# University of Florida/Egm4313/s12 Report 3, Problem 3.5

## Problem 3.5

Solved by: Andrea Vargas

### Problem Statement

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

1. Obtain the coefficients of ${\displaystyle x,x^{2},x^{3},x^{5}\!}$
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 ${\displaystyle c_{0},c_{1},c_{2},c_{3},c_{4},c_{5}\!}$ by using back substitution.

5. Using the initial conditions ${\displaystyle y(0)=1,y'(0)=0\!}$ find ${\displaystyle y(x)\!}$ and plot it

### Solution

1. To obtain the equations for the coefficients we use the following equation from (1) p7-11:

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

Finding the coefficients of ${\displaystyle x\!}$ where ${\displaystyle j=1\!}$:

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

Finding the coefficients of ${\displaystyle x^{2}\!}$ where ${\displaystyle j=2\!}$:

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

Finding the coefficients of ${\displaystyle x^{3}\!}$ where ${\displaystyle j=3\!}$:

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

Finding the coefficients of ${\displaystyle x^{5}\!}$ where ${\displaystyle j=5\!}$:

${\displaystyle c_{5}x^{5}\!}$

2. To verify all the above equations by long hand expansion we use the following equation from (4) p7-12:
${\displaystyle \sum _{j=2}^{5}c_{j}\times j\times (j-1)\times x^{j-2}-3\sum _{j=1}^{5}c_{j}\times j\times x^{j-1}+2\sum _{j=0}^{5}c_{j}\times x^{j}=4x^{2}-6x^{5}\!}$

Finding the coefficients when ${\displaystyle j=0\!}$:

${\displaystyle 2c_{0}\!}$

Finding the coefficients when ${\displaystyle j=1\!}$:

${\displaystyle -3c_{1}+2c_{1}x_{1}\!}$

Finding the coefficients when ${\displaystyle j=2\!}$:

${\displaystyle 2c_{2}-6c_{2}x+2c_{2}x^{2}\!}$

Finding the coefficients when ${\displaystyle j=3\!}$:

${\displaystyle 6c_{3}x-9c_{3}x^{2}+2c_{3}x^{3}\!}$

Finding the coefficients when ${\displaystyle j=5\!}$:

${\displaystyle 20c_{5}x^{3}-15c_{5}x^{4}+2c_{5}x^{5}\!}$

By collecting these terms we can compare them to the equations of part 1.

Coefficients of ${\displaystyle x\!}$:

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

Coefficients of ${\displaystyle x^{2}\!}$:

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

Coefficients of ${\displaystyle x^{3}\!}$:

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

Coefficients of ${\displaystyle x^{5}\!}$:

${\displaystyle c_{5}x^{5}\!}$

We can see that we obtain the same system of equations to solve for the coefficients with both methods.

3.Constructing the coefficients matrix:
${\displaystyle {\begin{bmatrix}2&-3&2&0&0&0\\0&2&-6&6&0&0\\0&0&2&-9&12&0\\0&0&0&2&-12&20\\0&0&0&0&2&-15\\0&0&0&0&0&2\end{bmatrix}}\!}$

Then, the system becomes:

${\displaystyle {\begin{bmatrix}2&-3&2&0&0&0\\0&2&-6&6&0&0\\0&0&2&-9&12&0\\0&0&0&2&-12&20\\0&0&0&0&2&-15\\0&0&0&0&0&2\end{bmatrix}}{\begin{bmatrix}c_{0}\\c_{1}\\c_{2}\\c_{3}\\c_{4}\\c_{5}\\\end{bmatrix}}={\begin{bmatrix}0\\0\\4\\0\\0\\-6\\\end{bmatrix}}\!}$

4. Solving for the coefficients:
${\displaystyle 2c_{5}=-6\rightarrow c_{5}=-3\!}$
${\displaystyle 2c_{4}-(15)(-3)=0\rightarrow c_{4}={\frac {-45}{2}}\!}$
${\displaystyle 2c_{3}-12(-{\frac {45}{2}})+20(-3)=0\rightarrow c_{3}=-105\!}$
${\displaystyle 2c_{2}-9(-105)+12(-{\frac {45}{2}})=4\rightarrow c_{2}=-{\frac {671}{2}}\!}$
${\displaystyle 2c_{1}-6(-{\frac {671}{2}})+6(-105)=0\rightarrow c_{1}=-{\frac {1383}{2}}\!}$
${\displaystyle 2c_{0}-3(-{\frac {1383}{2}})+2(-{\frac {671}{2}})=0\rightarrow c_{0}=-{\frac {2807}{4}}\!}$

#### Particular solution

This yields the particular solution:

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


#### Homogeneous Solution

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

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


#### General Solution

Using the given initial conditions ${\displaystyle y(0)=1,y'(0)=0\!}$ we find the overall solution:

${\displaystyle y(x)=y_{h}+y_{p}\!}$
${\displaystyle y(x)=C_{1}e^{2x}+C_{2}e^{x}-3x^{5}-{\frac {45}{2}}x^{4}-105x^{3}-{\frac {671}{2}}x^{2}-{\frac {1383}{2}}x-{\frac {2807}{4}}\!}$
${\displaystyle y'(x)=2C_{1}e^{2x}+C_{2}e^{x}-15x^{4}-90x^{3}-315x^{2}-671x-{\frac {1383}{2}}\!}$
Using the initial conditions to solve for ${\displaystyle C_{1}\!}$ and ${\displaystyle C_{2}\!}$

${\displaystyle 1=C_{1}+C_{2}-{\frac {2807}{4}}\!}$
${\displaystyle 0=C_{1}+C_{2}-{\frac {1383}{2}}\!}$
${\displaystyle C_{1}=-{\frac {45}{4}}\;\;\;C_{2}=714\!}$

The general solution becomes

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


### Plot

Below is a plot of the solution:

--Egm4313.s12.team11.vargas.aa 05:56, 21 February 2012 (UTC)