University of Florida/Egm6321/F10.TEAM1.WILKS/Mtg5

EGM6321 - Principles of Engineering Analysis 1, Fall 2009
Mtg 5: Thur, 03 Sept 09

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Note: [[media: Egm6321.f09.mtg4.djvu | Eq.(2)p.4-2 ]] and [[media: Egm6321.f09.mtg4.djvu | Eq.(4)p.4-2 ]] $$ y'(x)=p(x) \ $$ Integrate from a to x: $$ \int_{s=a}^{s=x} y'(s)\, ds = \int_{s=a}^{s=x} p(s)\, ds $$ $$ \left [ y(s) \right ]_{s=a}^{s=x} = y(x)-y(a) $$, where $$ y(a) = constant \ $$, $$ y(x)=\int_{s=a}^{s=x} p(s)\, ds +y(a)= \int_{}^{x} p(s)\, ds = \int_{}^{} p(x)\, dx+k $$ another way: $$ y(x)=\int_{}^{} p(x)\, dx +k $$ Where $$ \int_{}^{} p(x)\, dx = F(x) $$ and $$ k=constant \  $$ $$ \Rightarrow \ y(a)=F(a)+k \ $$ $$ \Rightarrow \ k=y(a)-F(a) \ $$ $$ \Rightarrow \ y(x)=F(x)-F(a)+y(a) \ $$

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But $$ F(x)-F(a)=\int_{s=a}^{s=x} p(s)\, ds \ $$ [[media: Egm6321.f09.mtg4.djvu | Eq.(3)p.4-2 ]]: Why this form of nonlinear 1st order ODE? Most general form:

Application:

Where $$ x^2y^5+6(y')^2 \ $$ is defined as $$ F(x,y,y') \ $$ HW: Show that $$ F(x,y,y')=0 \ $$ in Eq(3) is a nonlinear 1st order ODE. Hint: Define the differential operator $$ D(.) \ $$ associated with Eq(3).

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Form for exact nonlinear 1st order ODE: $$ F(x,y,y')=0 \ $$ is exact if $$ \exists \ $$ a function $$ \phi\ (x,y) $$ such that

where:

$$ \exists \ $$ is defined as "there exists"

$$ \phi\ _x = M(x,y) $$

$$ \phi\ _y = N(x,y) $$

Multiply Eq1) thru by $$ dx \ $$ to get:

[[media: Egm6321.f09.mtg4.djvu | Eq.(3)p.4-2 ]] : $$ M(x,y)dx + N(x,y)dy=0 \ $$

NOTE: If $$ F(X,y,y')=0 \ $$ does not have the form:

Then $$ F(.) \ $$ cannot be exact.

Application: [[media: Egm6321.f09.mtg5.djvu | Eq.(3)p.5-2 ]] is not exact because of the nonlinear term $$ (y')^2 \ $$

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Exactness test (continued)

p5-3 Eq(1):

$$ M(x,y)= \phi\ _x(x,y) \ $$

$$ N(x,y)= \phi\ _y(x,y) \ $$

Since $$ \phi\ _{xy} = \phi\ _{yx} \ $$

and $$ \phi\ _{xy} = \frac{\partial ^2 \phi\ }{\partial x \partial y} \ $$

and $$ \phi\ _{yx} = \frac{\partial ^2 \phi\ }{\partial y \partial x} \ $$

and $$ \frac{\partial ^2 \phi\ }{{\partial x} {\partial y} } = (\phi\ _{x})_y \ $$

and $$ \frac{\partial ^2 \phi\ }{{\partial y} {\partial x} } = (\phi\ _{y})_x \ $$

Application: [[media: Egm6321.f09.mtg4.djvu | Eq.(1)p.4-3 ]] Not Exact

$$ M(x,y)=2x^2 + \sqrt{y} \Rightarrow \ M_y= \frac{1}{2 \sqrt{y}} \ $$

$$ N(x,y)=x^5y^3 \Rightarrow \ N_x= 5x^4y^3 \ $$

$$ M_y \ne \ N_x \ $$