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

                    $y(x)=e^{5x}({\frac {496}{125}}-{\frac {3113}{125}}x+{\frac {7}{2}}x^{2})-{\frac {2}{25}}x^{2}+{\frac {8}{125}}x+{\frac {4}{125}}\!$

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

$(\lambda -\lambda _{1})(\lambda -(\lambda _{1}))=0\!$

$(\lambda -x)(\lambda -(x+\epsilon ))=0\!$

$\lambda ^{2}-\lambda x-\lambda \epsilon -\lambda x+x^{2}+x\epsilon =0\!$

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

                   $\therefore y''-y'(2\lambda +\epsilon )+y\lambda (\lambda +\epsilon )=0\!$ (1)

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:

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

$\therefore 0\equiv 0.\!$

      Since the left and right hand sides of the equation are zero, the solution is in fact a homogeneous equation.

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.

$\lim _{\epsilon \rightarrow 0}{\frac {f'(\epsilon )}{g'(\epsilon )}}=\lim _{\epsilon \rightarrow 0}{\frac {xe^{x(\lambda +\epsilon )}}{1}}\!$

$={\frac {xe^{(\lambda +0)x}}{1}}\!$

                                            $=xe^{x\lambda }\!$

Given

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

Solution

Taking the derivative with respect to lambda, we find that:

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

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:

                           ${\frac {d(e^{\lambda x})}{d\lambda }}=\lim _{e\rightarrow 0}{\frac {e^{(\lambda +\epsilon )x}-e^{\lambda x}}{\epsilon }}=xe^{\lambda x}\!$

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

             After performing these calculations, from (2) we get 148.478.

             And from the exact second homogeneous solution, we get 200.05.

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:

                                                       $y_{h}=c_{1}e^{x}+c_{2}e^{2x}\!$

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:

                                                       $y_{p}=2x^{2}+6x+7\!$ 

By superposition, we can find the complete general solution:

$y=y_{h}+y_{p}\!$

                                                  $y=c_{1}e^{x}+c_{2}e^{2x}+2x^{2}+6x+7\!$ 

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:

                                                  $y(x)=-8e^{x}+e^{2x}+2x^{2}+6x+7\!$

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

               $y_{p}=-3x^{5}-22.5x^{4}-105x^{3}-335.5x^{2}-691.5x-701.75\!$

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

         $y_{p}=-3x^{5}-22.5x^{4}-105x^{3}-335.5x^{2}-691.5x-701.75\!$

(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:

                                                            $y_{p}(x)=-3x^{5}-{\frac {45}{2}}x^{4}-105x^{3}-{\frac {671}{2}}x^{2}-{\frac {1383}{2}}x-{\frac {2807}{4}}\!$

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:

                                                                                      $y_{h}(x)=C_{1}e^{2x}+C_{2}e^{x}\!$

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

                                                     $y(x)=-{\frac {45}{4}}e^{2x}+714e^{x}-3x^{5}-{\frac {45}{2}}x^{4}-105x^{3}-{\frac {671}{2}}x^{2}-{\frac {1383}{2}}x-{\frac {2807}{4}}\!$

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

                                        $y_{p,2}=-{\frac {2835}{4}}-{\frac {1395}{2}}x-{\frac {675}{2}}x^{2}-105x^{3}-{\frac {45}{2}}x^{4}-3x^{5}\!$

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

                                         $y_{p}(x)=-{\frac {2807}{4}}-{\frac {1383}{2}}x-{\frac {671}{2}}x^{2}-105x^{3}-{\frac {45}{2}}x^{4}-3x^{5}\!$

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

                                                    $y_{h}(x)=C_{1}e^{2x}+C_{2}e^{x}\!$

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

                 $y(x)=-{\frac {45}{4}}e^{2x}+714e^{x}-{\frac {2807}{4}}-{\frac {1383}{2}}x-{\frac {671}{2}}x^{2}-105x^{3}-{\frac {45}{2}}x^{4}-3x^{5}\!$

--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:
$\sum _{j=2}^{5}C_{j}*j(j-1)x^{j-2}=C_{2}(2)+C_{3}6x+C_{4}12x^{2}+C_{5}20x^{3}$
$\sum _{j=0}^{3}C_{j}(j+2)(j+1)x^{j}=C_{2}(2)+C_{3}6x+C_{4}12x^{2}+C_{5}20x^{3}$

 The two sums are equal.

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

 The two sums are equal.

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

$y(x)=y_{h}(x)+y_{p}(x)\!$

$y(x)=c_{1}e^{-2x}+c_{2}xe^{-2x}+{\frac {1}{2}}e^{-x}sin(x)\!$

   $y(x)=e^{-2x}(c_{1}+c_{2}x)+{\frac {1}{2}}e^{-x}sin(x)\!$

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

$y=y_{h}+y_{p}\!$

$y=e^{-{\frac {x}{2}}}[Acos(\pi x)+Bsin(\pi x)]-{\frac {1}{2\pi }}xe^{-{\frac {x}{2}}}cos(\pi x)\!$

   $y=e^{-{\frac {x}{2}}}[Acos(\pi x)+Bsin(\pi x)-{\frac {1}{2\pi }}xcos(\pi x)]\!$

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.

(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:

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

(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:

                $y=(1-3.5x)e^{-2x}+e^{-2x}(-{\frac {4}{25}}xcos2x-{\frac {3}{25}}xsin2x)\!$

Egm4313.s12.team11.perez.gp 20:34, 22 February 2012 (UTC)