User:EGM6321.F10.TEAM1.WILKS/Mtg11

=EGM6321 - Principles of Engineering Analysis 1, Fall 2009= Mtg 11: Thur, 17Sept09

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Note: $$ f(x)= \alpha\ x^a+k $$ Let $$ \alpha\ = 0 $$ and $$ k=1 \Rightarrow \ f(x)=1 $$ $$ f(x=2)=f(x=11)=f(x=10^6)=1 \ $$ Thus $$ f(x)=y \ $$ and $$ f(x)=y' \ $$ is not possible Three arguments: $$ f(x,y,p)= \alpha\ x^a+ \beta\ y^b+ \gamma\ p^c+k $$ and k=5 $$ \Rightarrow \ f(x,y,p)=5 \ $$ $$ a = b =1 \ $$ $$ \gamma\ =0 \ $$ and $$ k=0 \ $$ $$ f(x,y,p)=xy \ $$ Thus $$ f(x,y,p)=y'' \ $$ is not possible

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Derive [[media: Egm6321.f09.mtg10.djvu | Eq.(4) P.10-2 ]]  and [[media: Egm6321.f09.mtg10.djvu  | Eq.(5) P.10-2 ]], second condition of exactness for N2_ODE Recall N1_ODE: $$ F(x,y,y')=0 \ $$ is exact if $$ \exists \ \phi\ (x,y) \ $$ such that $$ F=\frac{d}{dx} \phi\ (x,y)= \phi_x+\phi_y y' \ $$ Where $$ \phi_x = M(x,y) :=g(x,y) \ $$ and $$ \phi_y = N(x,y) :=f(x,y) \ $$ $$ F=f(x,y)y'+g(x,y)=0 \ $$ Exactness condition 2: $$ f_x=g_y \ $$ = [[media: Egm6321.f09.mtg5.djvu | Eq.(1) P.5-4 ]] $$ N_x=M_y \ $$ Now N2_ODE: HW [[media: Egm6321.f09.mtg10.djvu | Eq.(5) P.10-2 ]]Derive this relationship by differentiating [[media: Egm6321.f09.mtg10.djvu  | Eq.(3) P.10-1 ]] with respect to $$ p=y' \ $$ i.e., $$ g_{pp} \ $$ (2nd relationship in exactness condition 2.) END HW

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[[media: Egm6321.f09.mtg10.djvu | Eq.(4) P.10-2 ]] 1st relationship in exactness condition 2 Recall $$ \phi\ (x,y,p) \ $$ Recall from [[media: Egm6321.f09.mtg10.djvu | Eq.(2) P.10-1 ]]) and [[media: Egm6321.f09.mtg10.djvu  | Eq.(3) P.10-2 ]]: $$ f= \phi\ _p \ $$ and $$ g= \phi_x + \phi_y p \ $$ $$ \phi_{xp} = \phi_{px} \Rightarrow \ ( \phi\ )_p = ( \phi_p)_x  \ $$ Where $$( \phi_x )_p =  \frac{\partial}{\partial p} \left ( \frac{\partial \phi\ }{\partial x} \right )  \ $$ and $$( \phi_p )_x =  \frac{\partial}{\partial x} \left ( \frac{\partial \phi\ }{\partial p} \right )  \ $$ and $$\frac{\partial}{\partial p} \left ( \frac{\partial \phi\ }{\partial x} \right ) = (g- \phi_y p)_p  \ $$ and $$\frac{\partial}{\partial x} \left ( \frac{\partial \phi\ }{\partial p} \right ) = f_x  \ $$ and $$ (g- \phi_y p)_p = f_x  \ $$ Where $$ ( \phi_{py} = \phi_{yp} = ( \phi_p)_y = f_y $$

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$$ \phi_{yp} = \phi_{py} \Rightarrow \ ( \phi_y)_p = ( \phi_p)_y \ $$ Where $$ ( \phi_y)_p = \left [ \frac{1}{p} (g- \phi_x) \right ]_p \ $$ and $$ ( \phi_p)_y = f_y $$ and $$ \left [ \frac{1}{p} (g- \phi_x) \right ]_p = f_y \ $$ $$ \Rightarrow \ -\frac{1}{p^2} (g- \phi_x) + \frac{1}{p}(g_p- \phi_{xp}) = f_y \ $$