 differently as follows. Consider the following two L2-ODEs-CC (see p.7-2b):

$\displaystyle {y}_{p,1}''-3{y}_{p,1}'+2{{y}_{p,1}}={{r}_{1}}(x):=4{{x}^{2}}$

(6.0)

$\displaystyle {y}_{p,2}''-3{y}_{p,2}'+2{{y}_{p,2}}={{r}_{2}}(x):=-6{{x}^{5}}$

(6.1)

The particular solution $\displaystyle {y}_{p,1}$ had been found in R3.3 p.7-11. Find the particular solution $\displaystyle {y}_{p,2}$, and then obtain the solution $\displaystyle {y}$ for the L2-ODE-CC (2) p.7-11 with initial conditions (2b) p.3-6.

Compare the result with that obtained in R3.5.

Solution

Beginning with equation (6.1), we find that a particular solution has the form

$\displaystyle {{y}_{p,2}}(x)=\sum \limits _{j=0}^{n}{{{K}_{j}}{{x}^{j}}}$

(6.2)

where $\displaystyle {n=5}$. That is,

$\displaystyle {{y}_{p,2}}(x)={{K}_{0}}+{{K}_{1}}x+{{K}_{2}}{{x}^{2}}+{{K}_{3}}{{x}^{3}}+{{K}_{4}}{{x}^{4}}+{{K}_{5}}{{x}^{5}}$

(6.3)

Differentiating twice:

$\displaystyle {y}_{p,2}'(x)={{K}_{1}}+2{{K}_{2}}x+3{{K}_{3}}{{x}^{2}}+4{{K}_{4}}{{x}^{3}}+5{{K}_{5}}{{x}^{4}}$

(6.4)

$\displaystyle {y}_{p,2}''(x)=2{{K}_{2}}+6{{K}_{3}}x+12{{K}_{4}}{{x}^{2}}+20{{K}_{5}}{{x}^{3}}$

(6.5)

Substitute (6.3-5) into (6.1) to obtain

$\displaystyle (2{{K}_{2}}+6{{K}_{3}}x+12{{K}_{4}}{{x}^{2}}+20{{K}_{5}}{{x}^{3}})-3({{K}_{1}}+2{{K}_{2}}x+3{{K}_{3}}{{x}^{2}}+4{{K}_{4}}{{x}^{3}}+5{{K}_{5}}{{x}^{4}})$ $\displaystyle +2({{K}_{0}}+{{K}_{1}}x+{{K}_{2}}{{x}^{2}}+{{K}_{3}}{{x}^{3}}+{{K}_{4}}{{x}^{4}}+{{K}_{5}}{{x}^{5}})=-6{{x}^{5}}$

(6.6)

Rearranging terms with respect to $\displaystyle {x}$ power:

$\displaystyle (2{{K}_{2}}-3{{K}_{1}}+2{{K}_{0}})+x(6{{K}_{3}}-6{{K}_{2}}+2{{K}_{1}})+{{x}^{2}}(12{{K}_{4}}-9{{K}_{3}}+2{{K}_{2}})$ $\displaystyle +{{x}^{3}}(20{{K}_{5}}-12{{K}_{4}}+2{{K}_{3}})+{{x}^{4}}(-15{{K}_{5}}+2{{K}_{4}})+{{x}^{5}}(2{{K}_{5}})=-6{{x}^{5}}$

(6.7)

In matrix form:

$\left[{\begin{array}{cccccc}2&-3&2&0&0&0\\0&2&-6&6&0&0\\0&0&2&-9&12&0\\0&0&0&2&-12&20\\0&0&0&0&2&-15\\0&0&0&0&0&2\end{array}}\right]$ $\left({\begin{array}{cccccc}{K}_{0}\\{K}_{1}\\{K}_{2}\\{K}_{3}\\{K}_{4}\\{K}_{5}\end{array}}\right)\ =\left[{\begin{array}{cccccc}0\\0\\0\\0\\0\\-6\end{array}}\right]\$

(6.8)

Solving for $\displaystyle K$, using MATLAB, yields

$\displaystyle K=\{-708.75,-697.5,-337.5,-105,-22.5,-3\}$

(6.9)

which means that

$\displaystyle {y}_{p,2}=-708.75-697.5x-337.5{{x}^{2}}-105{{x}^{3}}-22.5{{x}^{4}}-3{{x}^{5}}$

(6.10)

From R3.3:

$\displaystyle {{y}_{p,1}}=32+12x+2{{x}^{2}}$

(6.11)

Summing for the final solution:

$\displaystyle y={{C}_{1}}{{y}_{p,1}}+{{C}_{2}}{{y}_{p,2}}$

(6.12)

$\displaystyle y={{C}_{1}}(32+12x+2{{x}^{2}})$ $\displaystyle +{{C}_{2}}(-708.75-697.5x-337.5{{x}^{2}}-105{{x}^{3}}-22.5{{x}^{4}}-3{{x}^{5}})$

(6.13)

For initial condition $\displaystyle y(0)=-5$

$\displaystyle y(0)=-5={{C}_{1}}(32)+{{C}_{2}}(-708.75)$

(6.14)

Similarly, for initial condition $\displaystyle y'(0)=2$

$\displaystyle y'(0)=2={{C}_{1}}(12)+{{C}_{2}}(-697.5)$

(6.15)

Equations (6.14) and (6.15) yield the following matrix equation:

$\left[{\begin{array}{cc}32&-708.75\\12&-697.5\end{array}}\right]$ $\left({\begin{array}{cc}{C}_{1}\\{C}_{2}\end{array}}\right)\ =\left[{\begin{array}{cc}-5\\2\end{array}}\right]\$

(6.16)

Solving (6.16) in MATLAB yields

$\displaystyle C=\{-0.355,-0.009\}$

(6.17)

Therefore the combined solution of $\displaystyle y$ is

$\displaystyle y(x)=-0.355(32+12x+2{{x}^{2}})$ $\displaystyle -0.009(-708.75-697.5x-337.5{{x}^{2}}-105{{x}^{3}}-22.5{{x}^{4}}-3{{x}^{5}})$

(6.18)

Which can be simplified to Final Equation

$\displaystyle y(x)=-4.98125+2.0175x+2.3275{{x}^{2}}+0.945{{x}^{3}}+0.2025{{x}^{4}}+0.027{{x}^{5}}$

(6.19)

Author

Solved and Typed By - Egm4313.s12.team1.essenwein 00:40, 18 February 2012 (UTC)
Reviewed By - Egm4313.s12.team1.silvestri 04:29, 22 February 2012 (UTC)

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Problem R3.7 : Verifying series representation for method of undetermined coefficients

Statement

Expand the series on both sides of (1)-(2) p.7-12b to verify these equalities.
These equalities are:

$\sum _{j=2}^{5}C_{j}j(j-1)x^{^{j-2}}=\sum _{j=0}^{3}C_{j+2}(j+2)(j+1)x^{^{j}}\!$

(7.0)

$\sum _{j=1}^{5}C_{j}jx^{^{j-1}}=\sum _{j=0}^{4}C_{j+1}(j+1)x^{^{j}}\!$

(7.1)

Solution

This transition occurs through a mid-level variable change. In this particular case, for (7.0), j-2 is represented by k:

$\sum _{k=0}^{3}C_{k+2}(k+2)(k+1)x^{^{k}}\!$

(7.3)

We can then represent the variable k with a j. This new summation looks like this:

$\sum _{j=0}^{3}C_{j+2}(j+2)(j+1)x^{^{j}}\!$

(7.4)

Expanding both sides, (7.0 and 7.4) yields the same result:

$C_{5}(5)(4)x^{^{3}}+C_{4}(4)(3)x^{^{2}}+C_{3}(3)(2)x^{^{1}}+C_{2}(2)(1)x^{^{0}}\!$

(7.5)

We follow the same process for equality (7.1):
This transition occurs through a mid-level variable change. In this particular case, for (7.1), j-1 is represented by k:

$\sum _{k=0}^{4}C_{k+1}(k+1)x^{^{k}}\!$

(7.6)

We can then represent the variable k with a j. This new summation looks like this:

$\sum _{j=0}^{4}C_{j+1}(j+1)x^{^{j}}\!$

(7.7)

Expanding both sides, (7.1 and 7.7)yields the same result:

$C_{5}(5)x^{^{4}}+C_{4}(4)x^{^{3}}+C_{3}(3)x^{^{2}}+C_{2}(2)x^{^{1}}+C_{1}\!$

(7.8)

Author

Solved and Typed By - Egm4313.s12.team1.silvestri 17:20, 19 February 2012 (UTC)
Reviewed By - --128.227.113.77 17:05, 22 February 2012 (UTC)

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Problem R3.8: Finding general solutions to Non-homogeneous Linear ODEs using the Method of Undetermined Coefficients and the Basic Rule

Statement

Find a general solution for the following two problems:

Kreyszig 2011 p.84 Problem 5

Given

Homogeneous solution: $y''+4y'+4y=e^{-x}cosx\!$

Solution

The solution $y(x)\!$ is composed of a general solution and a particular solution so that $y(x)=y_{g}(x)+y_{p}(x)\!$.
First, we will find the general solution, $y_{g}(x)\!$, by finding the roots of the characteristic equation:

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

(8.0)

Which means that the characteristic equation has a double root of $\lambda =-2\!$.
Based on the double root, then, the general solution is:

$\displaystyle {y_{g}(x)=(C_{1}+C_{2}x)e^{-{\frac {ax}{2}}}\rightarrow y_{g}(x)=(C_{1}+C_{2}x)e^{-2x}}$

(8.1)

Next, we need to find the particular solution, and based on analysis of the excitation, $r(x)=e^{-x}cosx\!$, we find that the particular solution is the following:

$\displaystyle {y_{p}(x)=e^{\alpha x}(Kcos\omega x+Msin\omega x)\rightarrow y_{p}(x)=e^{-x}Kcosx+e^{-x}Msinx}$

(8.2)

Now, we need to find the values for the constants $K,M\!$, by taking the first and second derivatives of the particular solution (8.2):

$\displaystyle {y_{p}(x)=e^{-x}Kcosx+e^{-x}Msinx}$

(8.2)

$\displaystyle {y_{p}(x)'=-e^{-x}Ksinx-e^{-x}Kcosx+e^{-x}Mcosx-e^{-x}Msinx}$

(8.3)

$\displaystyle {y_{p}(x)''=-e^{-x}Kcosx+e^{-x}Ksinx+e^{-x}Ksinx+e^{-x}Kcosx-e^{-x}Msinx-e^{-x}Mcosx-e^{-x}Mcosx+e^{-x}Msinx}$

(8.4)

Now, plug these into the homogeneous solution and simplify:

$\displaystyle {-e^{-x}(Kcosx-Ksinx-Ksinx-Kcosx+Msinx+Mcosx+Mcosx-Msinx)+4[-e^{-x}(Ksinx+Kcosx-Mcosx+Msinx)]+4[e^{-x}(Kcosx+Msinx)]=e^{-x}cosx}$

(8.5)

$\displaystyle {-e^{-x}[-2Ksinx+2Mcosx]-e^{-x}[4Ksinx+4Kcosx-4Mcosx+4Msinx]+e^{-x}[4Kcosx+4Msinx]=e^{-x}cosx}$

(8.6)

Pull $e^{-x}\!$ out of equation (8.6) and simplify:

$\displaystyle {e^{-x}[4Kcosx+4Msinx-4Ksinx-4Kcosx+4Mcosx-4Msinx+2Ksinx-2Mcosx]=e^{-x}cosx}$

(8.7)

$\displaystyle {e^{-x}[-2Ksinx+2Mcosx]=e^{-x}cosx}$

(8.8)

Now, solve for the unknown coefficients:

$\displaystyle {2M=1\rightarrow M={\frac {1}{2}}}$

(8.9)

$\displaystyle {-2K=0\rightarrow K=0}$

(8.10)

Plug into $y_{p}(x)\!$ (8.2):

$\displaystyle {y_{p}(x)={\frac {1}{2}}e^{-x}sinx}$

(8.11)

Now, we have both the particular solution and the general solution, which allows for us to solve for the final solution, $y(x)\!$

$\displaystyle y(x)=(C_{1}+C_{2}x)e^{-2x}+{\frac {1}{2}}e^{-x}sinx$

(8.12)

Kreyszig 2011 p.84 Problem 6

Given

Homogeneous solution: $y''+y'(\pi ^{2}+{\frac {1}{4}})=e^{-{\frac {x}{2}}}sin{\pi x}\!$

Solution

The solution $y(x)\!$ is composed of a general solution and a particular solution so that $y(x)=y_{g}(x)+y_{p}(x)\!$.
First, we will find the general solution, $y_{g}(x)\!$, by finding the roots of the characteristic equation, using the quadratic equation:

$\displaystyle {\lambda ^{2}+\lambda +(\pi ^{2}+{\frac {1}{4}})=0\rightarrow \lambda ={\frac {-1\pm {\sqrt {1-4(1)(\pi ^{2}+{\frac {1}{4}})}}}{2}}}$

(8.13)

$\displaystyle {\lambda =-{\frac {1}{2}}+i\pi }$

(8.14)

The roots of the characteristic equation are complex conjugates, meaning that the general solution, $y_{g}(x)\!$, is the following:

$\displaystyle {y_{g}(x)=e^{-{\frac {x}{2}}}(Acos{\pi x}+Bsin{\pi x})}$

(8.15)

Next, we need to find the particular solution, and based on analysis of the excitation, $r(x)=e^{-{\frac {x}{2}}}sin{\pi x}\!$, we find that the particular solution is the following:

$\displaystyle {y_{p}(x)=e^{\alpha x}(Kcos\omega x+Msin\omega x)\rightarrow y_{p}(x)=e^{-{\frac {x}{2}}}Kcos{\pi x}+e^{-{\frac {x}{2}}}Msin{\pi x}}$

(8.16)

Now, we need to find the values for the constants $K,M\!$, by taking the first and second derivatives of the particular solution (8.16):

$\displaystyle {y_{p}(x)=e^{\alpha x}(Kcos\omega x+Msin\omega x)\rightarrow y_{p}(x)=e^{-{\frac {x}{2}}}Kcos{\pi x}+e^{-{\frac {x}{2}}}Msin{\pi x}}$

(8.16)

$\displaystyle {y_{p}(x)'={-\pi e^{\frac {-x}{2}}Ksin{\pi x}-{\frac {{}e^{\frac {-x}{2}}}{2}}Kcos{\pi x}}+{\pi e^{\frac {-x}{2}}Mcos{\pi x}+{\frac {{}e^{\frac {-x}{2}}}{2}}Msin{\pi x}}}$

(8.17)

$\displaystyle {y_{p}(x)''={-\pi ^{2}e^{\frac {-x}{2}}Kcos{\pi x}+\pi {\frac {e^{\frac {-x}{2}}}{2}}Ksin{\pi x}}+\pi {\frac {e^{\frac {-x}{2}}}{2}}Ksin{\pi x}+{\frac {e^{\frac {-x}{2}}}{4}}Kcos{\pi x}-\pi ^{2}{e^{\frac {-x}{2}}}Msin{\pi x}-\pi {\frac {e^{\frac {-x}{2}}}{2}}{}Mcos{\pi x}-\pi {\frac {e^{\frac {-x}{2}}}{2}}{}Mcos{\pi x}+{\frac {e^{\frac {-x}{2}}}{4}}{}Msin{\pi x}}$

(8.18)

Now, we plug these into the homogeneous solution and simplify but we find that everything cancels out, meaning the unknown constants, $M,K\!$ are both equal to $0\!$. This means that the final solution is equal to just the general solution:

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

(8.19)

Author

Solved and Typed By ---Egm4313.s12.team1.wyattling 22:10, 20 February 2012 (UTC)
Reviewed By - Egm4313.s12.team1.rosenberg 03:07, 22 February 2012 (UTC)

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Problem R3.9 Finding general solutions to Non-homogeneous Linear ODEs

Statement

K 2011 page 85 problems 13 and 14
Problem 13: Find the complete solution for $8y''-6y'+y=6\cosh x\!$, with the initial conditions
$y(0)=0.2,y'(0)=0.05\!$

Problem 14: Find the complete solution for $y''+4y'+4y=e^{-2x}\sin 2x\!$, with the initial conditions
$y(0)=1,y'(0)=-1.5\!$

Solution

Problem 13:
In order to solve the equation we must use the definition $y=y_{h}+y_{p}\!$. We start with the given equation:

$\displaystyle {8y''-6y'+y=6\cosh x}$

(9.1)

From this we get the homogeneous characteristic equation:

$\displaystyle {8\lambda ^{2}-6\lambda +1=0x}$

(9.2)

Factoring the characteristic equation gives us:

$\displaystyle {(4\lambda -1)(2\lambda -1)=0}$

(9.3)

This give us the roots:

$\displaystyle {\lambda ={\frac {1}{4}},{\frac {1}{2}}}$

(9.4)

Plugging in the roots gives us the general homogeneous solution:

$\displaystyle {y_{h}=c_{1}e^{{\frac {1}{4}}x}+c_{2}e^{{\frac {1}{2}}x}}$

(9.5)

Now we must solve for the particular solution. So far we know:

$\displaystyle {\cosh x={\frac {e^{x}+e^{-x}}{2}}}$

(9.6)

Now using the sum rule, $\displaystyle {y_{p}=y_{p1}+y_{p2}}$, gives us:

$\displaystyle {y_{p}=Ce^{x}+Ke^{-x}}$

(9.7)

Now we must take the first and second derivatives of equation 9.7:

$\displaystyle {y_{p}'=Ce^{x}-Ke^{-x}}$

(9.8)

$\displaystyle {y_{p}''=Ce^{x}+Ke^{-x}}$

(9.9)

Substituting 9.8 and 9.9 back into our original equation gives us:

$\displaystyle {8(Ce^{x}+Ke^{-x})-6(Ce^{x}-Ke^{-x})+Ce^{x}+Ke^{-x}=6({\frac {e^{x}+e^{-x}}{2}})}$

(9.10)

Simplifying equation 9.10 give us:

$\displaystyle {3Ce^{x}+15Ke^{-x}=3(e^{x}+e^{-x})}$

(9.11)

From this we can deduce that:

$\displaystyle {3C=3,15K=3}$

(9.12)

Thus we get that:

$\displaystyle {C=1,K={\frac {1}{5}}}$

(9.13)

We now have our particular equation:

$\displaystyle {y_{p}=e^{x}+{\frac {1}{5}}e^{-x}}$

(9.14)

Thus:

$\displaystyle {y=c_{1}e^{{\frac {1}{4}}x}+c_{2}e^{{\frac {1}{2}}x}+e^{x}+{\frac {1}{5}}e^{-x}}$

(9.15)

Now we can solve for $c_{1}\!$ and $c_{2}\!$ using the given initial values.
We are given that at $y(0)=0.2\!$, we can now plug this into our equation to get:

$\displaystyle {0.2=c_{1}e^{{\frac {1}{4}}(0)}+c_{2}e^{{\frac {1}{2}}(0)}+e^{(0)}+{\frac {1}{5}}e^{-(0)}}$

(9.16)

This simplifies to:

$\displaystyle {0.2=c_{1}+c_{2}+1+{\frac {1}{5}}}$

(9.17)

Giving us that:

$\displaystyle {c_{1}+c_{2}=-1}$

(9.18)

For our second condition, $y'(0)=0.05\!$, we must we must take the derivative of our general solution:

$\displaystyle {y'={\frac {1}{4}}c_{1}e^{{\frac {1}{4}}x}+{\frac {1}{2}}c_{2}e^{{\frac {1}{2}}x}+e^{x}-{\frac {1}{5}}e^{-x}}$

(9.19)

At our initial condition gives us:

$\displaystyle {0.05={\frac {1}{4}}c_{1}+{\frac {1}{2}}c_{2}+1-{\frac {1}{5}}}$

(9.20)

Thus:

$\displaystyle {c_{1}+2c_{2}=-3}$

(9.21)

Combining 9.18 and 9.21 gives us that:

$\displaystyle {c_{1}=1,c_{2}=-2}$

(9.22)

Plugging these coefficients in gives us our final solution:

$\displaystyle {y=e^{{\frac {1}{4}}x}-2e^{{\frac {1}{2}}x}+e^{x}+{\frac {1}{5}}e^{-x}}$

(9.23)

Problem 14:
In order to solve the equation we must use the definition $y=y_{h}+y_{p}\!$. We start with the given equation:

$\displaystyle {y''+4y'+4y=e^{-2x}\sin 2x}$

(9.24)

From this we get the homogeneous characteristic equation:

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

(9.25)

Factoring the characteristic equation gives us:

$\displaystyle {(\lambda +2)(\lambda +2)=0}$

(9.26)

This give us the double root:

$\displaystyle {\lambda =-2}$

(9.27)

Plugging in the roots gives us the general homogeneous solution:

$\displaystyle {y_{h}=c_{1}e^{-2x}+c_{2}xe^{-2x}}$

(9.28)

Now we must solve for the particular solution. So far we know:

$\displaystyle {e^{-2x}\sin 2x=Cxe^{-2x}\cos 2x+Kxe^{-2x}\sin 2x}$

(9.29)

Now we must take the first and second derivatives of equation 9.29:

$\displaystyle {y_{p}'=-2e^{-2x}(Cx\cos 2x+Kx\sin 2x)+e^{-2x}(C\cos 2x-2Cx\sin 2x+K\sin 2x+2Kx\cos 2x)}$

(9.30)

$\displaystyle {y_{p}''=(-4C+4K)e^{-2x}\cos 2x+(-4k-4C)e^{-2x}\sin 2x}$

(9.31)

Substituting 9.30 and 9.31 back into our original equation gives us:

$\displaystyle {(-3C+4K)e^{-2x}\cos 2x+(-3k-4C)e^{-2x}\sin 2x=e^{-2x}\sin 2x}$

(9.32)

From this we can deduce that:

$\displaystyle {-3C+4K=0,-3K-4C=1}$

(9.33)

Thus we get that:

$\displaystyle {C={\frac {-4}{25}},K={\frac {-3}{25}}}$

(9.34)

We now have our particular equation:

$\displaystyle {y_{p}=e^{-2x}({\frac {-4}{25}}x\cos 2x-{\frac {3}{25}}x\sin 2x)}$

(9.35)

Thus:

$\displaystyle {y=c_{1}e^{-2x}+c_{2}xe^{-2x}+e^{-2x}({\frac {-4}{25}}x\cos 2x-{\frac {3}{25}}x\sin 2x)}$

(9.36)

Now we can solve for $c_{1}\!$ and $c_{2}\!$ using the given initial values.
We are given that at $y(0)=1\!$, we can now plug this into our equation to get:

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

(9.37)

Giving us that:

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

(9.38)

For our second condition, $y'(0)=-1.5\!$, we must we must take the derivative of our general solution:

$\displaystyle {y'=-2c_{1}e^{-2x}+c_{2}(-2xe^{-2x}+e^{-2x})+{\frac {-4}{25}}e^{-2x}\cos 2x-{\frac {3}{25}}e^{-2x}\sin 2x+{\frac {2}{25}}e^{-2x}x\cos 2x+{\frac {14}{25}}e^{-2x}x\sin 2x}$

(9.39)

At our initial condition we have:

$\displaystyle {y'(0)=-2+c_{2}-{\frac {4}{25}}-0+0+0=0}$

(9.40)

Thus:

$\displaystyle {c_{2}={\frac {54}{25}}}$

(9.41)

Plugging these coefficients in gives us our final solution:

$\displaystyle {y=1e^{-2x}+{\frac {54}{25}}xe^{-2x}+e^{-2x}({\frac {-4}{25}}x\cos 2x-{\frac {3}{25}}x\sin 2x)}$

(9.42)

Author

Solved and Typed By - Egm4313.s12.team1.rosenberg 03:04, 22 February 2012 (UTC)
Reviewed By - Egm4313.s12.team1.essenwein 02:39, 22 February 2012 (UTC)

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Contributing Members

Team Contribution Table
Problem Number	Lecture	Assigned To	Solved By	Typed By	Proofread By
3.1	R3.1 in Sec 3 p. 7-5	Wyatt Ling	Wyatt Ling	Wyatt Ling	George Armanious
3.2	R3.2 in Sec 3 p. 7-5	George Armanious	George Armanious	George Armanious	Jesse Durrance
3.3	R3.3 in Sec 3 p. 7-11	Emotion Silvestri	Emotion Silvestri	Emotion Silvestri	George Armanious
3.4	R3.4 in Sec 3 p. 7-11	Chris Stewart	Chris Stewart	Chris Stewart	Emotion Silvestri
3.5	R3.5 in Sec 7 p. 7-17 (Sakai)	Jesse Durrance	Jesse Durrance	Jesse Durrance	Emotion Silvestri
3.6	R3.6 in Sec 7 p. 7-17 (Sakai)	Eric Essenwein	Eric Essenwein	Eric Essenwein	Emotion Silvestri
3.7	R3.7 in Sec 7 p. 7-17 (Sakai)	Emotion	Emotion	Emotion	Wyatt Ling
3.8		Wyatt Ling	Wyatt Ling	Wyatt Ling	Steven Rosenberg
3.9		Steven Rosenberg	Steven Rosenberg	Steven Rosenberg	Eric Essenwein