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

Mtg 13: Tues, 22Sept09

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

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

(1)

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

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

Result: A first integration is:

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

(2)

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

HW: Solve Eq(2) for ${\displaystyle y(x)\ }$ 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 ${\displaystyle \phi \ \ }$ that yields the above class

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

Where ${\displaystyle p=y'\ }$ and ${\displaystyle p'=y''\ }$

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

Answer:

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

(1)

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

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

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

(2)

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

Condition 1 for exactness:

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

(1)

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

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

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

(2)

HW: Case ${\displaystyle n=1\ }$ (N1_ODE)

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

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

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

Case ${\displaystyle n=2\ }$ (N2_ODE)

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

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