# University of Florida/Egm4313/s12.team11.R3

Report 3
Intermediate Engineering Analysis Section 7566 Team 11 Due date: February 22, 2012.

## Problem 3.1

Solved by Luca Imponenti

### Problem Statement

Find the solution to the following L2-ODE-CC: $y''-10y'+25y=r(x)\!$
With the following excitation: $r(x)=7e^{5x}-2x^{2}\!$
And the following initial conditions: $y(0)=4,y'(0)=-5\!$
Plot this solution and the solution in the example on p.7-3

### Homogeneous Solution

To find the homogeneous solution we need to find the roots of our equation
$\lambda ^{2}-10\lambda +25=0\!$ $(\lambda -5)(\lambda -5)=0\!$ $\lambda =5\!$
We know the homogeneous solution for the case of a real double root with $\lambda =5\!$ to be
$y_{h}=c_{1}e^{5x}+c_{2}xe^{5x}\!$

### Particular Solution

For the given excitation we must use the Sum Rule to the particular solution as follows
$y_{p}=y_{p1}+y_{p2}\!$ where $y_{p1}\!$ and $y_{p2}\!$ are the solutions to $r_{1}(x)=7e^{5x}\!$ and $r_{2}(x)=-2x^{2}\!$, respectively

#### First Particular Solution

$r_{1}(x)=7e^{5x}\!$,
from table 2.1, K 2011, pg. 82 we have
$y_{p1}=Ce^{5x}\!$
but this corresponds to one of our homogeneous solutions so we must use the modification rule to get
$y_{p1}=Cx^{2}e^{5x}\!$
Plugging this into the original L2-ODE-CC then substituting;
$y_{p1}''-10y_{p1}'+25y_{p1}=r_{1}(x)\!$
$(Cx^{2}e^{5x})''-10(Cx^{2}e^{5x})'+25(Cx^{2}e^{5x})=r_{1}(x)\!$
$25Cx^{2}e^{5x}+10Cxe^{5x}+10Cxe^{5x}+2Ce^{5x}-10(5Cx^{2}e^{5x}+2Cxe^{5x})+25Cx^{2}e^{5x}=7e^{5x}\!$
$e^{5x}[25Cx^{2}+10Cx+10Cx+2C-50Cx^{2}-20Cx+25Cx^{2}]=7e^{5x}\!$
$e^{5x}2C=7e^{5x}\!$
so $C={\frac {7}{2}}\!$ and the first particular solution is,
       $y_{p1}={\frac {7}{2}}x^{2}e^{5x}\!$

#### Second Particular Solution

$r_{2}(x)=-2x^{2}\!$,
from table 2.1, K 2011, pg. 82 we have
$y_{p2}=a_{2}x^{2}+a_{1}x+a_{0}\!$
Plugging this into the original L2-ODE-CC then substituting;
$y_{p2}''-10y_{p2}'+25y_{p2}=r_{2}(x)\!$
$(a_{2}x^{2}+a_{1}x+a_{0})''-10(a_{2}x^{2}+a_{1}x+a_{0})'+25(a_{2}x^{2}+a_{1}x+a_{0})=-2x^{2}\!$
$2a_{2}-10(2a_{2}x+a_{1})+25(a_{2}x^{2}+a_{1}x+a_{0})=-2x^{2}\!$
grouping like terms we get three equations to solve for the three unknowns, these are written in matrix form
$25a_{2}x^{2}+(25a_{1}-20a_{2})x+(25a_{0}-10a_{1}+2a_{2})=-2x^{2}\!$
       ${\begin{bmatrix}2&-10&25\\-20&25&0\\25&0&0\end{bmatrix}}{\begin{bmatrix}a_{2}\\a_{1}\\a_{0}\end{bmatrix}}={\begin{bmatrix}0\\0\\-2\end{bmatrix}}\!$
solving by back subsitution leads to $a_{2}=-{\frac {2}{25}},a_{1}={\frac {8}{125}},a_{0}={\frac {4}{125}}\!$
so the second particular solution is,
       $y_{p2}=-{\frac {2}{25}}x^{2}+{\frac {8}{125}}x+{\frac {4}{125}}\!$

### General Solution

The general solution is the summation of the homogeneous and particular solutions
$y=y_{h}+y_{p1}+y_{p2}\!$
$y=c_{1}e^{5x}+c_{2}xe^{5x}+{\frac {7}{2}}x^{2}e^{5x}-{\frac {2}{25}}x^{2}+{\frac {8}{125}}x+{\frac {4}{125}}\!$
$y=e^{5x}(c_{1}+c_{2}x+{\frac {7}{2}}x^{2})-{\frac {2}{25}}x^{2}+{\frac {8}{125}}x+{\frac {4}{125}}\!$
Applying the first initial condition $y(0)=4\!$
$y(0)=c_{1}+{\frac {4}{125}}=4\!$
$c_{1}={\frac {496}{125}}\!$
Second initial condition $y'(0)=-5\!$
$y'={\frac {d}{dx}}y=e^{5x}[5(c_{1}+c_{2}x+{\frac {7}{2}}x^{2})+c_{2}+7x]-{\frac {4}{25}}x+{\frac {8}{125}}\!$
$y'=e^{5x}[{\frac {35}{2}}x^{2}+(5c_{2}+7)x+5c_{1}+c_{2}]-{\frac {4}{25}}x+{\frac {8}{125}}\!$
$y'(0)=5c_{1}+c_{2}+{\frac {8}{125}}=-5\!$
$c_{2}=-{\frac {3113}{125}}\!$
The general solution to the differential equation is therefore

### Plot

Below is a plot of this solution and the solution to in the example on p.7-3
our solution $y(x)=e^{5x}({\frac {496}{125}}-{\frac {3113}{125}}x+{\frac {7}{2}}x^{2})-{\frac {2}{25}}x^{2}+{\frac {8}{125}}x+{\frac {4}{125}}\!$ (shown in red)
example on p.7-3 $y(x)=e^{5x}(4-25*x+*{\frac {7}{2}}x^{2})\!$ (shown in blue)
Egm4313.s12.team11.imponenti 05:55, 20 February 2012 (UTC)

## Problem 3.2

Solved by Gonzalo Perez

#### Problem Statement

Developing the second homogeneous solution for the case of double real root as a limiting case of distinct roots.

#### Given

Consider two distinct roots of the form:
$\lambda _{1}=x\!$ and $\lambda _{2}=x+\epsilon \!$
(where $\epsilon \!$ is perturbation).

#### Given

Find the homogeneous L2-ODE-CC having the above distinct roots.

#### Solution

$(\lambda -\lambda _{1})(\lambda -(\lambda _{1}))=0\!$
$(\lambda -x)(\lambda -(x+\epsilon ))=0\!$
$\lambda ^{2}-\lambda x-\lambda \epsilon -\lambda x+x^{2}+x\epsilon =0\!$
$\lambda ^{2}-\lambda (2x+\epsilon )+x(x+\epsilon )=0\!$

#### Given

Show that ${\frac {e^{(\lambda +\epsilon )x}-e^{\lambda x}}{\epsilon }}\!$ is a homogeneous solution. (2)

#### Solution

Let's find the corresponding derivatives:
$y={\frac {e^{(\lambda +\epsilon )x}-e^{\lambda x}}{\epsilon }}\!$
$y'={\frac {(\lambda +\epsilon )e^{\lambda +\epsilon x}-\lambda e^{\lambda x}}{\epsilon }}\!$
$y''={\frac {(\lambda +\epsilon )^{2}e^{(\lambda +\epsilon )x}-\lambda ^{2}e^{\lambda x}}{\epsilon }}\!$
If we now take these three equations and plug them into the homogeneous L2-ODE-CC (1), we get:
$e^{(\lambda +\epsilon )x}(\lambda ^{2}+2\lambda \epsilon +\epsilon ^{2}-2\lambda ^{2}-2\lambda \epsilon -\lambda \epsilon -\epsilon ^{2}+\lambda ^{2}+\epsilon \lambda )+e^{\lambda x}(-\lambda ^{2}+2\lambda ^{2}+\lambda \epsilon -\lambda ^{2}-\lambda \epsilon )=0\!$
$\therefore 0\equiv 0.\!$

#### Given

Find the limit of the homogeneous solution in (2) as epsilon approaches zero (think l'Hopital's Rule).

#### Solution

Using l'Hopital's Rule,
$\lim _{\epsilon \rightarrow 0}{\frac {e^{(\lambda +\epsilon )x}-e^{\lambda x}}{\epsilon }}={\frac {0}{0}}\!$
(this is an indeterminate form).
L'Hopital's Rule states that we can divide this function into two functions, $f(\epsilon )\!$ and $g(\epsilon )\!$, and then find their derivatives and attempt to find the limit of ${\frac {f'(\epsilon )}{g'(\epsilon )}}\!$. If a limit exists for this, then a limit exists for our original function.
$\lim _{\epsilon \rightarrow 0}{\frac {f'(\epsilon )}{g'(\epsilon )}}=\lim _{\epsilon \rightarrow 0}{\frac {xe^{x(\lambda +\epsilon )}}{1}}\!$
$={\frac {xe^{(\lambda +0)x}}{1}}\!$

#### Given

Take the derivative of $e^{\lambda x}$ with respect to lambda.

#### Solution

Taking the derivative with respect to lambda, we find that:
It is important to remember that we must hold $x\!$ as a constant when finding this derivative.

#### Given

Compare the results in parts (3) and (4), and relate to the result by using variation of parameters

#### Solution

Taking a closer look at Parts 3 and 4 of this problem, we discover that they're in fact equal:

#### Given

Numerical experiment: Compute (2) setting lambda equal to 5 and epsilon equal to 0.001, and compare to the value obtained from the exact second homogeneous solution.

#### Solution

Egm4313.s12.team11.perez.gp 20:32, 22 February 2012 (UTC)

## Problem 3.3

Solved by Jonathan Sheider

#### Problem Statement

Find the complete solution to the ODE with the given initial conditions. Plot the solution y(x).

#### Given

${y}''-3{y}'+2y=4x^{2}\!$
$y(0)=0,{y}'(0)=0\!$

#### Solution

First let us analyze the homogenous solution to the given ODE:
${y}''-3{y}'+2y=0\!$
The characteristic equation for this ODE is therefore:
$\lambda ^{2}-3\lambda +2=0\!$
Solving for $\lambda \!$:
$(\lambda -2)(\lambda -1)=0\!$
$\lambda ={1,2}\!$
$\lambda _{1}=1\!$ and $\lambda _{2}=2\!$
Therefore the equation has two real roots and a homogenous solution of the following form (from Kreyszig 2011, p.54-57):
$y_{h}=c_{1}e^{\lambda _{1}x}+c_{2}e^{\lambda _{2}x}\!$
And finally we find the general homogenous solution:
Next let us evaluate the particular solution to the given ODE:
${y}''-3{y}'+2y=4x^{2}\!$
The function on the right hand side of this equation implies that the particular solution has the following form based on Table 2.1 (from Kreyszig 2011, p.82):
$y_{p}=K_{n}x^{n}+K_{n-1}x^{n-1}+...+K_{1}x+K_{0}\!$
Therefore for this ODE we have:
$y_{p}=K_{2}x^{2}+K_{1}x+K_{0}\!$
Differentiating $y_{p}\!$ to obtain ${y_{p}}'\!$ and ${y_{p}}''\!$ respectively:
${y}'=2K_{2}x+K_{1}\!$
${y}''=2K_{2}\!$
Substituting these equations into the original ODE yields:
$(2K_{2})-3(2K_{2}x+K_{1})+2(K_{2}x^{2}+K_{1}x+K_{0})=4x^{2}\!$
$2K_{2}-6K_{2}x-3K_{1}+2K_{2}x^{2}+2K_{1}x+2K_{0}=4x^{2}\!$
And we know that the coefficients of the variables on each side of the equation must be equal:
$(2K_{2})x^{2}+(2K_{1}-6K_{2})x+(2K_{2}-3K_{1}+2K_{0})=4x^{2}\!$
Therefore we find:
$2K_{2}=4\!$
$K_{2}=2\!$
Now, solving for $K_{1}\!$ and $K_{0}\!$:
$2K_{1}-6K_{2}=0\!$
$2K_{1}=6(2)\!$
$K_{1}=6\!$
And also: $2K_{2}-3K_{1}+2K_{0}=0\!$
$2K_{0}=3(6)-2(2)\!$
$K_{0}=7\!$
Finally we arrive at the particular solution:
By superposition, we can find the complete general solution:
$y=y_{h}+y_{p}\!$
Using the given initial conditions, we can solve for $c_{1}\!$ and $c_{2}\!$, first by using the initial conditions for $y\!$:
$y(0)=c_{1}+c_{2}+7=0\!$
Differentiating the complete general solution and using the given initial condition for ${y}'\!$:
${y}'=c_{1}e^{x}+2c_{2}e^{2x}+4x+6\!$
${y}'(0)=c_{1}+2c_{2}+6=0\!$
Solving this system of equations, by solving the first equation for $c_{1}\!$:
$c_{1}=-7-c_{2}\!$
Plugging this into the second equation yields:
$(-7-c_{2})+2c_{2}+6=-1+c_{2}=0\!$
$c_{2}=1\!$
And therefore:
$c_{1}=-7-(1)\!$
$c_{1}=-8\!$
Finally we have the complete general solution that evaluates the given initial conditions:
A plot of $y(x)\!$ from $x=-3\!$ to $x=3\!$ is shown using MatLAB:

#### Checking

Differentiating the solution to find ${y}'\!$ and ${y}''\!$ respectively:
$y=-8e^{x}+e^{2x}+2x^{2}+6x+7\!$
${y}'=-8e^{x}+2e^{2x}+4x+6\!$
${y}''=-8e^{x}+4e^{2x}+4\!$
Substituting into the original ODE yields:
${y}''-3{y}'+2y=4x^{2}\!$
$(-8e^{x}+4e^{2x}+4)-3(-8e^{x}+2e^{2x}+4x+6)+2(-8e^{x}+e^{2x}+2x^{2}+6x+7)=4x^{2}\!$
$-8e^{x}+4e^{2x}+4+24e^{x}-6e^{2x}-12x-18-16e^{x}+2e^{2x}+4x^{2}+12x+14=4x^{2}\!$
$(-8+24-16)e^{x}+(4-6+2)e^{2x}+(-12+12)x+(-18+4+14)+4x^{2}=4x^{2}\!$
$4x^{2}=4x^{2}\!$
Therefore this solution is correct.
--Egm4313.s12.team11.sheider 21:30, 21 February 2012 (UTC)

## Problem 3.4

Solved by Daniel Suh

### Problem Statement

Use Basic Rule (1) and Sum Rule (3) to show that the appropriate particular solution for $y''-3y'+2y=4x^{2}-6x^{5}\!$ is of the form $\sum _{j=0}^{n}c_{j}x^{j}\!$, with n = 5.

### Solution
#### Finding the Particular Solution with Basic Rule and Sum Rule

We know that in standard form, for a particular solution, $y''_{p}+ay'_{p}+by_{p}=r(x)\!$ $y_{p}=y_{p1}+y_{p2}\!$
Using Basic Rule (1) and Sum Rule (3), we know that
$y''_{p}+ay'_{p}+by_{p}=\sum _{i}r_{i}(x)\!$
Additionally, we choose $r_{1}(x)\rightarrow y_{p1}(x)\!$ $r_{2}(x)\rightarrow y_{p2}(x)\!$

##### Solving for Particular Solution 1

Using the Method of Undetermined Coefficients, we find that $y_{p1}=C_{2}x^{2}+C_{1}x+C_{0}\!$ $y'_{p1}=2C_{2}x+C_{1}\!$ $y''_{p1}=2C_{2}\!$
Plugging $y_{p1}\!$ into $y''-3y'+2y=4x^{2}\!$ gives
$(2C_{2})-3(2C_{2}x+C_{1})+2(C_{2}x^{2}+C_{1}x+C_{0})=4x^{2}\!$
Solving for coefficients $2C_{2}=4\!$ $-6C_{2}+2C_{1}=0\!$ $-3C_{1}+2C_{0}+2C_{2}=0\!$
Results in $C_{2}=2\!$ $C_{1}=6\!$ $C_{0}=7\!$
$y_{p1}=2x^{2}+6x+7\!$

##### Solving for Particular Solution 2

Using the Method of Undetermined Coefficients, we find that $y_{p2}=K_{5}x^{5}+K_{4}x^{4}+K_{3}x^{3}+K_{2}x^{2}+K_{1}x+K_{0}\!$ $y'_{p2}=5K_{5}x^{4}+4K_{4}x^{3}+3K_{3}x^{2}+2K_{2}x+K_{1}\!$ $y''_{p2}=20K_{5}x^{3}+12K_{4}x^{2}+6K_{3}x+2K_{2}\!$
Plugging $y_{p2}\!$ into $y''-3y'+2y=-6x^{5}\!$ gives
$(20K_{5}x^{3}+12K_{4}x^{2}+6K_{3}x+2K_{2})-3(5K_{5}x^{4}+4K_{4}x^{3}+3K_{3}x^{2}+2K_{2}x+K_{1})+2(K_{5}x^{5}+K_{4}x^{4}+K_{3}x^{3}+K_{2}x^{2}+K_{1}x+K_{0})=-6x^{5}\!$
Solving for coefficients results in $K_{5}=-3\!$ $K_{4}=-22.5\!$ $K_{3}=-105\!$ $K_{2}=-337.5\!$ $K_{1}=-697.5\!$ $K_{0}=-708.75\!$
$y_{p2}=-3x^{5}-22.5x^{4}-105x^{3}-337.5x^{2}-697.5x-708.75\!$

##### Combining Particular Solutions

Plugging $y_{p1}\!$ and $y_{p2}\!$ into $y_{p}=y_{p1}+y_{p2}\!$ gives

#### Comparing to Summation Form

The particular solution in the summation form is $y_{p}=\sum _{j=0}^{n}c_{j}x^{j}\!$
if n=5, then $y_{p}=\sum _{j=0}^{5}c_{j}x^{j}\!$ $y'_{p}=\sum _{j=0}^{5}c_{j}*j*x^{(j-1)}\!$ $y''_{p}=\sum _{j=0}^{5}c_{j}*j*(j-1)*x^{(j-2)}\!$
Plugging the particular solution into $y''-3y'+2y=-6x^{5}\!$, the final solution gives
(check Problem 3.5 for explanation on how to get this answer)

#### Result

As you can see, the two particular solutions of $y_{p}\!$ are equal. Thus, using the Basic Rule (1) and Sum rule (3) does give you the correct particular solution for $y''-3y'+2y=4x^{2}-6x^{5}\!$, in the form of $y_{p}=\sum _{j=0}^{n}c_{j}x^{j}\!$
Created by [Daniel Suh] 20:57, 21 February 2012 (UTC)

## Problem 3.5

Solved by: Andrea Vargas

### Problem Statement

Given $y''-3y'+2y=4x^{2}-6x^{5}\!$
1. Obtain the coefficients of $x,x^{2},x^{3},x^{5}\!$ 2.Verify all equations by long-hand expansion. Use the series before adjusting the indexes.
3. Put the system of equations in an upper triangular matrix.
4. Solve for $c_{0},c_{1},c_{2},c_{3},c_{4},c_{5}\!$ by using back substitution.
5. Using the initial conditions $y(0)=1,y'(0)=0\!$ find $y(x)\!$ and plot it

### Solution

1. To obtain the equations for the coefficients we use the following equation from (1) p7-11:
$\sum _{j=0}^{3}[c_{j+2}(j+2)(j+1)-3c_{j+1}(j+1)+2c_{j}]x^{j}-3c_{5}(5)x^{4}+2[c_{4}x^{4}+c^{5}x^{5}]=4x^{2}-6x^{5}\!$
Finding the coefficients of $x\!$ where $j=1\!$:
$[c_{3}(3)(2)-3(c_{2})(2)+2c_{1}]x^{1}=[6c_{3}-6c_{2}+2c_{1}]x\!$
Finding the coefficients of $x^{2}\!$ where $j=2\!$:
$[c_{4}(4)(3)-3(c_{3})(3)+2c_{2}]x^{2}=[12c_{4}-9c_{3}+2c_{2}]x^{2}\!$
Finding the coefficients of $x^{3}\!$ where $j=3\!$:
$[c_{5}(5)(4)-3(c_{4})(4)+2c_{3}]x^{3}=[20c_{4}-12c_{3}+2c_{2}]x^{3}\!$
Finding the coefficients of $x^{5}\!$ where $j=5\!$:
$c_{5}x^{5}\!$
2. To verify all the above equations by long hand expansion we use the following equation from (4) p7-12: $\sum _{j=2}^{5}c_{j}\times j\times (j-1)\times x^{j-2}-3\sum _{j=1}^{5}c_{j}\times j\times x^{j-1}+2\sum _{j=0}^{5}c_{j}\times x^{j}=4x^{2}-6x^{5}\!$
Finding the coefficients when $j=0\!$:
$2c_{0}\!$
Finding the coefficients when $j=1\!$:
$-3c_{1}+2c_{1}x_{1}\!$
Finding the coefficients when $j=2\!$:
$2c_{2}-6c_{2}x+2c_{2}x^{2}\!$
Finding the coefficients when $j=3\!$:
$6c_{3}x-9c_{3}x^{2}+2c_{3}x^{3}\!$
Finding the coefficients when $j=5\!$:
$20c_{5}x^{3}-15c_{5}x^{4}+2c_{5}x^{5}\!$
By collecting these terms we can compare them to the equations of part 1.
Coefficients of $x\!$:
$[c_{3}(3)(2)-3(c_{2})(2)+2c_{1}]x^{1}=[6c_{3}-6c_{2}+2c_{1}]x\!$
Coefficients of $x^{2}\!$:
$[c_{4}(4)(3)-3(c_{3})(3)+2c_{2}]x^{2}=[12c_{4}-9c_{3}+2c_{2}]x^{2}\!$
Coefficients of $x^{3}\!$:
$[c_{5}(5)(4)-3(c_{4})(4)+2c_{3}]x^{3}=[20c_{4}-12c_{3}+2c_{2}]x^{3}\!$
Coefficients of $x^{5}\!$:
$c_{5}x^{5}\!$
We can see that we obtain the same system of equations to solve for the coefficients with both methods.
3.Constructing the coefficients matrix: ${\begin{bmatrix}2&-3&2&0&0&0\\0&2&-6&6&0&0\\0&0&2&-9&12&0\\0&0&0&2&-12&20\\0&0&0&0&2&-15\\0&0&0&0&0&2\end{bmatrix}}\!$
Then, the system becomes:
${\begin{bmatrix}2&-3&2&0&0&0\\0&2&-6&6&0&0\\0&0&2&-9&12&0\\0&0&0&2&-12&20\\0&0&0&0&2&-15\\0&0&0&0&0&2\end{bmatrix}}{\begin{bmatrix}c_{0}\\c_{1}\\c_{2}\\c_{3}\\c_{4}\\c_{5}\\\end{bmatrix}}={\begin{bmatrix}0\\0\\4\\0\\0\\-6\\\end{bmatrix}}\!$
4. Solving for the coefficients: $2c_{5}=-6\rightarrow c_{5}=-3\!$ $2c_{4}-(15)(-3)=0\rightarrow c_{4}={\frac {-45}{2}}\!$ $2c_{3}-12(-{\frac {45}{2}})+20(-3)=0\rightarrow c_{3}=-105\!$ $2c_{2}-9(-105)+12(-{\frac {45}{2}})=4\rightarrow c_{2}=-{\frac {671}{2}}\!$ $2c_{1}-6(-{\frac {671}{2}})+6(-105)=0\rightarrow c_{1}=-{\frac {1383}{2}}\!$ $2c_{0}-3(-{\frac {1383}{2}})+2(-{\frac {671}{2}})=0\rightarrow c_{0}=-{\frac {2807}{4}}\!$

#### Particular solution

This yields the particular solution:

#### Homogeneous Solution

$y''_{h}-y'_{h}+2y_{h}=0\!$ $\lambda ^{2}-3\lambda +2=0\!$ Then, we can find the characteristic equation: $(\lambda -2)(\lambda -1)\!$ $\lambda =2,1\!$ Then the solution for the homogeneous equation becomes:

#### General Solution

Using the given initial conditions $y(0)=1,y'(0)=0\!$ we find the overall solution:
$y(x)=y_{h}+y_{p}\!$ $y(x)=C_{1}e^{2x}+C_{2}e^{x}-3x^{5}-{\frac {45}{2}}x^{4}-105x^{3}-{\frac {671}{2}}x^{2}-{\frac {1383}{2}}x-{\frac {2807}{4}}\!$ $y'(x)=2C_{1}e^{2x}+C_{2}e^{x}-15x^{4}-90x^{3}-315x^{2}-671x-{\frac {1383}{2}}\!$ Using the initial conditions to solve for $C_{1}\!$ and $C_{2}\!$
$1=C_{1}+C_{2}-{\frac {2807}{4}}\!$ $0=C_{1}+C_{2}-{\frac {1383}{2}}\!$ $C_{1}=-{\frac {45}{4}}\;\;\;C_{2}=714\!$
The general solution becomes

### Plot

Below is a plot of the solution:
--Egm4313.s12.team11.vargas.aa 05:56, 21 February 2012 (UTC)

## Problem 3.6

Solved by Francisco Arrieta

### Problem 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-ODE-CC (see p. 7-2b)
       $y_{p,1}''-3y_{p,1}'+2y_{p,1}=r_{1}(x)=4x^{2}\!$
       $y_{p,2}''-3y_{p,2}'+2y_{p,2}=r_{2}(x)=-6x^{5}\!$
    The particular solution to $y_{p,1}\!$ had been found in R3.3 p.7-11.
    Find the particular solution $y_{p,2}\!$ , and then obtain the solution $y\!$ for the L2-ODE-CC (2) p.7-11 with initial conditions (2b) p.3-6.

### Given

    First particular solution
       $y_{p,1}=2x^{2}+6x+7\!$
    Initial Conditions
       $y(0)=1\!$
       $y'(0)=0\!$

### Solution
#### Particular Solution

    Since the specific excitation $r_{2}(x)=-6x^{5}\!$ , using table 2.1 from K 2011 p.82, the choice for the particular solution is $y_{p}(x)=\sum _{j=0}^{n}K_{j}x^{j}\!$
    Then
       $y_{p,2}=c_{0}+c_{1}x+c_{2}x^{2}+c_{3}x^{3}+c_{4}x^{4}+c_{5}x^{5}\!$
       $y_{p,2}'=c_{1}+2c_{2}x+3c_{3}x^{2}+4c_{4}x^{3}+5c_{5}x^{4}\!$
       $y_{p,2}''=2c_{2}+6c_{3}x+12c_{4}x^{2}+20c_{5}x^{3}\!$
    Plugging these equations back in the original $y_{p,2}''-3y_{p,2}'+2y_{p,2}=-6x^{5}\!$
       $(2c_{2}+6c_{3}x+12c_{4}x^{2}+20c_{5}x^{3})-3(c_{1}+2c_{2}x+3c_{3}x^{2}+4c_{4}x^{3}+5c_{5}x^{4})+2(c_{0}+c_{1}x+c_{2}x^{2}+c_{3}x^{3}+c_{4}x^{4}+c_{5}x^{5})=-6x^{5}\!$
    Using coefficient matching of left hand side to the right hand side of the equation, the following equations are obtained
       $c:2c_{2}-3c_{1}+2c_{0}=0\!$
       $x:6c_{3}-6c_{2}+2c_{1}=0\!$
       $x^{2}:12c_{4}-9c_{3}+2c_{2}=0\!$
       $x^{3}:20c_{5}-12c_{4}+2c_{3}=0\!$
       $x^{4}:-15c_{5}+2c_{4}=0\!$
       $x^{5}:2c_{5}=-6\!$
    Which is equivalent to the following matrix
       ${\begin{bmatrix}2&-3&2&0&0&0\\0&2&-6&6&0&0\\0&0&2&-9&12&0\\0&0&0&2&-12&20\\0&0&0&0&2&-15\\0&0&0&0&0&2\end{bmatrix}}{\begin{bmatrix}c_{0}\\c_{1}\\c_{2}\\c_{3}\\c_{4}\\c_{5}\end{bmatrix}}={\begin{bmatrix}0\\0\\0\\0\\0\\-6\end{bmatrix}}\!$
    Using back substitution method to solve for every coefficient, starting with $c_{5}\!$
       $2c_{5}=-6\rightarrow c_{5}=-3\!$
       $2c_{4}-15(-3)=0\rightarrow c_{4}=-{\frac {45}{2}}\!$
       $2c_{3}-12(-{\frac {45}{2}})+20(-3)=0\rightarrow c_{3}=-105\!$
       $2c_{2}-9(-105)+12(-{\frac {45}{2}})=0\rightarrow c_{2}=-{\frac {675}{2}}\!$
       $2c_{1}-6(-{\frac {675}{2}})+6(-105)=0\rightarrow c_{1}=-{\frac {1395}{2}}\!$
       $2c_{0}-3(-{\frac {1395}{2}})+2(-{\frac {675}{2}})=0\rightarrow c_{0}=-{\frac {2835}{4}}\!$
    Plugging these values back into $y_{p,2}=c_{0}+c_{1}x+c_{2}x^{2}+c_{3}x^{3}+c_{4}x^{4}+c_{5}x^{5}\!$ gives
    Since these equations are L2-ODE-CC, the superposition principle applies and $y_{p}=y_{p,1}+y_{p,2}\!$ and the general particular solution becomes

#### Homogeneous Solution

       $y_{h}''-3y_{h}'+2y_{h}=0\!$
    Due to the linearity of this homogeneous equation, a linear combination of two linear independent solutions is also a solution
       $y_{h,1}''-3y_{h,1}'+2y_{h,1}=0\!$
       $y_{h,2}''-3y_{h,2}'+2y_{h,2}=0\!$
    With solutions
       $y_{h,1}=e^{\lambda _{1}x}\!$
       $y_{h,2}=e^{\lambda _{2}x}\!$
    In order to determine the value of $\lambda _{1,2}\!$, the characteristic equation must be determine from the homogeneous equation
       $\lambda ^{2}-3\lambda +2=0\!$
             $\lambda _{1,2}={\frac {-a\pm {\sqrt {a^{2}-4b}}}{2}}\!$
             $\lambda _{1,2}={\frac {-(-3)\pm {\sqrt {(-3)^{2}-4(2)}}}{2}}\!$
                   $\lambda _{1}=2\;\;\;\;\;\;\;\lambda _{2}=1\!$
    Then the solutions for each distinct linearly independent homogeneous equation becomes
       $y_{h,1}=e^{2x}\!$
       $y_{h,2}=e^{x}\!$
    Because of linearity, the linear combination of the previous two equations times two constants that satisfy 2 initial conditions, is also a solution

#### General Solution

    The overall solution for the L2-ODE-CC
       $y(x)=y_{h}+y_{p}\!$
       $y(x)=C_{1}e^{2x}+C_{2}e^{x}-{\frac {2807}{4}}-{\frac {1383}{2}}x-{\frac {671}{2}}x^{2}-105x^{3}-{\frac {45}{2}}x^{4}-3x^{5}\!$
       $y'(x)=2C_{1}e^{2x}+C_{2}e^{x}-{\frac {1383}{2}}-671x-315x^{2}-90x^{3}-15x^{4}\!$
    Using the initial conditions to solve for $C_{1}\!$ and $C_{2}\!$
       $1=C_{1}+C_{2}-{\frac {2807}{4}}\!$
       $0=C_{1}+C_{2}-{\frac {1383}{2}}\!$
             $C_{1}=-{\frac {45}{4}}\;\;\;C_{2}=714\!$
    The general solution becomes
--Egm4313.s12.team11.arrieta 20:26, 19 February 2012 (UTC)

## Problem 3.7

Solved By Kyle Gooding

### Problem Statement

Expand the series on both sides of (1),(2) pg. 7-12b to verify these equalities. (1)

### Given

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

### Solutions

Expanding both sides of (1) results in: $\sum _{j=2}^{5}C_{j}*j(j-1)x^{j-2}=C_{2}(2)(2-1)x^{0}+C_{3}(3)(3-1)x^{1}+C_{4}(4)(4-1)x^{2}+C_{5}(5)(5-1)x^{3}$ $\sum _{j=0}^{3}C_{j}(j+2)(j+1)x^{j}=C_{0+2}(0+2)(0+1)x^{0}+C_{1+2}(1+2)(1+1)x^{1}+C_{2+2}(2+2)(2+1)x^{2}+C_{3+2}(3+2)(3+1)x^{3}$
Simplifying: $\sum _{j=2}^{5}C_{j}*j(j-1)x^{j-2}=C_{2}(2)+C_{3}6x+C_{4}12x^{2}+C_{5}20x^{3}$ $\sum _{j=0}^{3}C_{j}(j+2)(j+1)x^{j}=C_{2}(2)+C_{3}6x+C_{4}12x^{2}+C_{5}20x^{3}$
Expanding both sides of (2) results in: $\sum _{j=1}^{5}C_{j}*jx^{j-1}=C_{1}(1)x^{0}+C_{2}(2)x^{1}+C_{3}(3)x^{2}+C_{4}(4)x^{3}+C_{5}(5)x^{4}$ $\sum _{j=0}^{4}C_{j+1}*(j+1)x^{j}=C_{0+1}(0+1)x^{0}+C_{1+1}(1+1)x^{1}+C_{1+2}(2+1)x^{2}+C_{2+2}(3+1)x^{3}+C_{3+2}(4+1)x^{4}$
Simplifying: $\sum _{j=1}^{5}C_{j}*jx^{j-1}=C_{1}(1)+C_{2}(2)x^{1}+C_{3}(3)x^{2}+C_{4}(4)x^{3}+C_{5}(5)x^{4}$ $\sum _{j=1}^{5}C_{j}*jx^{j-1}=C_{1}(1)+C_{2}(2)x^{1}+C_{3}(3)x^{2}+C_{4}(4)x^{3}+C_{5}(5)x^{4}$
Egm4313.s12.team11.gooding 23:49, 19 February 2012 (UTC)

## Problem 3.8

solved by Luca Imponenti

### Kreyszig 2011 pg.84 problem 5
#### Problem Statement

Find a (real) general solution. State which rule you are using. Show each step of your work.
       $y''+4y'+4y=e^{-x}cos(x)\!$

#### Homogeneous Solution

To find the homogeneous solution, $y_{h}\!$, we must find the roots of the equation
$\lambda ^{2}+4\lambda +4=0\!$
$(\lambda +2)(\lambda +2)=0\!$
$\lambda =-2\!$
We know the homogeneous solution for the case of a double root to be
$y_{h}=c_{1}e^{\lambda x}+c_{2}xe^{\lambda x}\!$
       $y_{h}=c_{1}e^{-2x}+c_{2}xe^{-2x}\!$

#### Particular Solution

We have the following excitation
       $r(x)=e^{-x}cos(x)\!$
From table 2.1, K 2011, pg. 82, we have
       $y_{p}(x)=e^{-x}[Kcos(x)+Msin(x)]\!$
Since this does not correspond to our homogeneous solution we can use the Basic Rule (a), K 2011, pg. 81 to solve for the particular solution
$y_{p}''+4y_{p}'+4y_{p}=r(x)\!$
where
$y_{p}'=e^{-x}[-Ksin(x)+Mcos(x)]-e^{-x}[Kcos(x)+Msin(x)]\!$
$y_{p}'=e^{-x}[-Ksin(x)+Mcos(x)-Kcos(x)-Msin(x)]\!$
and
$y_{p}''=e^{-x}[-Kcos(x)-Msin(x)+Ksin(x)-Mcos(x)]-e^{-x}[Ksin(x)+Mcos(x)-Kcos(x)-Msin(x)]\!$
$y_{p}''=e^{-x}[-Kcos(x)-Msin(x)+Ksin(x)-Mcos(x)+Ksin(x)-Mcos(x)+Kcos(x)+Msin(x)]\!$
$y_{p}''=e^{-x}[2Ksin(x)-2Mcos(x)]\!$
Plugging these equations back into the differential equation
$y_{p}''+4y_{p}'+4y_{p}=r(x)\!$
$e^{-x}[2Ksin(x)-2Mcos(x)]+4e^{-x}[-Ksin(x)+Mcos(x)-Kcos(x)-Msin(x)]+4e^{-x}[Kcos(x)+Msin(x)]=e^{-x}cos(x)\!$
$e^{-x}[2Ksin(x)-2Mcos(x)-4Ksin(x)+4Mcos(x)-4Kcos(x)-4Msin(x)+4Kcos(x)+4Msin(x)]=e^{-x}cos(x)\!$
$2Mcos(x)-2Ksin(x)=cos(x)\!$
from the above equation it is obvious that $K=0\!$ and $M={\frac {1}{2}}\!$
therefore the particular solution to the differential equation is
       $y_{p}(x)={\frac {1}{2}}e^{-x}sin(x)\!$

#### General Solution

The general solution will be the summation of the homogeneous and particular solutions
$y(x)=y_{h}(x)+y_{p}(x)\!$
$y(x)=c_{1}e^{-2x}+c_{2}xe^{-2x}+{\frac {1}{2}}e^{-x}sin(x)\!$
The coefficients $c_{1}\!$ and $c_{2}\!$ can be readily solved for given either initial conditions or boundary value conditions.

### Kreyszig 2011 pg.84 problem 6
#### Problem Statement

Find a (real) general solution. State which rule you are using. Show each step of your work.
       $y''+y'+(\pi ^{2}+{\frac {1}{4}})y=e^{-{\frac {x}{2}}}sin(\pi x)\!$

#### Homogeneous Solution

To find the homogeneous solution, $y_{h}\!$, we must find the roots of the equation
$\lambda ^{2}+\lambda +(\pi ^{2}+{\frac {1}{4}})=0\!$
$\lambda ={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}\!$ with $a=1,b=1,c=(\pi ^{2}+{\frac {1}{4}})\!$
$\lambda ={\frac {-1\pm {\sqrt {1^{2}-4*1*(\pi ^{2}+{\frac {1}{4}})}}}{2*1}}\!$
$\lambda =-\alpha \pm i\omega =-{\frac {1}{2}}\pm \pi i\!$
We know the homogeneous solution for the case of a double root to be
$y_{h}=e^{-\alpha x}[Acos(\omega x)+Bsin(\omega x)]\!$
       $y_{h}=e^{-{\frac {x}{2}}}[Acos(\pi x)+Bsin(\pi x)]\!$

#### Particular Solution

We have the following excitation
       $r(x)=e^{-{\frac {x}{2}}}sin(\pi x)\!$
From table 2.1, K 2011, pg. 82, we have
       $y_{p}(x)=e^{-{\frac {x}{2}}}[Kcos(\pi x)+Msin(\pi x)]\!$
Since this corresponds to our homogeneous solution we must use the Modification Rule (b), K 2011, pg. 81 to solve for the particular solution
so $y_{p}(x)=xe^{-{\frac {x}{2}}}[Kcos(\pi x)+Msin(\pi x)]\!$
differentiating
$y_{p}'=\pi xe^{-{\frac {x}{2}}}[-Ksin(\pi x)+Mcos(\pi x)]+(e^{\frac {x}{2}}-{\frac {1}{2}}xe^{-{\frac {x}{2}}})[Kcos(\pi x)+Msin(\pi x)]\!$
$y_{p}'=e^{-{\frac {x}{2}}}(\pi x[-Ksin(\pi x)+Mcos(\pi x)]+(1-{\frac {1}{2}}x)[Kcos(\pi x)+Msin(\pi x)])\!$
$y_{p}'=e^{-{\frac {x}{2}}}[-\pi xKsin(\pi x)+\pi xMcos(\pi x)+Kcos(\pi x)+Msin(\pi x)-{\frac {1}{2}}xKcos(\pi x)-{\frac {1}{2}}xMsin(\pi x)]\!$
and
$y_{p}''=e^{-{\frac {x}{2}}}[-\pi ^{2}xKcos(\pi x)-\pi Ksin(\pi x)-\pi ^{2}xMsin(\pi x)+\pi Mcos(\pi x)-\pi Ksin(\pi x)+\pi Mcos(\pi x)+{\frac {1}{2}}\pi xKsin(\pi x)-{\frac {1}{2}}Kcos(\pi x)-{\frac {1}{2}}\pi xMcos(\pi x)-{\frac {1}{2}}Msin(\pi x)]-{\frac {1}{2}}e^{-{\frac {x}{2}}}[-\pi xKsin(\pi x)+\pi xMcos(\pi x)+Kcos(\pi x)+Msin(\pi x)-{\frac {1}{2}}xKcos(\pi x)-{\frac {1}{2}}xMsin(\pi x)]\!$
$y_{p}''=e^{-{\frac {x}{2}}}[-\pi ^{2}xKcos(\pi x)-\pi Ksin(\pi x)-\pi ^{2}xMsin(\pi x)+\pi Mcos(\pi x)-\pi Ksin(\pi x)+\pi Mcos(\pi x)+{\frac {1}{2}}\pi xKsin(\pi x)-{\frac {1}{2}}Kcos(\pi x)-{\frac {1}{2}}\pi xMcos(\pi x)-{\frac {1}{2}}Msin(\pi x)+{\frac {1}{2}}\pi xKsin(\pi x)-{\frac {1}{2}}\pi xMcos(\pi x)-{\frac {1}{2}}Kcos(\pi x)-{\frac {1}{2}}Msin(\pi x)+{\frac {1}{4}}xKcos(\pi x)+{\frac {1}{4}}xMsin(\pi x)]\!$
grouping cosine and sine terms we get
$y_{p}''=e^{-{\frac {x}{2}}}[(2\pi M-\pi xM+({\frac {1}{4}}-\pi ^{2})xK-K)cos(\pi x)+(({\frac {1}{4}}-\pi ^{2})xM-M+\pi xK-2\pi K)sin(\pi x)]\!$
and
$y_{p}'=e^{-{\frac {x}{2}}}[(\pi xM+K-{\frac {1}{2}}xK)cos(\pi x)+(-\pi xK+M-{\frac {1}{2}}xM)sin(\pi x)]\!$
next we substitute the above equations into the ODE
$y_{p}''+y_{p}'+(\pi ^{2}+{\frac {1}{4}})y_{p}=r(x)\!$
$e^{-{\frac {x}{2}}}[(2\pi M-\pi xM+({\frac {1}{4}}-\pi ^{2})xK-K)cos(\pi x)+(({\frac {1}{4}}-\pi ^{2})xM-M+\pi xK-2\pi K)sin(\pi x)]+e^{-{\frac {x}{2}}}[(\pi xM+K-{\frac {1}{2}}xK)cos(\pi x)+(-\pi xK+M-{\frac {1}{2}}xM)sin(\pi x)]+(\pi ^{2}+{\frac {1}{4}})xe^{-{\frac {x}{2}}}[Kcos(\pi x)+Msin(\pi x)]=e^{-{\frac {x}{2}}}sin(\pi x)\!$
$e^{-{\frac {x}{2}}}[(2\pi M-\pi xM+({\frac {1}{4}}-\pi ^{2})xK-K)cos(\pi x)+(({\frac {1}{4}}-\pi ^{2})xM-M+\pi xK-2\pi K)sin(\pi x)+(\pi xM+K-{\frac {1}{2}}xK)cos(\pi x)+(-\pi xK+M-{\frac {1}{2}}xM)sin(\pi x)+(\pi ^{2}+{\frac {1}{4}})x(Kcos(\pi x)+Msin(\pi x))]=e^{-{\frac {x}{2}}}sin(\pi x)\!$
$e^{-{\frac {x}{2}}}[(2\pi M-\pi xM+({\frac {1}{4}}-\pi ^{2})xK-K+\pi xM+K-{\frac {1}{2}}xK+(\pi ^{2}+{\frac {1}{4}})xK)cos(\pi x)+(({\frac {1}{4}}-\pi ^{2})xM-M+\pi xK-2\pi K-\pi xK+M-{\frac {1}{2}}xM+(\pi ^{2}+{\frac {1}{4}})xM)sin(\pi x)]=e^{-{\frac {x}{2}}}sin(\pi x)\!$
$2\pi Mcos(\pi x)-2\pi Ksin(\pi x)=sin(\pi x)\!$
after cancelling terms; we can equate cosine and sine coefficients to get two equations
$2\pi M=0\!$
$-2\pi K=1\!$
so $M=0\!$ and $K=-{\frac {1}{2\pi }}\!$
and the particular solution to the ODE is
       $y_{p}(x)=-{\frac {1}{2\pi }}xe^{-{\frac {x}{2}}}cos(\pi x)\!$

#### General Solution

The general solution will be the summation of the homogeneous and particular solutions
$y=y_{h}+y_{p}\!$
$y=e^{-{\frac {x}{2}}}[Acos(\pi x)+Bsin(\pi x)]-{\frac {1}{2\pi }}xe^{-{\frac {x}{2}}}cos(\pi x)\!$
Egm4313.s12.team11.imponenti 04:28, 21 February 2012 (UTC)

## Problem 3.9

Solved by Gonzalo Perez

#### Problem Statement

Solve the initial value problem. State which rule you are using. Show each step of your calculation in detail.

#### Given
#### (K 2011 pg.85 #13)

$8y''-6y'+y=6\cosh x\!$ (1)
Initial conditions are:
$y(0)=0.2,y'(0)=0.05\!$

#### Solution

The general solution of the homogeneous ordinary differential equation is
$8y''-6y'+y=0\!$
We can use this information to determine the characteristic equation:
$8\lambda ^{2}-6\lambda +1=0\!$
And proceeding to find the roots,
$4\lambda (2\lambda -1)-1(2\lambda -1)=0\!$
Thus, $(4\lambda -1)(2\lambda -1)=0\!$.
Solving for the roots, we find that $\lambda ={\frac {1}{4}},{\frac {1}{2}},\!$
where the general solution is
$y_{k}=c_{1}e^{{\frac {1}{4}}x}+c_{2}e^{{\frac {1}{2}}x}\!$.
The solution of $y_{p}\!$ of the non-homogeneous ordinary differential equation is
$x={\frac {e^{x}+e^{-x}}{2}}\!$.
Using the Sum rule as described in Section 2.7, the above function translates into the following:
$y_{p}=y_{p1}+y_{p2}\!$, where Table 2.1 tells us that:
$y_{p1}=Ae^{x}\!$ and $y_{p2}=Be^{x}\!$.
Therefore, $y_{p}=Ae^{x}+Be^{-x}\!$.
Now, we can substitute the values ($y_{p},y_{p}',y_{p}''\!$) into (1) to get:
$8(Ae^{x}+Be^{-x})-6(Ae^{x}-Be^{-x})+Ae^{x}+Be^{-x}=6({\frac {e^{x}+e^{-x}}{2}})\!$
$=3Ae^{x}+15Be^{-x}\!$
$=3(e^{x}+e^{-x})\!$
Now that we have this equation, we can equate coefficients to find that:
$3A=3\!$
$\therefore A=1\!$
$B={\frac {3}{15}}={\frac {1}{5}}\$
and thus, $y_{p}=e^{x}+{\frac {1}{5}}e^{-x}\!$
We find that the general solution is in fact:
$y=y_{k}+y_{p}\!$
$y=c_{1}e^{\frac {1}{4}}x+c_{2}e^{\frac {1}{2}}x+e^{x}+3e^{-x}\!$
whereas the general solution of the given ordinary differential equation is actually:
$y=c_{1}e^{\frac {1}{4}}x+c_{2}e^{\frac {1}{2}}x+e^{x}+{\frac {1}{5}}e^{-x}\!$
Solving for the initial conditions given and first plugging in $y(0)=0.2\!$, we get that:
$0.2=c_{1}e^{{\frac {1}{4}}(0)}+c_{2}e^{{\frac {1}{2}}(0)}+e^{0}+3e^{(0)}\!$
$0.2=c_{1}e^{(0)}+c_{2}e^{(0)}+e^{(0)}+3e^{(0)}\!$
$0.2=c_{1}+c_{2}+1+{\frac {1}{5}}\!$
$\therefore c_{1}+c_{2}=-1\!$. (2)
And now we can determine the first order ODE :
$y'={\frac {1}{4}}c_{1}e^{{\frac {1}{4}}(x)}+{\frac {1}{2}}c_{2}e^{{\frac {1}{2}}(x)}+e^{x}-{\frac {1}{5}}e^{x}\!$
The second initial condition that was given to us, $y'(0)=0.05\!$ can now be plugged in:
$0.05={\frac {1}{4}}c_{1}e^{{\frac {1}{4}}(0)}+{\frac {1}{2}}c_{2}e^{{\frac {1}{2}}(0)}+e^{(0)}-{\frac {1}{5}}e^{(0)}\!$
$0.05={\frac {1}{4}}c_{1}e^{(0)}+{\frac {1}{2}}c_{2}e^{(0)}+e^{(0)}-{\frac {1}{5}}e^{(0)}\!$
$0.05={\frac {1}{4}}c_{1}+{\frac {1}{2}}c_{2}+1-{\frac {1}{5}}\!$
${\frac {1}{4}}c_{1}+{\frac {1}{2}}c_{2}=-0.75\!$
$\therefore c_{1}+2c_{2}=-3\!$ (3)
Once we solve (2) and (3), we can get the values:
$c_{1}=1,c_{2}=-2\!$.
And once we substitute these values, we get the following solution for this IVP:

#### Given
#### (K 2011 pg.85 #14)

$y''+4y'+4y=e^{-2x}sin2x\!$ (1)
Initial conditions are:
$y(0)=1,y'(0)=-1.5\!$

#### Solution

The general solution of the homogeneous ordinary differential equation is
$y''+4y'+4y=0\!$
We can use this information to determine the characteristic equation:
$\lambda ^{2}+4\lambda +4=0\!$
And proceeding to find the roots,
$(\lambda +2)(\lambda +2)=0\!$
Solving for the roots, we find that $\lambda =-2,-2\!$
where the general solution is:
$y_{k}=c_{1}e^{-2x}+c_{2}e^{-2x}x\!$, or:
$y_{k}=(c_{1}+c_{2}x)e^{-2x}\!$
Now, according to the Modification Rule and Table 2.1 in Section 2.7, we know that we have to multiply by x to get:
$y_{p}=e^{-2x}(Kxcos2x+Mxsin2x)\!$, since the solution of $y_{k}\!$ is a double root of the characteristic equation.
We can then derive to get $y_{p}'\!$:
$y_{p}'=-2e^{-2x}(Kxcos2x+Mxsin2x)+e^{-2x}(Kcos2x-2Kxsin2x+Msin2x+2Mxcos2x)\!$
$y_{p}'=(-2K+2M)e^{-2x}xcos2x+(-2M-2K)e^{-}2xxsin2x+Ke^{-2x}cos2x+Me^{-2x}sinx\!$
Deriving once again to solve for $y_{p}''\!$, we get the following:
$y_{p}''=4e^{-2x}(Kxcos2x+Mxsin2x)-2e^{-2x}(Kcos2x-2Kxsin2x+Msin2x+2Mxcos2x)-2e^{-2x}(Kcos2x-2Kxsin2x+Msin2x+2Mxcos2x)\!$
$y_{p}''=4e^{-2x}(Kxcos2x+Mxsin2x)-4e^{-2x}(Kcos2x-2Kxsin2x+Msin2x+2Mxcos2x)+e^{-2x}(-4Ksin2x-4Kxcos2x+4Mcos2x-4Mxsin2x)\!$
$y_{p}''=(-4K+4M)e^{-2x}cos2x+(-4K-4M)e^{-2x}sin2x\$
Now, we can substitute the values ($y_{p},y_{p}',y_{p}''\!$) into (1) to get:
$(-4K+4M)e^{-2x}cos2x+(-4M-4K)e^{-2x}sin2x+4((-2K+2M)e^{-2x}xcos2x+(-2M-2K)e^{-2x}xsin2x+Ke^{-2x}cos2x+Me^{-2x}sinx)+4(e^{-2x}(Kxcos2x+Mxsin2x))=e^{2x}sin2x\!$
$\therefore (-3K+4M)e^{-2x}cos2x+(-3M-4K)e^{-2x}sin2x=e^{-2x}sin2x\!$
Now that we have this equation, we can equate coefficients to find that:
$-3K+4M=0\!$ and $-4K-3M=1\!$
and finally discover that:
$M=-{\frac {3}{25}}\!$ and $K=-{\frac {4}{25}}\!$.
Plugging in these values in $y_{p}\!$, we find that:
$y_{p}=e^{-2x}(-{\frac {4}{25}}xcos2x-{\frac {3}{25}}xsin2x)\!$
And finally, we arrive at the general solution of the given ordinary differential equation:
$y=y_{k}+y_{p}\!$
$y=(c_{1}+c_{2}x)e^{-2x}+e^{-2x}(-{\frac {4}{25}}xcos2x-{\frac {3}{25}}xsin2x)\!$
Solving for the initial conditions given and first plugging in $y(0)=1\!$, we get that:
$1=(c_{1}+c_{2}(0))e^{-2(0)}+e^{-2(0)}(-{\frac {4}{25}}(0)cos2(0)-{\frac {3}{25}}(0)sin2(0))\!$
$1=(c_{1}+c_{2}(0))e^{0}\!$
$\therefore c_{1}=1\!$
The second initial condition that was given to us, $y'(0)=-1.5\!$ can now be plugged in:
$y'=-{\frac {1}{5}}e^{-}2x(10c_{1}+10c_{2}x-5c_{2}+(3-14x)sin2x+(4-2x)cos(2x))\!$
$-1.5=-{\frac {1}{5}}(10c_{1}-5c_{2}+4)\!$
$\therefore c_{2}=-3.5\!$
And once we substitute these values, we get the following solution for this IVP:
Egm4313.s12.team11.perez.gp 20:34, 22 February 2012 (UTC)