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

$y^{\prime }(x)=p(x)$

Integrate from a to x:

$\underset{s=a}{\overset{s=x}{\int }}{y}^{\prime }(s)ds=\underset{s=a}{\overset{s=x}{\int }}p(s)ds$

${[y(s)]}_{s=a}^{s=x}=y(x)-y(a)$, where $y(a)=constant$,

$y(x)=\underset{s=a}{\overset{s=x}{\int }}p(s)ds+y(a)=\underset{}{\overset{x}{\int }}p(s)ds=\underset{}{\overset{}{\int }}p(x)dx+k$
another way: $y(x)=\underset{}{\overset{}{\int }}p(x)dx+k$

Where $\underset{}{\overset{}{\int }}p(x)dx=F(x)$ 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 $F(x)-F(a)=\underset{s=a}{\overset{s=x}{\int }}p(s)ds$

$\begin{array}{c}\Rightarrow y(x)=\underset{s=a}{\overset{s=x}{\int }}p(s)ds+y(a)=\underset{}{\overset{x}{\int }}p(s)ds=\underset{}{\overset{}{\int }}p(x)dx+k\end{array}$

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

Most general form:

$\begin{array}{c}F(x,y,y^{\prime })=0\end{array}$

Application:

$\begin{array}{c}{x}^2{y}^5+6(y^{\prime }{)}^2=0\end{array}$

Where $x^2{y}^5+6(y^{\prime }{)}^2$ is defined as $F(x,y,y^{\prime })$

HW: Show that $F(x,y,y^{\prime })=0$ 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:

$F(x,y,y^{\prime })=0$ is exact if $\mathrm{\exists }$ a function $\varphi (x,y)$ such that

$\begin{array}{c}F=\frac{d\varphi }{dx}=\frac{d}{dx}\varphi (x,y(x))=\frac{\partial \varphi }{\partial x}(x,y)+\frac{\partial \varphi }{\partial y}(x,y)\frac{dy}{dx}=0\end{array}$

where:

$\mathrm{\exists }$ is defined as "there exists"

$\varphi {}_x=M(x,y)$

$\varphi {}_y=N(x,y)$

Multiply Eq1) thru by $dx$ to get:

Eq.(3)p.4-2  : $M(x,y)dx+N(x,y)dy=0$

NOTE: If $F(X,y,y^{\prime })=0$ does not have the form:

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

Then $F(.)$ cannot be exact.

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

Page 5-4

Exactness test (continued)

p5-3 Eq(1):

$M(x,y)=\varphi {}_x(x,y)$

$N(x,y)=\varphi {}_y(x,y)$

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

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

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

and $\frac{{\partial }^2\varphi }{\partial x\partial y}=(\varphi {}_x{)}_y$

and $\frac{{\partial }^2\varphi }{\partial y\partial x}=(\varphi {}_y{)}_x$

$\begin{array}{c}{M}_y=N_x\end{array}$

Application: Eq.(1)p.4-3 Not Exact

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

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

$M_y\ne N_x$

References

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