University of Florida/Egm6321/F10.TEAM1.WILKS/Mtg17

EGM6321 - Principles of Engineering Analysis 1, Fall 2009

Mtg 17: Thur, 10Oct09

Page 17-1

Linearity $\Rightarrow$ superposition $y=y_H+y_P$

Homogeneous Solution $y_H$
- Euler Equations
- Trial solution (undefined coefficient)
- Reduction of order method 2: Undetermined factor

Homogeneous L2_ODE_VC: cf. Eq.(1) P.3-1

$\begin{array}{c}{y}^{″}+a_1(x)y^{\prime }+a_{0}y=0\end{array}$

Where $a_0(x)$ can be substituted for $a_1(x)$ or $a_{0}y$ in Eq(1)

Given one homogeneous solution $u_1(x)$ known

Find second homogeneous solution $u_2(x)$ such that

$\begin{array}{c}{y}_H(x)=k_1{u}_1(x)+k_2{u}_2(x)\end{array}$

Where$k_1,k_2$ are constants

Assume full homogeneous solution

$\begin{array}{c}y(x)=U(x)u_1(x)\end{array}$

Where$U(x)$ is an unknown to be determined

Where$u_1(x)$ is known

Page 17-2

"Full" = includes $u_2(x)$

Add the following: $a_0(x)[y=U{u}_1]$

and $a_1(x)[y^{\prime }=U{u_1}^{\prime }+U^{\prime }{u}_1]$

and $[y^{″}=U{u_1}^{″}+2U^{\prime }{u_1}^{\prime }+U^{″}{u}_1]$

To get $a_{0}y+a_1{y}^{\prime }+y^{″}=U[a_0{u}_1+a_1{u_1}^{\prime }+{u_1}^{″}]+U^{\prime }[a_1{u}_1+2{u_1}^{\prime }]+U^{″}{u}_1=0$ by Eq(1) p17-1

Reduce to ${u_1}^{″}+a_1{u_1}^{\prime }+a_0{u}_1=0$

Since $u_1$ is a homogeneous solution $\Rightarrow 0=U^{\prime }(a_1{u}_1+2{u_1}^{\prime })+U^{″}{u}_1$, NOTE missing dependent variable U in front of $U^{\prime }$ term

Let $Z:=U^{\prime }\Rightarrow$ homogeneous L1_ODE_VC for Z

Page 17-3

$\begin{array}{c}\Rightarrow u_1(x)Z^{\prime }+(a_1(x)u_1(x)+2{u_1}^{\prime }(x)Z=0\end{array}$

Solve for Z,
- integration factorial method (HW)
- Direct integration (because Eq(1) is homogeneous)

$\Rightarrow \frac{Z^{\prime }}{Z}+(a_1+\frac{2{u_1}^{\prime }}{u_1})=0$

Where $a_1,\frac{2{u_1}^{\prime }}{u_1}$ are known

Integrate $\log \left|Z\right|+2\log \left|u_1\right|+\underset{}{\overset{x}{\int }}{a}_1(s)ds=k$, where k is a constant

$\begin{array}{c}Z(x)=\frac{c}{(u_1{)}^2}{e}^{-\underset{}{\overset{x}{\int }}{a}_1(s)ds}=U^{\prime }(x)\end{array}$

where $c=e^k$

$\Rightarrow U(x)=\underset{}{\overset{x}{\int }}\frac{c}{(u_1(t){)}^2}{e}^{-\underset{}{\overset{t}{\int }}{a}_1(s)ds}dt+\tilde{c}$

where $\overline{a}{}_1(t)=\underset{}{\overset{t}{\int }}{a}_1(s)ds$

Page 17-4

Homogeneous solution

$\begin{array}{c}y(x)=U(x)u_1(x)=c{u}_1\underset{}{\overset{x}{\int }}\frac{1}{(u_1(t){)}^2}{e}^{-\overline{a}{}_1(t)}dt+\tilde{c}{u}_1=\tilde{c}{u}_1+c{u}_2\end{array}$

where $\tilde{c}=k_1$ and $c=k_2$

$\begin{array}{c}\Rightarrow u_2=u_1\underset{}{\overset{x}{\int }}\frac{1}{(u_1(t){)}^2}{e}^{-\overline{a}{}_1(t)}dt\end{array}$

HW: obtain Eq.(2) P.17-3 $Z(x)$ using the integrating factor method

References

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