User:Egm6321.f11.team1.steele.m/Report5

Homework 5 Problems Assigned to Manuel Steele (in pairs)

R5.6 Question from meeting 31, page 31-6: Given the trial solution (5) from p. 30-4 in Method 2 $$y = x^{r},\ r=\text{constant}$$ $$\displaystyle (Equation\;5.6.1)$$

Find that the trial solution is equivalent to the combined trial solutions (3)-(4) p.30-4 in Method 1.

$$x=e^t$$ $$\displaystyle (Equation\;5.6.2)$$ $$y=e^{rt},\ r=\text{constant}$$ $$\displaystyle (Equation\;5.6.3)$$

Solution
Equations 5.6.2 and 5.6.3 are used in method 1 to solve the Euler Ln-ODE-VC below. Equation 5.6.1 represents method 2 for solving the same equation.

$$a_nx^ny^{(n)} + a_{n-1}x^{(n-1)}y^{(n-1)} + \cdots + a_1xy' + a_0y = 0$$ $$\displaystyle (Equation\;5.6.4)$$ Note that $$y' = y^{(1)}$$ and $$y = y^{(0)}$$.

'''Work on a solution is still in progress as of 2:00 AM CST (11/1/11). Team members can update as needed.'''

R5.8 Question from meeting 32, page 32-2: Given: recall p. 12-2, R*2.17 $$y' + a_0(x)y = 0$$ $$\displaystyle (Equation\;5.8.1)$$

Find: use the same idea of variation of constants (parameters) to find the particular solution $$Y_P(x)$$ after knowing the homogeneous solution $$Y_H(x)$$, i.e., let y(x) = A(x)$$y_H(x)$$, with A(x) being the unknown to be found.

From lecture notes [[media:pea1.f11.mtg32.djvu|Mtg 32]]

Solution
The key is to start with equation (3) on p. 11-3 below. $$\underbrace{\color{blue}{1}}_{\color{blue}{a_1(x)}} \cdot \underbrace{y'}_{\color{blue}{y^{(1)}}} + \underbrace{\frac{Q(x)}{P(x)}}_{\color{blue}{a_0(x)}}y^{\color{blue}{(0)}}=\underbrace{\frac{R(x)}{P(x)}}_{\color{blue}{b(x)}}$$ $$\displaystyle (Equation\;5.8.2)$$ The equation is an L1-ODE-VC. Note that the non-homogeneous equation above utilizes the following integration factor for the solution of y. But we must find the homogenous counterpart to this solution

$$y(x) = \frac{1}{h(x)} \left[\int^x h(s)b(s)ds + k_2 \right]$$ $$\displaystyle (Equation\;5.8.4)$$

The homogeneous solution of the L1-ODE-VC was derived in R2.17 by team 1. The solution summary is repeated for clarity. Note that team 1 has already submitted this solution for R2.17 back in Report 2. The steps are from a simple cross-reference used to define the homogeneous solution of 5.8.2  $$\displaystyle \frac{y'}{y} = -{a}_{0} $$ $$\displaystyle (Eq 2.17.1) $$

Integration of both sides of this equation gives

<p style="text-align:center;"> $$ \ln \left( y\right)= -\int^{x} {a}_{0} \left( s\right)ds + C$$   <p style="text-align:right;">$$\displaystyle (Eq 2.17.2)$$

Solving for $$y$$ gives

<p style="text-align"center;">$$ y_H= \exp\left[ -\int^{x} {a}_{0} \left( s\right)ds + C \right]= A\exp\left[ -\int^{x} {a}_{0} \left( s\right)ds\right]$$

Now, the homogenous solution is known from 2.17. The next step is to find the complete solution for the homogenous equation- $$y(x) = A(x)y_H(x)$$. The technique demonstrated in the lecture is to substitute the chosen form of y(x) into the original Euler equation to find the complete solution below. This is a classic application of the variation of parameters: (1) find homogenous solution, (2) find complete solution of Euler form - the L1-ODE-VC, and (3) find the particular solution by using (1) and (2).

<p style="font-size:125%" align="center">$$y' + a_0(x)y = b(x) = 0$$ <p style="text-align:right;">$$\displaystyle (Equation\;5.8.4)$$ The derivative simplifies the Euler equation. First, it must be derived. <p style="font-size:125%" align="center">$$y = A(x)y_H$$ <p style="font-size:125%" align="center">$$y' = A_xy_H + Ay_{Hx}$$ The values can be substituted into the L1-ODE-VC to give the form below. <p style="font-size:125%" align="center">$$A_xy_H + Ay_{Hx} + a_0(x)Ay_H = 0$$ The equation can be rerranged to give a simple form for integration by natural logarithms and subsequent exponentiation for the complete solution of the L1-ODE-VC. <p style="font-size:125%" align="center">$$A_xy_H = -A(y_{Hx} + a_0(x)y_H$$ <p style="font-size:125%" align="center">$$\frac{A_x}{A} = -\frac{y_{Hx} + a_0(x)y_H}{y_H} = -\frac{y_{Hx}}{y_H} + a_0(x)$$ The rearranged equation can be integrated by standard application of natural logarithms. <p style="font-size:125%" align="center">$$\int^x \frac{A_x}{A} = -\int^x \frac{y_{Hx}}{y_H} + \int^x a_0(s)ds$$

Use the identity for logarithms: $$\ln \frac{a}{b} = \ln{a} - \ln{b}$$, which gives $$\ln \frac{A(x)}{y_H} = \int^x a_0(s)ds$$. The equation then gives the next output form. <p style="font-size:125%" align="center">$$A(x) = y_H{\exp \int^x a_0(s)ds } = {y_H}^2$$ <p style="text-align:right;">$$\displaystyle (Equation\;5.8.5)$$ The final complete solution is y(x) = A(x)$$y_H$$, which gives <p style="font-size:125%" align="center">$$y(x) = {y_H}^3$$ <p style="text-align:right;">$$\displaystyle (Equation\;5.8.6)$$ The particular solution is simply found by substration the homogenous solution from the complete solution. <p style="font-size:125%" align="center">$$y_p = y(x) - y_H = {y_H}^3 - y_H$$ <p style="text-align:right;">$$\displaystyle (Equation\;5.8.7)$$ References for R5.6 and R*5.8 Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 11) [[media:pea1.f11.mtg11.djvu|Mtg 11]] 13 Sep 2011. Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 12) [[media:pea1.f11.mtg12.djvu|Mtg 12]] 15 Sep 2011. Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 30) [[media:pea1.f11.mtg30.djvu|Mtg 30]] 25 Oct 2011. Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 31) [[media:pea1.f11.mtg31.djvu|Mtg 31]] 25 Oct 2011. Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 32) [[media:pea1.f11.mtg32.djvu|Mtg 32]] 27 Oct 2011.