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

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

${\displaystyle y'(x)=p(x)\ }$

Integrate from a to x:

${\displaystyle \int _{s=a}^{s=x}y'(s)\,ds=\int _{s=a}^{s=x}p(s)\,ds}$

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

${\displaystyle y(x)=\int _{s=a}^{s=x}p(s)\,ds+y(a)=\int _{}^{x}p(s)\,ds=\int _{}^{}p(x)\,dx+k}$
another way: ${\displaystyle y(x)=\int _{}^{}p(x)\,dx+k}$

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

${\displaystyle \Rightarrow \ y(a)=F(a)+k\ }$

${\displaystyle \Rightarrow \ k=y(a)-F(a)\ }$

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

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

${\displaystyle \displaystyle {\begin{aligned}\Rightarrow \ y(x)=\int _{s=a}^{s=x}p(s)\,ds+y(a)=\int _{}^{x}p(s)\,ds=\int _{}^{}p(x)\,dx+k\end{aligned}}}$	(1)

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

Most general form:

${\displaystyle \displaystyle {\begin{aligned}F(x,y,y')=0\end{aligned}}}$	(2)

Application:

${\displaystyle \displaystyle {\begin{aligned}x^{2}y^{5}+6(y')^{2}=0\end{aligned}}}$	(3)

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

HW: Show that ${\displaystyle F(x,y,y')=0\ }$ in Eq(3) is a nonlinear 1st order ODE.

Hint: Define the differential operator ${\displaystyle D(.)\ }$ associated with Eq(3).

Form for exact nonlinear 1st order ODE:

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

${\displaystyle \displaystyle {\begin{aligned}F={\frac {d\phi \ }{dx}}={\frac {d}{dx}}\phi \ (x,y(x))={\frac {\partial \phi \ }{\partial x}}(x,y)+{\frac {\partial \phi \ }{\partial y}}(x,y){\frac {dy}{dx}}=0\end{aligned}}}$	(1)

where:

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

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

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

Multiply Eq1) thru by ${\displaystyle dx\ }$ to get:

Eq.(3)p.4-2  : ${\displaystyle M(x,y)dx+N(x,y)dy=0\ }$

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

${\displaystyle \displaystyle {\begin{aligned}M(x,y)dx+N(x,y)dy=0\end{aligned}}}$	(2)

Then ${\displaystyle F(.)\ }$ cannot be exact.

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

Exactness test (continued)

p5-3 Eq(1):

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

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

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

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

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

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

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

${\displaystyle \displaystyle {\begin{aligned}M_{y}=N_{x}\end{aligned}}}$	(1)

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

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

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

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