User:EGM6341.s11.TEAM1.WILKS/Mtg17

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

[[media: 2010_09_29_07_17_06.djvu | Mtg 17:]] Wed, 29 Sep 10

[[media: 2010_09_29_07_17_06.djvu | Page 17-1]]
Recall: [[media: 2010_09_28_15_08_25.djvu | Eq.(1)p.16-1]] :

Find $$\displaystyle h$$.

[[media: 2010_09_28_15_08_25.djvu | p.16-1]]: If $$ h=k_1 \ $$ (constant) $$ \Rightarrow \ \ $$ Eq.(1) is satisfied, i.e., h = constant is one solution  of Eq.(1). HW Question: Without assuming a priority that h=constant, discuss the search for the solution of Eq.(1). END HW

= Application: Generate exact N2-ODEs (reverse engineering) =

Method:

1) Select:

where $$ p=y' \ $$.

Then

2) Differentiate $$\displaystyle \phi(x,y,p) = k$$ obtain an exact N2-ODE:

End Method

Examples
Example 1) $$ \phi\ (x,y,p) = 4x^5p^2+2x^2y \ $$, see [[media: Egm6321.f09/Lecture_plan | F09]] for details Example 2) $$ \phi\ (x,y,p) = 3p^5 \cos x^2+2xy^3=k \ $$ $$ \phi_x = 3p^5 (-\sin x^2) (2x)+2y^3 \ $$ $$ \phi_y = 6xy^2 \ $$

[[media: 2010_09_29_07_17_06.djvu | Page 17-2]]
N2-ODE (exact): Eq.(3)p.17-1 $$ \Rightarrow \ \ $$ [[media: 2010_09_23_14_52_54.djvu | Eq.(2)p.15-2]], [[media: 2010_09_23_14_52_54.djvu| Eq.(3)p.15-2]] , [[media: 2010_09_23_14_52_54.djvu | Eq.(4)p.15-2]]

where $$ \phi_p = f \ $$ and $$ \phi_yy'+ \phi_x =g \ $$

where $$ (15p^4 \cos x^2) = \phi_p \ $$ and $$ (6xy^2) = \phi_y \ $$ and $$ \left [ -6xp^5 \sin x^2+2y^3 \right ] = \phi_x \ $$

Show that the above N2-ODE (Eq.(3)) is exact.

Exact condition 1:

$$ f= \phi_p=15p^4 \cos x^2 \ $$

$$ g= \phi_y y'+ \phi\ _x = (6xy^2)y'+ \left [ -6xp^5 \sin x^2 + 2y^3 \right ] \ $$

HW Exact condition 2: END HW

[[media: 2010_09_29_07_17_06.djvu | Page 17-3]]
Find first integral $$ \phi\ \ $$ :

since $$ f(x,y,p) = 15p^4 \cos x^2\ $$, and $$ h(x,y) \ $$ can be found using the definition of $$ g \ $$. $$ g= \phi_yy'+ \phi_x \ $$, with $$ y'=p \ $$

2 choices