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:

$$\sum_{j=0}^{3}[c_{j+2}(j+2)(j+1)-3c_{j+1}(j+1)+2c_j]x^j-3c_5(5)x^4+2[c_4x^4+c^5x^5]=4x^2-6x^5 \!$$

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 $$j=5\!$$:

$$c_5x^5\!$$

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

Finding the coefficients when $$j=0\!$$:

$$2c_0\!$$

Finding the coefficients when $$j=1\!$$:

$$-3c_1+2c_1x_1\!$$

Finding the coefficients when $$j=2\!$$:

$$2c_2-6c_2x+2c_2x^2\!$$

Finding the coefficients when $$j=3\!$$:

$$6c_3x-9c_3x^2+2c_3x^3\!$$

Finding the coefficients when $$j=5\!$$:

$$20c_5x^3-15c_5x^4+2c_5x^5\!$$

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 $$x^5\!$$:

$$c_5x^5\!$$

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: $$y_h(x)=C_1e^{2x}+C_2e^{x}\!$$

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

$$ y(x)=y_h+y_p \!$$ $$ y(x)=C_1e^{2x}+C_2e^x-3x^5-\frac{45}{2}x^4-105x^3-\frac{671}{2}x^2-\frac{1383}{2}x-\frac{2807}{4} \!$$ $$ y'(x)=2C_1e^{2x}+C_2e^x-15x^4-90x^3-315x^2-671x-\frac{1383}{2}\!$$ 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)