User:EGM6321.f09.team1.Zhichao Gong/Mtg10

Mtg 10: Tue, 15 Sep 09

page10-1

Note: HW p.9-3, p.9-2: "exact"

means "either exact or exactly inte-" grable by int. fact. method" p.9-4: The exact func. phi is called a first intergral of ODE F(x, y, y' ...)=0

$${\color{blue}p.9-4} \ {\color{red}(4):} \ F(x, \ y, \ {y}^{'}, \ {y}^{''})$$

$$=\frac{d}{dx}\phi(x, \ y, \ )={\phi}^{x}+{\phi}^{y}{\color{blue} \underset{p}{ \underbrace}}+{\phi}^{p}{\color{blue} \underset{{p}^{'}}{ \underbrace}}$$

$$F=f(x, \ y, \ p){y}^{''}+g(x, \ y, \ p) \ {\color{red}(1)}$$

$$f(x, \ y, \ p):={\phi}^{p}(x, \ y, \ p) \ {\color{red}(2)}$$

$$g(x, \ y, \ p):={\phi}^{x}+{\phi}^{y}p \ {\color{red}(3)}$$

Eq(1): 1st condition of exactness.

page10-2

Application: N2_ODE not exact

$${\color{blue} \underset{f}{ \underbrace}}{{y}^{''}}^{\frac{3}{2}}+{\color{blue} \underset{g}{ \underbrace}}=0$$

$$2nd. \ condition \ of \ exactness:$$

If both exactness condition hold, then

$$obtained \ from \ {\color{red}(2)} \ {\color{blue}p.10-1}$$

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Application: Consider

$${\color{blue} \underset{f(x, \ y, \ p)}{ \underbrace}}{y}^{''}+{\color{blue} \underset{g(x, \ y, \ p)}{ \underbrace}}=0$$

<P>1st exactness condition. satisfied.</P> <P>HW: 2nd exactness condition also satisfied:</P> <P> </P>

$${f}_{x}=y \ \Rightarrow \ {f}_{xx}=0$$

$${g}_{y}=p \ \Rightarrow \ {g}_{yp}=1$$

$$A \ first \ int. \ of \ ODE \ is: \ f=\frac{ \partial \phi}{ \partial p}$$

$$\Rightarrow \ \phi(x, \ y, \ p)=h(x, \ y)+\int_{}^{}(x, \ y)dp$$

$$=h(x, \ y)+xyp$$

Use defn of gp.10-1 Eq(3):

$$g(x, \ y, \ p)={\phi}_{x}+{\phi}_{y}p$$

$$={h}_{x}+{h}_{y}p$$

$$observe \ if \ {\color{red}\underline{{\color{black}h={k}_{{\color{blue}1}}}const}}\Rightarrow \ {h}_{x}={h}_{y}=0$$

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$$1st \ int. \ \underset{ \overset{||}{{k}_{\color{blue}1}+xyp}}{\phi(x, \ y, \ p)}={\color{blue} \underset{ \overset{||}{{k}_{2}}}}$$

$$\Rightarrow \ (xy){y}^{'}=c-k$$

$$\Rightarrow$$

$$\frac{{y}^{2}}{2}=({\color{red} \underset{ \overset{\uparrow}{Constant}}}-{\color{red} \underset{ \overset{\uparrow}{Constant}}})logx+ {\color{blue} \underset{ \overset{|| \ {\color{red}constant}}{{k}_{3}}}}$$

$$\frac{{y}^{2}}{2}=({k}_{2}-{k}_{1})logx+{k}_{3}$$