User:Egm6321.f10.team4.Yoon/Mtg16

= EGM6321 - Principles of Engineering Analysis 1, Fall 2010 =

Mtg 16: Tue, 28 Sep 10

[[media: 2010_09_28_15_08_25.djvu | Page 16-1]]
[[media: 2010_09_23_14_52_54.djvu |Eqn.(1) and Eqn.(2) p.15-4:]]

If $$\displaystyle h=k_1 $$ (constant) $$\displaystyle \ \ \Rightarrow h_x=h_y = 0 $$ 1st integral $$\displaystyle \underbrace{\Phi(x,y,p)}_{\color{blue}{k_1 + xyp}} = k_2$$ k3 and k4 are two constants of integration(2nd order equation.)

End Application

Note: Recall N1-ODE $$\displaystyle F(x,y,y')=0$$ is exact if $$\displaystyle \exists \underbrace{\Phi(x,y)}_{=k}$$ such that $$\displaystyle F= \frac{d\Phi}{dx}(x,y)=0$$

[[media: 2010_09_28_15_08_25.djvu | Page 16-2]]
2nd exactness condition: $$\displaystyle \underbrace{N_x}_{\color{blue}{f_x}}=\underbrace{M_y}_{\color{blue}{g_y}}$$

cf.[[media: 2010_09_23_14_52_54.djvu | Eqn.(1) and Eqn.(2) p.15-3:]]

Clairaut 1739-40 proved $$\displaystyle \Phi_{xy}=\Phi_{yx}$$.

End Note

N2-ODE: 2nd exactness condition
Back to N2-ODE: 2nd relation of exactness condition 2, i.e., [[media: 2010_09_23_14_52_54.djvu |Eqn.(2) p.15-3:]] is easy to prove:

=> Just find $$\displaystyle g_{pp}$$ using definition of g in [[media: 2010_09_23_14_52_54.djvu | Eqn.(3) p.15-2:]]. see F09

1st relation of exactness condition 2: [[media: 2010_09_23_14_52_54.djvu | Eqn.(1) p.15-3:]] $$\displaystyle \Phi(x,y,p)$$ - 3 arguments

Recall $$\displaystyle f=\Phi_p $$[[media: 2010_09_23_14_52_54.djvu |Eqn.(4) p.15-2]], $$\displaystyle g=\Phi_x + \Phi_y p $$[[media: 2010_09_23_14_52_54.djvu |Eqn.(3) p.15-2]]

[[media: 2010_09_28_15_08_25.djvu | Page 16-4]]
F09: Use $$\displaystyle \Phi_{xy} = \Phi_{yx} $$, and $$\displaystyle \underbrace{_{\color{blue}{Eqn.(2)p.16-3}}}_{\color{blue}{\Phi_x}} $$ and $$\displaystyle \underbrace{\color{blue}{_{Eqn.(1)}}}_{\color{blue}{\Phi_y}}$$, to obtain the 1st relation for exactness condition 2, i.e., [[media: 2010_09_28_15_08_25.djvu |Eqn.(1) p.15-3]].