User:Egm6321.f09.Team1.vasquez/HW2

= Problem 7: Second condition of exactness, relation (1) for a N2-ODE =

Given
A general N2-ODE is given as,


 * {| style=";width:70%" border="0" align="center"



F(x,y,y',y'') = 0 $$
 * $$\displaystyle


 * }
 * }

and satisfies exactness condition (1) if it is in the form,


 * {| style=";width:70%" border="0" align="center"



f(x,y,p)y'' + g(x,y,p) = 0 $$
 * $$\displaystyle


 * }
 * }

where $$\displaystyle p:=y' $$

Find
To derive 2nd condition of exactness, relation (1) of a N2-ODE from $$\displaystyle \phi_{xy}=\phi_{yx}$$, which is given by


 * {| style=";width:70%" border="0" align="center"



f_{xx} + 2p f_{xy} + p^2 f_{yy} = g_{xp} + p g_{yp} - g_{y} $$ $$
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (34)


 * }
 * }

Solution
We know that $$\displaystyle \phi_{xy}=\phi_{yx}$$

where,


 * {| style=";width:70%" border="0" align="center"



\phi_{x} = g - p ( g_p - f_x) + p^2 f_y $$ $$
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (35)


 * }
 * }


 * {| style=";width:70%" border="0" align="center"



\phi_{y} = g_p - f_y p - f_x $$ $$
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (36)


 * }
 * }

Taking the partial derivative of (35) with respect to $$\displaystyle y$$ and the partial derivative of (36) with respect to $$\displaystyle x$$ gives


 * {| style=";width:70%" border="0" align="center"



\phi_{xy} = g_y - p_y ( g_p - f_x) - p (g_{py} - f_{xy}) + 2 p_y f_y + p^2 f_{yy} $$ $$
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (37)


 * }
 * }


 * {| style=";width:70%" border="0" align="center"



\phi_{yx} = g_{px} - f_{yx} p - f_y p_x - f_{xx} $$ $$
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (38)


 * }
 * }

Since (37) and (38) are equal


 * {| style=";width:70%" border="0" align="center"



g_y - \cancelto{0}{p_y} g_p + \cancelto{0}{p_y} f_x - p g_{py} + p f_{xy} + 2 \cancelto{0}{p_y} f_y + p^2 f_{yy} = g_{px} - f_{yx} p - f_y \cancelto{0}{p_x} - f_{xx} $$
 * $$\displaystyle


 * }
 * }

Re-arranging terms yields


 * {| style=";width:70%" border="0" align="center"



$$\displaystyle f_{xx} + 2 p f_{xy} + p^2 f_{yy}  = g_{xp} + p g_{yp} - g_y $$
 * style="width:45%; padding:10px; border:2px solid #8888aa" |
 * style= |


 * }
 * }

= Problem 8: Second condition of exactness, relations (1) and (2) for a N2-ODE =

Given
The following N2-ODE,


 * {| style=";width:70%" border="0" align="center"

(8 x^5y')y''+2x^2y'+20x^4(y')2+4xy=0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (39)
 * }
 * }

Find
Verify the second condition of exactness, relations (1) and (2)

Solution
Let $$\displaystyle p=y'$$, then (39) can be written as follows:


 * {| style=";width:70%" border="0" align="center"



(8 x^5p)y''+2x^2p+20x^4(p)2+4xy=0 $$ $$
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (40)


 * }
 * }

From (40) we know that


 * {| style=";width:70%" border="0" align="center"



f(x,y,p)=8 x^5p $$ $$
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (41)


 * }
 * }


 * {| style=";width:70%" border="0" align="center"

g(x,y,p)=2x^2p+20x^4(p)2+4xy $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (42)
 * }
 * }

Taking the partial derivatives of (41) and (42) gives


 * {| style=";width:70%" border="0" align="center"

\begin{array}{*{20}l} {f_x = 40x^4 p} & {f_{xx}  = 160x^3 p} & {g_x  = 4xp + 80x^3 p^2  + 4y} & {g_{xp}  = 4x + 160x^3 p}  \\ {} & {f_{xy} = 0} & {g_y  = 4x} & {g_{yp}  = 0}  \\ {} & {f_{xp} = 40x^4 } & {g_p  = 2x^2  + 40x^4 p} & {g_{pp}  = 40x^4 }  \\ {f_y = 0} & {f_{yy}  = 0} & {} & {}  \\ {} & {f_{yp} = 0} & {} & {}  \\ \end{array} $$
 * $$\displaystyle
 * $$\displaystyle


 * }
 * }

The first relation of the second exactness condition is given by


 * {| style=";width:70%" border="0" align="center"

f_{xx} + 2 p f_{xy} + p^2 f_{yy}  = g_{xp} + p g_{yp} - g_y $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * style= |
 * $$\displaystyle \longrightarrow (43)
 * }
 * }

Substituting the elements of (43) gives


 * {| style=";width:70%" border="0" align="center"

(160x^3 p) + 2 p (0) + p^2 (0)  = (4x + 160x^3 p) + p (0) - (4x) $$
 * $$\displaystyle
 * $$\displaystyle
 * style= |
 * }
 * }

Then,


 * {| style=";width:70%" border="0" align="center"

160x^3 p = \cancel{4x} + 160x^3 p - \cancel{4x} $$
 * $$\displaystyle
 * $$\displaystyle
 * style= |
 * }
 * }

Therefore, the first relation is satisfied

The second relation of the second exactness condition is given by


 * {| style=";width:70%" border="0" align="center"

f_{xp} + pf_{yp} + 2f_y = g_{pp} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * style= |
 * $$\displaystyle \longrightarrow (44)
 * }
 * }

Substituting the elements of (44) gives


 * {| style=";width:70%" border="0" align="center"

(0) + p(0) + 2(0) = (0) $$
 * $$\displaystyle
 * $$\displaystyle
 * style= |
 * }
 * }

Then,


 * {| style=";width:70%" border="0" align="center"

0 = 0 $$
 * $$\displaystyle
 * $$\displaystyle
 * style= |
 * }
 * }

Therefore, the second relation is satisfied

Since both relations (43) and (44) are satisfied we conclude that the given N2_ODE is exact.

= Problem 9: Exactness of a N2_ODE =

Given
The following differential equation,


 * {| style=";width:70%" border="0" align="center"

\sqrt{x} y'' + 2xy' + 3y=0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (45)
 * }
 * }

Find
Verify the exactness of the given ODE

Solution
Let $$\displaystyle p=y'$$, then (45) can be written as follows:


 * {| style=";width:70%" border="0" align="center"

x^{\frac{1}{2}} y'' + 2xp + 3y=0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (46)


 * }
 * }

From (46) we verify the first condition of exactness, where


 * {| style=";width:70%" border="0" align="center"



f(x,y,p)=x^{\frac{1}{2}} $$ $$
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (47)


 * }
 * }


 * {| style=";width:70%" border="0" align="center"

g(x,y,p)=2xp + 3y $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (48)
 * }
 * }

Taking the partial derivatives of (47) and (48) gives


 * {| style=";width:70%" border="0" align="center"

\begin{array}{*{20}l} {f_x = \frac{1}{2}x^{-\frac{1}{2}}} & {f_{xx}  = -\frac{1}{4}x^{-\frac{3}{2}}} & {g_x  = 2p} & {g_{xp}  = 2}  \\ {} & {f_{xy} = 0} & {g_y  = 3} & {g_{yp}  = 0}  \\ {} & {f_{xp} = 0 } & {g_p  = 2x} & {g_{pp}  = 0 }  \\ {f_y = 0} & {f_{yy}  = 0} & {} & {}  \\ {} & {f_{yp} = 0} & {} & {}  \\ \end{array} $$
 * $$\displaystyle
 * $$\displaystyle


 * }
 * }

The first relation of the second exactness condition is given by


 * {| style=";width:70%" border="0" align="center"

f_{xx} + 2 p f_{xy} + p^2 f_{yy}  = g_{xp} + p g_{yp} - g_y $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * style= |
 * $$\displaystyle \longrightarrow (49)
 * }
 * }

Substituting the elements of (49) gives


 * {| style=";width:70%" border="0" align="center"

(-\frac{1}{4}x^{-\frac{3}{2}}) + 2 p (0) + p^2 (0)  = (2) + p (0) - (3) $$
 * $$\displaystyle
 * $$\displaystyle
 * style= |
 * }
 * }

Then,


 * {| style=";width:70%" border="0" align="center"

-\frac{1}{4}x^{-\frac{3}{2}} \ne -1 $$
 * $$\displaystyle
 * $$\displaystyle
 * style= |
 * }
 * }

Therefore, the first relation is not satisfied.

The second relation of the second exactness condition is given by


 * {| style=";width:70%" border="0" align="center"

f_{xp} + pf_{yp} + 2f_y = g_{pp} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * style= |
 * <p style="text-align:right;">$$\displaystyle \longrightarrow (50)
 * }
 * }

Substituting the elements of (50) gives


 * {| style=";width:70%" border="0" align="center"

(0) + p(0) + 2(0) = (0) $$
 * $$\displaystyle
 * $$\displaystyle
 * style= |
 * }
 * }

Then,


 * {| style=";width:70%" border="0" align="center"

0 = 0 $$
 * $$\displaystyle
 * $$\displaystyle
 * style= |
 * }
 * }

Therefore, the second relation is satisfied.

Since the first relation of the second exactness condition is not satisfied we conclude that the given N2_ODE is not exact.