User:EGM6321.F10.TEAM1.WILKS/Mtg5

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

[[media: Egm6321.f09.mtg5.djvu | Page 5-1]]
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 $$ y(a) = constant \ $$,

another way: $$ y(x)=\int_{}^{} p(x)\, dx +k $$

Where $$ \int_{}^{} p(x)\, dx = F(x) $$ and $$ k=constant \  $$

$$ \Rightarrow \ y(a)=F(a)+k \ $$

$$ \Rightarrow \ k=y(a)-F(a) \ $$

$$ \Rightarrow \ y(x)=F(x)-F(a)+y(a) \ $$

[[media: Egm6321.f09.mtg5.djvu | Page 5-2]]
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). END HW

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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 $$ (y')^2 \ $$

[[media: Egm6321.f09.mtg5.djvu | Page 5-4]]
Exactness test (continued)

[[media: Egm6321.f09.mtg5.djvu | Eq.(1)p.5-3 ]]:

$$ M(x,y)= \phi_x(x,y) \ $$

$$ N(x,y)= \phi_y(x,y) \ $$

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

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

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

$$ M_y \ne \ N_x \ $$