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

$\displaystyle {\begin{aligned}y''+a_{1}(x)y'+a_{0}y=0\end{aligned}}$

(1)

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

$\displaystyle {\begin{aligned}y_{H}(x)=k_{1}u_{1}(x)+k_{2}u_{2}(x)\end{aligned}}$

(3)

Where$k_{1},k_{2}\$ are constants

Assume full homogeneous solution

$\displaystyle {\begin{aligned}y(x)=U(x)u_{1}(x)\end{aligned}}$

(2)

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)\left[y=Uu_{1}\right]\$

and $a_{1}(x)\left[y'=Uu_{1}'+U'u_{1}\right]$

and $\left[y''=Uu_{1}''+2U'u_{1}'+U''u_{1}\right]$

To get $a_{0}y+a_{1}y'+y''=U\left[a_{0}u_{1}+a_{1}u_{1}'+u_{1}''\right]+U'\left[a_{1}u_{1}+2u_{1}'\right]+U''u_{1}=0\$ by Eq(1) p17-1

Reduce to $u_{1}''+a_{1}u_{1}'+a_{0}u_{1}=0\$

Since $u_{1}\$ is a homogeneous solution $\Rightarrow \ 0=U'(a_{1}u_{1}+2u_{1}')+U''u_{1}\$, NOTE missing dependent variable U in front of $U'\$ term

Let $Z:=U'\Rightarrow \ \$ homogeneous L1_ODE_VC for Z

Page 17-3

$\displaystyle {\begin{aligned}\Rightarrow \ u_{1}(x)Z'+(a_{1}(x)u_{1}(x)+2u_{1}'(x)Z=0\end{aligned}}$

(1)

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

$\Rightarrow \ {\frac {Z'}{Z}}+(a_{1}+{\frac {2u_{1}'}{u_{1}}})=0\$

Where $a_{1},{\frac {2u_{1}'}{u_{1}}}\$ are known

Integrate $\log \left|Z\right\vert +2\log \left|u_{1}\right\vert +\int _{}^{x}a_{1}(s)\,ds=k$, where k is a constant

$\displaystyle {\begin{aligned}Z(x)={\frac {c}{(u_{1})^{2}}}e^{-\int _{}^{x}a_{1}(s)\,ds}=U'(x)\end{aligned}}$

(2)

where $c=e^{k}\$

$\Rightarrow \ U(x)=\int _{}^{x}{\frac {c}{(u_{1}(t))^{2}}}e^{-\int _{}^{t}a_{1}(s)\,ds}\,dt+{\tilde {c}}\$

where ${\bar {a}}\ _{1}(t)=\int _{}^{t}a_{1}(s)\,ds\$

Page 17-4

Homogeneous solution

$\displaystyle {\begin{aligned}y(x)=U(x)u_{1}(x)=cu_{1}\int _{}^{x}{\frac {1}{(u_{1}(t))^{2}}}e^{-{\bar {a}}\ _{1}(t)}\,dt+{\tilde {c}}u_{1}={\tilde {c}}u_{1}+cu_{2}\end{aligned}}$

(1)

where ${\tilde {c}}=k_{1}\$ and $c=k_{2}\$

$\displaystyle {\begin{aligned}\Rightarrow u_{2}=u_{1}\int _{}^{x}{\frac {1}{(u_{1}(t))^{2}}}e^{-{\bar {a}}\ _{1}(t)}\,dt\end{aligned}}$

(2)

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