User:EGM6341.s11.TEAM1.WILKS/Mtg31

=EGM6321 - Principles of Engineering Analysis 1, Fall 2010=

[[media: 2010_10_28_14_54_20.djvu | Mtg 31:]] Thu, 28 Oct 10

= Application: King 2003 p.28 (Pb.1.1 ab) cont'd =

[[media: 2010_10_28_14_54_20.djvu | Page 31-1]]
Wolfram Alpha $$ \Rightarrow \ \ $$ $$ r_1=1 \ $$ $$ r_2=\frac{1}{x-1} \ $$ See F09 [[media: Egm6321.f09.mtg21.djvu |p.21-4]] for the problem statement on the method of finding $$ u_2(x) \ $$.

= Application: Obtaining L2-ODE-VC from characteristic equation =

Reverse engineering.

Consider two roots: $$ r_1=2 \ $$ and $$ r_2=\frac{1}{x+1} \ $$. Thus the characteristic function is

$$ \ (r-r_1)(r-r_2)=0 \ $$ $$ \Rightarrow \ r^2 - 2r - \frac{r}{x+1} + \frac{2}{x+1}=0 \ $$

So even though $$\displaystyle r_2(x) = \frac{1}{x+1}$$ is not a valid root, and since $$\displaystyle r_1 = 2 = \text{constant}$$ is a valid root, we still have $$\displaystyle y(x) = e^{r_1 x}$$ as a valid trial solution, which leads to the above L2-ODE-VC. On the other hand, the above L2-ODE-VC for sure will lead to the above characteristic equation.

The point is there must be at least one valid root, i.e., constant root, for the above procedure of generating L2-ODE-VC to be valid. In other words, the following two roots $$ r_3 = 2x \ $$ and $$ r_4 = \frac{1}{x+1} \ $$ are not valid, and cannot be used to construct an L2-ODE-VC with the characteristic function being

$$ \ (r-r_3)(r-r_4)=0 \ $$

HW5.7

Find $$ u_1, u_2 \ $$ (2 homogeneous solutions) of Eq.(1) using trial solution $$ y=e^{rx} \ $$

END HW5.7

= Application: Trial solution with singularity =

Consider: (see also F09 [[media: Egm6321.f09.mtg22.djvu | p.22-2]])

Given:

[[media: 2010_10_28_14_54_20.djvu | Page 31-2]]
1) Verify exactness of [[media: 2010_10_28_14_54_20.djvu | Eq.(2)p.31-1]] 2) Can you use the Integrating Factor Method ? 3) Trial Solution 3.1) $$ y=e^{rx} \ $$, with r=constant 3.2) $$ y=xe^{rx} \ $$, with r=constant 3.3) $$ y=\frac{e^{rx}}{x} \ $$ , with r=constant Find $$ u_1 \ $$ in [[media: 2010_10_28_14_54_20.djvu | Eq.(3)p.31-1]] 4) Find $$ u_2 \ $$ using variation of parameters. Wolfram Alpha for $$ y=\frac{e^{rx}}{x} \Rightarrow \ r_1=i \ $$ and $$ r_2=-i \ $$, where $$ r_1, r_2 \ $$ are constant

$$ \Rightarrow \ u_1=\frac{e^{ix}}{x} \ $$ $$ \Rightarrow \ u_2=\frac{e^{-ix}}{x} \ $$ $$ \Rightarrow \ \bar u \ _1 =\frac{\cos x}{x} \ $$ $$ \Rightarrow \ \bar u \ _2 =\frac{\sin x}{x} \ $$

Obtain L2-ODE-VC from a known trial solution and a characteristic equation
Reverse engineering

Add the following equations: $$ \left [ \frac{e^{rx}}{x} \right ] \times a_0(x) \ $$ $$ \left [ \frac{d}{dx} \left ( \frac{e^{rx}}{x} \right ) = \left ( r-\frac{1}{x} \right ) \frac{e^{rx}}{x} \right ] \times a_1(x) \ $$ $$ \left [ \frac{d^2}{dx^2} \left ( \frac{e^{rx}}{x} \right ) = \left ( r^2+\frac{2}{x^2}-\frac{2r}{x} \right ) \frac{e^{rx}}{x} \right ] \times a_2(x) \ $$ Obtain the following: $$ \frac{e^{rx}}{x} \left [ a_2 \left ( r^2+\frac{2}{x^2}-\frac{2r}{x} \right ) +a_1 \left ( r-\frac{1}{x} \right ) +a_0 \right ]=\frac{e^{rx}}{x}(r^2+1) =0 \ $$, where $$ r=\pm i \ $$ for the last term $$ (r^2+1) \ $$ 3 equations for $$ a_2, a_1, a_0 \ $$ $$ r^2(a_2)+r \left ( \frac{-2}{x}a_2+a_1 \right ) +1 \left ( \frac{2}{x^2}a_2-\frac{a_1}{x}+a_0 \right )=r^2+0r+1 \ $$ $$ \Rightarrow \ a_2=1 \ $$ $$ \Rightarrow \ a_1= \frac{2}{x} \ $$ $$ \Rightarrow \ a_0=1 \ $$ The previous three equations lead to $$ \Rightarrow \ y''+\frac{2}{x}y'+y=0 \ $$ which is the same as [[media: 2010_10_28_14_54_20.djvu | Eq.(2)p.31-1]]

[[media: 2010_10_28_14_54_20.djvu | Page 31-4]]
HW5.8 Find L2-ODE-VC (homogeneous) such that trial solution $$ \frac{e^{rx}}{x^2} \ $$ and such that $$ r^2+3=0 \ $$ is the characteristic equation. END HW5.8