User:Egm6321.f2010.team5.riveros/hw3p9

=Problem 9 - Show the given equation satisfies the second condition for exactness= From Meeting 17, p. 17-2

Given
We are given a nonlinear second order ODE,
 * $$ \displaystyle

\underbrace{\phi_p}_{f}y + \underbrace{\phi_yy' + \phi_x}_{g} =\underbrace{(15p^4cos (x^2))}_{\phi_p}y + \underbrace{(6xy^2)}_{\phi_y}y' + \underbrace{(-6xp^5sin(x^2) + 2y^3)}_{\phi_x} = 0 $$ Thus,
 * $$ \displaystyle

f = \phi_p = 15p^4cos (x^2) $$ and,
 * $$ \displaystyle

g = \phi_yy' + \phi_x = (6xy^2)y' + (-6xyp^5sin(x^2) + 2y^3) $$

Find
The second condition of exactness for nonlinear second order ODEs states that the ODE must satisfy the following two relations:
 * 1) $$ \displaystyle f_{xx}+2pf_{xy}+2p^2f_{yy} = g_{xp}+pg_{yp}-g_{y}$$
 * 2) $$ \displaystyle f_{xp}+pf_{yp}+2f_{y}=g_{pp}$$

Show that the given equation satisfies the second condition of exactness.

First relation
The terms on the left hand side are found,
 * $$\displaystyle

f_{xx} = -60\,{p}^{4}\cos({x}^{2}){x}^{2}-30\,{p}^{4}\sin({x}^{2}) $$


 * $$\displaystyle

f_{xy} = 0 $$


 * $$\displaystyle

f_{yy} = 0 $$ Thus, the left hand side is,
 * $$\displaystyle

f_{xx}+2pf_{xy}+2p^2f_{yy} = -60\,{p}^{4}\cos({x}^{2}){x}^{2}-30\,{p}^{4}\sin({x}^{2}) $$ The individual terms on the right hand side are found,
 * $$\displaystyle

g_{xp} = 6\,{y}^{2}-30\,{p}^{4}\sin({x}^{2})-60\,{p}^{4}\cos({x}^{2}){x}^{2} $$


 * $$\displaystyle

g_{yp} = 12\,xy $$


 * $$\displaystyle

g_{y} = 12\,xyp+6\,{y}^{2} $$ Thus, the right hand side is,
 * $$\displaystyle

g_{xp}+pg_{yp}-g_{y} = -60\,{p}^{4}\cos({x}^{2}){x}^{2}-30\,{p}^{4}\sin({x}^{2}) $$ We then check for equality,
 * $$\displaystyle f_{xx}+2pf_{xy}+2p^2f_{yy} =g_{xp}+pg_{yp}-g_{y} \implies -60\,{p}^{4}\cos({x}^{2}){x}^{2}-30\,{p}^{4}\sin({x}^{2})=-60\,{p}^{4}\cos({x}^{2}){x}^{2}-30\,{p}^{4}\sin({x}^{2})$$

Which is true.

Second relation
The individual terms on the left hand side are found,
 * $$\displaystyle

f_{xp} = -120\,{p}^{3}\sin({x}^{2})x $$


 * $$\displaystyle

f_{yp} = 0 $$


 * $$\displaystyle

f_{y} = 0 $$ Thus, the left hand side is,
 * $$\displaystyle

f_{xp}+pf_{yp}+2f_{y}=-120\,{p}^{3}\sin({x}^{2})x $$ The right hand side is then found,
 * $$\displaystyle

g_{pp}=-120\,{p}^{3}\sin({x}^{2})x $$

We check for equality,
 * $$\displaystyle f_{xp}+pf_{yp}+2f_{y}=g_{pp} \implies -120\,{p}^{3}\sin({x}^{2})x=-120\,{p}^{3}\sin({x}^{2})x$$

Which is also true; therefore, the given equation satisfies the second condition of exactness.

[Author]

Egm6321.f2010.team5.riveros 06:20, 5 October 2010 (UTC)