User:Egm4313.s12.team11.arrieta/Report 3

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


 * $$ y(0)=1 \!$$
 * $$ y'(0)=0 \!$$

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 $$ y_p(x)=\sum_{j=0}^{n}K_jx^j \!$$


 * Then


 * $$ y_{p,2}=c_0+c_1x+c_2x^2+c_3x^3+c_4x^4+c_5x^5 \!$$
 * $$ y_{p,2}'=c_1+2c_2x+3c_3x^2+4c_4x^3+5c_5x^4 \!$$
 * $$ y_{p,2}''=2c_2+6c_3x+12c_4x^2+20c_5x^3 \!$$


 * Plugging these equations back in the original $$ y_{p,2}''-3y_{p,2}'+2y_{p,2}=-6x^5 \!$$


 * $$ (2c_2+6c_3x+12c_4x^2+20c_5x^3)-3(c_1+2c_2x+3c_3x^2+4c_4x^3+5c_5x^4)+2(c_0+c_1x+c_2x^2+c_3x^3+c_4x^4+c_5x^5)=-6x^5 \!$$


 * 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 $$ c_5 \!$$


 * $$ 2c_5=-6 \rightarrow c_5=-3 \!$$
 * $$ 2c_4-15(-3)=0 \rightarrow c_4=-\frac{45}{2} \!$$
 * $$ 2c_3-12(-\frac{42}{2})+20(-3)=0 \rightarrow c_3=-105 \!$$
 * $$ 2c_2-9(-105)+12(-\frac{42}{2})=0 \rightarrow c_2=-\frac{675}{2} \!$$
 * $$ 2c_1-6(-\frac{675}{2})+6(-105)=0 \rightarrow c_1=-\frac{1395}{2} \!$$
 * $$ 2c_0-3(-\frac{1395}{2})+2(-\frac{675}{2})=0 \rightarrow c_0=-\frac{2835}{4} \!$$


 * Plugging these values back into $$ y_{p,2}=c_0+c_1x+c_2x^2+c_3x^3+c_4x^4+c_5x^5 \!$$ 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


 * $$ y_{h,1}=e^{\lambda_1x} \!$$
 * $$ y_{h,2}=e^{\lambda_2x} \!$$


 * 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

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

General Solution

 * The overall solution for the L2-ODE-CC


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


 * 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)