User:EGM6341.s11.TEAM1.WILKS/s11Mtg17

=EGM6321 - Principles of Engineering Analysis 1, Fall 2010= Mtg 17: Wed,29Sept10

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Recall: [[media: Egm6321.f09.mtg16.djvu | Eq.(1)p.16-1]] : [[media: Egm6321.f09.mtg16.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). Application: Generate exact N2_ODEs (reverse engineering) Method: 1) Select: Where $$ p=y' \ $$ Then 2): Example 1) $$ \phi\ (x,y,p) = 4x^5p^2+2x^2y \ $$, see [[media: Egm6321.f09/Lecture_plan | F09]] for details Example 2) $$ \phi\ (x,y,p) = 3p^5 \cos x^2+2xy^3=k \ $$ $$ \phi\ _x = 3p^5 (-\sin x^2) (2x)+2y^3 \ $$ $$ \phi\ _y = 6xy^2 \ $$

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N2_ODE (exact): Eq.(3)p.17-1 $$ \Rightarrow \ \ $$ [[media: Egm6321.f09.mtg15.djvu | Eq.(2)p.15-2]], [[media: Egm6321.f09.mtg15.djvu | Eq.(3)p.15-2]] , [[media: Egm6321.f09.mtg15.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 above N2_ODE (Eq.(3)) is exact. Exact condition 1: $$ f= \phi\ _p=15p^4 \cos x^2 \ $$ $$ g= \phi\ _y y'+ \phi\ _x = (6xy^2)y'+ \left [ -6xp^5 \sin x^2 + 2y^3 \right ] \ $$ Exact condition 2: HW

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Find first integral $$ \phi\ \ $$ :

Where you must find $$ h(x,y) \ $$ using $$ g \ $$ and $$ f(x,y,p) = 15p^4 \cos x^2=3p^5 \cos x^2 \ $$ $$ g= \phi\ _yy'+ \phi\ _x \Rightarrow \ \ $$, where $$ y'=p \ $$ 2 choices