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

EGM6321 - Principles of Engineering Analysis 1, Fall 2009

Mtg 5: Thur, 03 Sept 09

Page 5-1

Note: Eq.(2)p.4-2 and 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)\$

Page 5-2

But $F(x)-F(a)=\int _{s=a}^{s=x}p(s)\,ds\$

$\displaystyle {\begin{aligned}\Rightarrow \ y(x)=\int _{s=a}^{s=x}p(s)\,ds+y(a)=\int _{}^{x}p(s)\,ds=\int _{}^{}p(x)\,dx+k\end{aligned}}$

(1)

Eq.(3)p.4-2 : Why this form of nonlinear 1st order ODE?

Most general form:

$\displaystyle {\begin{aligned}F(x,y,y')=0\end{aligned}}$

(2)

Application:

$\displaystyle {\begin{aligned}x^{2}y^{5}+6(y')^{2}=0\end{aligned}}$

(3)

Where $x^{2}y^{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).

Page 5-3

Form for exact nonlinear 1st order ODE:

$F(x,y,y')=0\$ is exact if $\exists \$ a function $\phi \ (x,y)$ such that

$\displaystyle {\begin{aligned}F={\frac {d\phi \ }{dx}}={\frac {d}{dx}}\phi \ (x,y(x))={\frac {\partial \phi \ }{\partial x}}(x,y)+{\frac {\partial \phi \ }{\partial y}}(x,y){\frac {dy}{dx}}=0\end{aligned}}$

(1)

where:

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

$\phi \ _{x}=M(x,y)$

$\phi \ _{y}=N(x,y)$

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

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:

$\displaystyle {\begin{aligned}M(x,y)dx+N(x,y)dy=0\end{aligned}}$

(2)

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

Application: Eq.(3)p.5-2 is not exact because of the nonlinear term $(y')^{2}\$

Page 5-4

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}\$

$\displaystyle {\begin{aligned}M_{y}=N_{x}\end{aligned}}$

(1)

Application: 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^{5}y^{3}\Rightarrow \ N_{x}=5x^{4}y^{3}\$

$M_{y}\neq \ N_{x}\$