University of Florida/Egm6321/F10.TEAM1.WILKS/Mtg13

EGM6321 - Principles of Engineering Analysis 1, Fall 2009

Mtg 13: Tues, 22Sept09

Page 13-1

Euler integrating factor method Eq.(1) P.12-3

$\displaystyle {\begin{aligned}(x^{m}y^{n})\left[{\sqrt {x}}y''+2xy'+3y\right]=0\end{aligned}}$

(1)

Where: $(x^{w}y^{n})=h(x,y)\$

HW: Find (m,n) such that Eq(1) is exact.

Result: A first integration is:

$\displaystyle {\begin{aligned}\phi \ (x,y,p)=xp+(2x^{\frac {3}{2}}-1)y+k_{1}=k_{2}\end{aligned}}$

(2)

Where $p=y'\$ and $k_{1}\$ and $k_{2}\$ are constants

HW: Solve Eq(2) for $y(x)\$ HINT: L1_ODE_VC (integrating factor)

A class of exact L2_ODE_VC (how to invent more exact L2_ODE_VC)

HW: Find mathematical structure of $\phi \ \$ that yields the above class

Page 13-2

$F={\frac {d\phi \ }{dx}}=\phi \ _{x}(x,y,p)+\phi \ _{y}p+\phi \ _{p}p'=P(x)y''+Q(x)y'+R(x)y\$

Where $p=y'\$ and $p'=y''\$

and $P(x)=\phi \ _{p}\$ and $Q(x)=\phi \ _{y}\$ and $R(x)y=\phi \ _{x}\$

Answer:

$\displaystyle {\begin{aligned}\phi \ (x,y,p)=P(x)p+T(x)y+k\end{aligned}}$

(1)

Eq.(2) P.13-1  : $P(x)=x\$ and $T(x)=2x^{\frac {3}{2}}-1\$ and $k=k_{1}\$

Exact Nn_ODE's, where N means nonlinear and n means nth order

$\displaystyle {\begin{aligned}F(x,y^{(0)}...y^{(n)}=0\end{aligned}}$

(2)

Where $y^{(0)}=y\$ and $y^{(1)}=y'\$ and $y^{(n)}={\frac {d^{n}y}{dx^{n}}}\$

Page 13-3

Condition 1 for exactness:

$\displaystyle {\begin{aligned}F={\frac {d\phi \ }{dx}}(x,y^{(0)},...,y^{n-1})=\phi \ _{x}+\phi \ _{y(0)}y^{(1)}+...+\phi \ _{y(n-1)}y^{(n)}\end{aligned}}$

(1)

Where $\phi \ _{x}+\phi \ _{y(0)}y^{(1)}+...+\phi \ _{y(n-1)}y^{(n)}\$ is related term by term to $x,y^{(0)},...,y^{n-1}\$

Condition 2 of exactness: $f_{i}:={\frac {\partial F}{\partial y^{(i)}}}\$, where $i=1,...,n\$

$\displaystyle {\begin{aligned}f_{0}-{\frac {df_{1}}{dx}}+{\frac {d^{2}f_{2}}{dx^{2}}}-...+(-1)^{n}{\frac {d^{n}f_{n}}{dx^{n}}}=0\end{aligned}}$

(2)

HW: Case $n=1\$ (N1_ODE)

$F(x,y,y')=0={\frac {d}{dx}}\phi \ (x,y)\$

$f_{0}-{\frac {df_{1}}{dx}}=0\Leftrightarrow \phi \ _{xy}=\phi \ _{yx}\$

HINT: $f_{1}=\phi \ _{y}\$

Page 13-4

Case $n=2\$ (N2_ODE)

$f(x,y,y',y'')=0={\frac {d\phi \ }{dx}}(x,y,y')\$

$f_{0}-{\frac {df_{1}}{dx}}+{\frac {d^{2}f_{2}}{dx^{2}}}=0\$