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

## 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)