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

Problem 3.5

Solved by: Andrea Vargas

Problem Statement

Given $y^{″}-3y^{\prime }+2y=4x^2-6x^5$

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 $y(0)=1,y^{\prime }(0)=0$ 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:
$\left[\begin{array}{cccccc}2& -3& 2& 0& 0& 0\\ 0& 2& -6& 6& 0& 0\\ 0& 0& 2& -9& 12& 0\\ 0& 0& 0& 2& -12& 20\\ 0& 0& 0& 0& 2& -15\\ 0& 0& 0& 0& 0& 2\end{array}\right]$

Then, the system becomes:

$\left[\begin{array}{cccccc}2& -3& 2& 0& 0& 0\\ 0& 2& -6& 6& 0& 0\\ 0& 0& 2& -9& 12& 0\\ 0& 0& 0& 2& -12& 20\\ 0& 0& 0& 0& 2& -15\\ 0& 0& 0& 0& 0& 2\end{array}\right]\left[\begin{array}{c}{c}_0\\ c_1\\ c_2\\ c_3\\ c_4\\ c_5\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 4\\ 0\\ 0\\ -6\end{array}\right]$

4. Solving for the coefficients:
$2c_5=-6\to c_5=-3$
$2c_4-(15)(-3)=0\to c_4=\frac{-45}{2}$
$2c_3-12(-\frac{45}{2})+20(-3)=0\to c_3=-105$
$2c_2-9(-105)+12(-\frac{45}{2})=4\to c_2=-\frac{671}{2}$
$2c_1-6(-\frac{671}{2})+6(-105)=0\to c_1=-\frac{1383}{2}$
$2c_0-3(-\frac{1383}{2})+2(-\frac{671}{2})=0\to c_0=-\frac{2807}{4}$

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

$y{\prime }^{}{\text{'}}_h-y{\text{'}}_h+2y_h=0$
${\lambda }^2-3\lambda +2=0$
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 $y(0)=1,y^{\prime }(0)=0$ we find the overall solution:

$y(x)=y_h+y_p$
$y(x)=C_1{e}^{2x}+C_2{e}^x-3x^5-\frac{45}{2}{x}^4-105x^3-\frac{671}{2}{x}^2-\frac{1383}{2}x-\frac{2807}{4}$
$y^{\prime }(x)=2C_1{e}^{2x}+C_2{e}^x-15x^4-90x^3-315x^2-671x-\frac{1383}{2}$
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:

File:R3 5.jpg

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