# University of Florida/Egm6321/F10.TEAM1.WILKS/Mtg9

## EGM6321 - Principles of Engineering Analysis 1, Fall 2009

Mtg 9: Thur, 10Sept09

### Page 9-1

 {\displaystyle \displaystyle {\begin{aligned}c(x):=\int _{}^{x}b(s)\,ds=:{\bar {b}}\ (s)\end{aligned}}} (1)

Note: Symbol notations:

${\displaystyle \Delta \ }$ defined as equal by definition

Non-symmetric notation ${\displaystyle :=}$ means equal by definition as well, better notation than ${\displaystyle \Delta \ }$

Goal: Derive mathematical structure of a class of exact N1_ODEs

Exact L1_ODE_VC:

 {\displaystyle \displaystyle {\begin{aligned}{\bar {b}}\ (x)y'+\left[a(x)y+k\right]=0\end{aligned}}} (2)

Application: Just invent any ${\displaystyle a(x),b(x)\ }$
Let ${\displaystyle a(x)=x^{4}\ }$

${\displaystyle b(x)=x\Rightarrow \ {\bar {b}}\ (x)={\frac {1}{2}}x^{2}\ }$

${\displaystyle k=10\ }$

### Page 9-2

 {\displaystyle \displaystyle {\begin{aligned}{\frac {1}{2}}x^{2}y'+\left[x^{4}y+10\right]=0\end{aligned}}} (3)

HW: Show that Eq(3) L1_ODE_VC is "exact". See Note P.10-1

Question: How about N1_ODEs? Eq.(2)P.6-4

N1 means Nonlinear, 1st Order

 {\displaystyle \displaystyle {\begin{aligned}M_{y}(x,y)=a(x)c(y)\end{aligned}}} (4)

 {\displaystyle \displaystyle {\begin{aligned}N_{x}(x,y)=b(x)c(y)\Rightarrow \ N(x,y)=\left(\int _{}^{x}b(s)\,ds\right)c(y)\end{aligned}}} (5)

Where ${\displaystyle \int _{}^{x}b(s)={\bar {b}}\ (x)\ }$

Eq.(2) P.6-4:

${\displaystyle {\frac {1}{N}}(N_{x}-M_{y})={\frac {1}{{\bar {b}}\ (x)c(y)}}\left[b(x)c(y)-a(x)c(y)\right]=-f(x)\ }$

### Page 9-3

Eq.(4) P.9-2 ${\displaystyle M(x,y)=a(x)\int _{}^{y}c(s)\ }$

Where ${\displaystyle \int _{}^{y}c(s)={\bar {c}}\ (y)\ }$

 {\displaystyle \displaystyle {\begin{aligned}M(x,y)=a(x){\bar {c}}\ (y)\end{aligned}}} (6)

From Eq.(6) P.9-3 , Eq.(4) P.9-2 and Eq.(3) P.4-2 obtain:

 {\displaystyle \displaystyle {\begin{aligned}{\bar {b}}\ (x)c(y)y'+a(x){\bar {c}}\ (y)=0\end{aligned}}} (7)

Application: Consider the following

${\displaystyle a(x)=5x^{3}+2\ }$

${\displaystyle b(x)=x^{2}\Rightarrow \ {\bar {b}}\ (x)={\frac {1}{3}}x^{3}\ }$

${\displaystyle c(y)=y^{4}\Rightarrow \ {\bar {c}}\ (y)={\frac {1}{5}}y^{5}\ }$

 {\displaystyle \displaystyle {\begin{aligned}\left({\frac {1}{3}}s^{3}\right)\left(y^{4}\right)y'+\left(5x^{3}+2\right)\left({\frac {1}{5}}y^{5}\right)=0\end{aligned}}} (8)

HW: Show Eq(8) is "exact" N1_ODE. Note P.10-1

### Page 9-4

L2_ODE_VC with missing dependant variable

${\displaystyle P(x)y''+Q(x)y'+R(x)y=S(x)\ }$

Where ${\displaystyle R(x)\rightarrow \ 0\ }$ due to missing dependant variable y.

Eq.(1)P.2-3 ${\displaystyle p:=y'\rightarrow \ P(x)p'+Q(x)p=S(x)\ }$ is a L1_ODE_VC

Solution: Eq.(4)P.8-2

Exact N2_ODEs:

General N2_ODEs: ${\displaystyle F(x,y,y',y'')=0\ }$

Application: ${\displaystyle (x^{3}+2x^{5}y^{2})y''+(x^{\frac {3}{2}}+10){\sqrt {y}}y'+y^{100}=0\ }$

${\displaystyle F=0\ }$ is exact means ${\displaystyle \exists \phi \ (x,y,y')\ }$ such that ${\displaystyle F(x,y,y',y'')={\frac {d}{dx}}\phi \ (x,y,y')\ }$