User:Egm6321.f09.Team1/HW2

= Problem 1 - Integrating Factor Method =

Given
General form of a First Order Nonlinear Ordinary Differential Equations(N1-ODE) is given by,
 * {| style="width:100%" border="0" align="left"

F(x, y(x),\dot y(x)) = 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1)
 * }
 * }

By using $$\displaystyle F=\frac{d}{dx} \phi(x,y(x))$$ we get


 * {| style="width:100%" border="0" align="left"

F = \underbrace{\frac{\partial \phi}{\partial x}}_{M(x,y)} + \underbrace{\frac{\partial \phi}{\partial y}}_{N(x,y)} \frac{dy}{dx} = 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 2)
 * }
 * }

We can multiply this by an Integrating Factor, $$\displaystyle h(x,y)$$, to get


 * {| style="width:100%" border="0" align="left"

h F = h M + h N \frac{dy}{dx} = 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 3)
 * }
 * }

Then it follows that an exactness condition for an N1ODE's is, with equation 2 as the special case where $$h=1$$


 * {| style="width:100%" border="0" align="left"

(hM)_y = (hN)_x
 * $$\displaystyle
 * $$\displaystyle

$$ $$
 * $$\displaystyle (Eq. 4)
 * }
 * }

Given that $$\displaystyle h_x N = 0$$, it is found that
 * {| style="width:100%" border="0" align="left"

\Rightarrow -h_y M + h(N_x - M_y) = 0 $$ $$ or
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 5)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\Rightarrow \frac{h_y}{h} = \frac{1}{M}(N_x - M_y) = -g(y) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 6)
 * }
 * }

Find
Complete the details to obtain Euler's Integrating factor $$\displaystyle h(y)$$.

Solution
Integrate equation 6
 * {| style="width:100%" border="0" align="left"

\int \frac{h_y}{h} dy = - \int g(y) dy $$ $$ This yields
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 7)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\ln h = - \int^y g(y) dy $$ $$ The final equation is
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 8)
 * }
 * }
 * {| style="width:100%" border="0" align="left"


 * style="width:50%; padding:10px; border:2px solid #8888aa" |
 * $$ \displaystyle
 * $$ \displaystyle

h = e^{- \int^y g(y) dy} $$ $$
 * $$\displaystyle (Eq. 9)
 * }
 * }

= Problem 2 - Finding Integrating Factor =

Given

 * {| style="width:100%" border="0" align="left"

y' + \frac{1}{x}y = x^2 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1)
 * }
 * }

Find
1. Show that the solution for the integrating factor is $$h(x) = x$$ 2. Solve for $$y$$

Solution
1. With an integrating factor, $$h(x)$$, we should get the equation
 * {| style="width:100%" border="0" align="left"

(hy)' = h x^2 $$ $$ Using $$h(x) = x$$ yields
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 2)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

xy' + y = x^3 $$ $$ Dividing the above equation with $$x$$ yields the original equation
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 3)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

y' + \frac{1}{x} y = x^2 $$ $$ Hence, $$h(x) = x$$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

2. Substituting $$h(x) = x$$ into equation 2 yields
 * {| style="width:100%" border="0" align="left"

(xy)' = x^3 $$ $$ Next, we integrate the above equation
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 4)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

xy = \frac{x^4}{4} + C $$ $$ where $$C$$ is a constant. The final form of this equation is
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 5)
 * }
 * }
 * {| style="width:100%" border="0" align="left"


 * style="width:50%; padding:10px; border:2px solid #8888aa" |
 * $$ \displaystyle
 * $$ \displaystyle

y = \frac{x^3 }{4} + \frac{C}{x} $$ $$
 * $$\displaystyle (Eq. 6)
 * }
 * }

= Problem 3 - L1-ODE-VC Exactness =

Given

 * {| style="width:100%" border="0" align="left"

\frac{1}{2} x^2 y' + \left[ x^4 y + 10 \right] = 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 1)
 * }
 * }

Find
Show that equation 1 has an exact solution

Solution
The first exactness condition for an N1-ODE is satisfied of the equation can be written in the form
 * {| style="width:100%" border="0" align="left"

M(x,y) + N(x,y) y'= 0 $$ $$ This is fulfilled if
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 2)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

M(x,y) =  x^4 y + 10 $$ $$ and
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 3)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

N(x,y) = \frac{1}{2} x^2 $$ $$ The second exactness condition requires that
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 4)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

(hM)_y = (hN)_x $$ $$ where $$h = h(x,y)$$ is an integrating factor. Equation 5 expands to
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 5)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

h_y M + h M_y = h_x N + h N_x $$ $$ If $$h = h(y)$$, equation 6 can be written
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 6)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\frac{h_y}{h}  = \frac{1}{M} \left(N_x - M_y\right) $$ $$ Substituting in $$M$$ and $$N$$ yields
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 7)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\frac{h_y}{h}  = \frac{1}{x^4 y + 10} \left(x - x^4\right) $$ $$ Integrating both sides with respect to $$y$$ yields
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 8)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\ln h  = \ln \left( y + \frac{10}{x^4}\right) \frac{1}{x^4}\left(x - x^4\right) + C_1 $$ $$ or
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 9)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

h = \exp\left[\ln \left( y + \frac{10}{x^4}\right) \left(x^{-3} - 1\right) + C_1\right] $$ $$ or
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 10)
 * }
 * }


 * {| style="width:100%" border="0" align="left"

h = C_2 \left( y + \frac{10}{x^4}\right)^{(x^{-3} - 1)} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 10)
 * }
 * }

Hence, both exactness criteria are fulfilled and there is an exact solution to the given equation.

= Problem 4 - L1-ODE-VC Exactness =

Given

 * {| style="width:100%" border="0" align="left"

\frac{1}{3} x^3 y^4 y' + \left(5x^3 + 2 \right) \left(\frac{1}{5} y^5 \right) = 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 1)
 * }
 * }

Find
Show that equation 1 is an exact nonlinear, first order ODE.

Solution
With


 * {| style="width:100%" border="0" align="left"

M = \left(5x^3 + 2 \right) \left(\frac{1}{5} y^5 \right) $$ $$ and
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 2)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

N = \frac{1}{3} x^3 y^4 $$ $$ we get
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 3)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

M + N y'= 0 $$ $$ Hence, the first exactness condition is satisfied. The second exactness condition is
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 4)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

(hM)_y = (hN)_x $$ $$ Expanding equation 5 yields
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 5)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

h_y (5 x^3 + 2) \frac{y^5}{5} + h (5 x^3 + 2) y^4= h_x \frac{1}{3} x^3 y^4 + h x^2 y^4 $$ $$ If we choose $$h = x^{-3} y^{-5}$$, both sides of equation 6 equals zero, and hence the second exactness condition is fulfilled. Equation 5 becomes
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 6)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\left[(5 + 2 x^{-3}) \frac{1}{5}\right]_y = \left[\frac{1}{3} y^{-1} \right]_x $$ $$ Hence, equation 1 is an exact non-linear first order ODE.
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 5)
 * }
 * }

= Problem 5 - N2-ODE-VC Exactness =

Given
Consider the following equation
 * {| style="width:100%" border="0" align="left"

xy y'' + x(y')^2 + y y' = 0 $$ $$ Using
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 1)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

f(x,y,p) = xy $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 2)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

g(x,y,p) = x(y')^2 + y y' = 0 $$ $$ Equation 1 can be written
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 3)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

f(x,y,p) y'' + g(x,y,p) = 0 $$ $$ Hence, the first exactness condition is fulfilled.
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 4)
 * }
 * }

Find
Show that the second exactness condition is satisfied for equation 1.

Solution
Note: Equation 1 can be rewritten
 * {| style="width:100%" border="0" align="left"

(xyy')' = 0 $$ $$ First, we write $$f$$ and $$g$$ as
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 5)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

f(x,y,p) = xy $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 2)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

g(x,y,p) = x p^2 + y p = 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 3)
 * }
 * }

The second condition of exactness is given by
 * {| style="width:100%" border="0" align="left"

f_{xx} + 2p f_{xy} + p^2 f_{yy} = g_{xp} + p g_{yp} - g_y $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 6)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

f_{xp} + p f_{yp} + 2 f_y = g_pp $$ $$ Substituting $$f$$ and $$g$$ into equation 6 and 7 yields
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 7)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

0 + 2p \cdot 1 + p^2 \cdot 0 = 2p + p \cdot 1 - p $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 8)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

0 + p \cdot 0 + 2 x = 2x $$ $$ or
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 9)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

2p = 2p $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 8)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

2 x = 2x $$ $$ Hence, the second exactness condition is fulfilled.
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 9)
 * }
 * }

= Problem 6: Derive the second exactness condition for N2-ODE=

Given
A general N2-ODE is given in the form:


 * {| style="width:100%" border="0" align="left"

F(x,y,y',y) = 0 = f(x,y,p)y + g(x,y,p) $$ $$ where
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 1)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

f(x,y,p):=\phi_p(x,y,p) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 2)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

g(x,y,p):=\phi_x + \phi_y(x,y,p)p $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 3)
 * }
 * }

and
 * {| style="width:100%" border="0" align="left"

p:=y'. $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 4)
 * }
 * }

Find
Derive the second condition of exactness (given in Eq. 5)


 * {| style="width:100%" border="0" align="left"

f_{xp}+pf_{yp} + 2f_y = g_{pp} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 5)
 * }
 * }

Solution
Differentiate Eq. 3 with respect to p
 * {| style="width:100%" border="0" align="left"

g_p = \phi_{xp} + \phi_y + p\phi_{yp} $$ $$ Note that $$\displaystyle f = \phi_p$$, so
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 6)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

g_p = f_{x} + \phi_y + p f_{y} $$ $$ Differentiate a second time
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 7)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

g_{pp} = f_{xp} + \phi_{py} + f_{y} + p f_{yp} $$ $$ Substitute $$\displaystyle f = \phi_p$$ again
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 8)
 * }
 * }
 * {| style="width:100%" border="0" align="left"


 * style="width:50%; padding:10px; border:2px solid #8888aa" |
 * $$ \displaystyle
 * $$ \displaystyle

g_{pp} = f_{xp} + p f_{yp} + 2f_{y} $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 9)
 * }
 * }

= Problem 7: Deriving First Condition for N2-ODE Exactness Criterion =

Given
A general N2-ODE is given in the form:


 * {| style="width:100%" border="0" align="left"

F(x,y,y',y) = 0 = f(x,y,p)y + g(x,y,p) $$ $$ where
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 1)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

f(x,y,p):=\phi_p(x,y,p) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 2)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

g(x,y,p):=\phi_x + \phi_y(x,y,p)p $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 3)
 * }
 * }

and
 * {| style="width:100%" border="0" align="left"

p:=y'. $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 4)
 * }
 * }

Find
Derive the first exactness condition given by Eq 5.


 * {| style="width:100%" border="0" align="left"



f_{xx} + 2p f_{xy} + p^2 f_{yy} = g_{xp} + p g_{yp} - g_{y} $$ $$
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 5)


 * }
 * }

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

where,


 * {| style="width:100%" border="0" align="left""



\phi_{x} = g - p ( g_p - f_x) + p^2 f_y $$ $$
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 6)


 * }
 * }


 * {| style="width:100%" border="0" align="left"



\phi_{y} = g_p - f_y p - f_x $$ $$
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 7)


 * }
 * }

Taking the partial derivative of (6) with respect to $$\displaystyle y$$ and the partial derivative of (7) with respect to $$\displaystyle x$$ gives


 * {| style="width:100%" border="0" align="left"



\phi_{xy} = g_y - p_y ( g_p - f_x) - p (g_{py} - f_{xy}) + 2 p_y f_y + p^2 f_{yy} $$ $$
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 8)


 * }
 * }


 * {| style="width:100%" border="0" align="left"



\phi_{yx} = g_{px} - f_{yx} p - f_y p_x - f_{xx} $$ $$
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 9)


 * }
 * }

Since (8) and (9) are equal


 * {| style="width:100%" border="0" align="left"



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} $$ $$
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 10)
 * }
 * }

Re-arranging terms yields


 * {| style="width:100%" border="0" align="left"

$$\displaystyle f_{xx} + 2 p f_{xy} + p^2 f_{yy}  = g_{xp} + p g_{yp} - g_y $$ $$
 * style="width:45%; padding:10px; border:2px solid #8888aa" |
 * style="width:45%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right;">$$\displaystyle (Eq. 11)
 * }
 * }

= Problem 8: Applying N2-ODE Exactness Criterion=

Given
The following N2-ODE,


 * {| style="width:100%" border="0" align="left""

(8 x^5y')y''+2x^2y'+20x^4(y')2+4xy=0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 1) )
 * }
 * }

Find
Is this ODE exact?

Solution
The 2 exactness conditions are given by,


 * {| style="width:100%" border="0" align="left"



1: f_{xx} + 2pf_{xy} + p^2f_{yy} = g_{xp} + pg_{yp} - g_{y} $$ $$
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 2)


 * }
 * }

and


 * {| style="width:100%" border="0" align="left"



2: f_{xp} + pf_{yp} + 2f_{y} = g_{pp} $$ $$
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 3)


 * }
 * }

By substituting $$\displaystyle p=y'$$, into Eq. 1, the result is:


 * {| style="width:100%" border="0" align="left"



(8 x^5p)y''+2x^2p+20x^4(p)2+4xy=0 $$ $$
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 4)


 * }
 * }

From (4) and knowing the form of a N2-ODE (See Problem 7 Eq. 1),


 * {| style="width:100%" border="0" align="left"



f(x,y,p)=8 x^5p $$ $$
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 5)
 * }
 * }


 * {| style="width:100%" border="0" align="left"

g(x,y,p)=2x^2p+20x^4(p)2+4xy $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 6)
 * }
 * }

Taking the partial derivatives of (5) and (6) respectively yields


 * {| style="width:100%" border="0" align="left"

\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 p^2  + 40x^4 p} & {g_{pp}  = 40x^4 }  \\ {f_y = 0} & {f_{yy}  = 0} & {} & {}  \\ {} & {f_{yp} = 0} & {} & {}  \\ \end{array} $$
 * $$\displaystyle
 * $$\displaystyle


 * }
 * }

Substituting the elements of Eq. 2 results in


 * {| style="width:100%" border="0" align="left"

(160x^3 p) + 2 p (0) + p^2 (0)  = (4x + 160x^3 p) + p (0) - (4x) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 7)
 * }
 * }

Applying algebra yields:


 * {| style="width:100%" border="0" align="left

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


 * {| style="width:100%" border="0" align="left"


 * style="width:30%; padding:10px; border:2px solid #8888aa" |
 * $$ \displaystyle
 * $$ \displaystyle

1=1 $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 8)
 * }
 * }

Therefore, the first relation is satisfied

Substituting derived element into Eq. 3 results in


 * {| style="width:100%" border="0" align="left"r"

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

Then,


 * {| style="width:100%" border="0" align="left"


 * style="width:30%; padding:10px; border:2px solid #8888aa" |
 * $$ \displaystyle
 * $$ \displaystyle

0=0 $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 9)
 * }
 * }

Therefore, the second relation is satisfied

Because the 2 exactness conditions are met ( Eq. 8 and Eq. 9), the N2-ODE is exact.

= Problem 9: Exactness of N2-ODE=

Given
The following N2-ODE,


 * {| style="width:100%" border="0" align="left""

\sqrt{x}y''+2xy'+3y=0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 1) )
 * }
 * }

Find
Is this ODE exact?

Solution
The 2 exactness conditions are given by,


 * {| style="width:100%" border="0" align="left"



1: f_{xx} + 2pf_{xy} + p^2f_{yy} = g_{xp} + pg_{yp} - g_{y} $$ $$
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 2)


 * }
 * }

and


 * {| style="width:100%" border="0" align="left"



2: f_{xp} + pf_{yp} + 2f_{y} = g_{pp} $$ $$
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 3)


 * }
 * }

By substituting $$\displaystyle p=y'$$, into Eq. 1, the result is:


 * {| style="width:100%" border="0" align="left"



(\sqrt{x})y''+(2xp+3y)=0 $$ $$
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 4)


 * }
 * }

From Eq. 4 and knowing the form of a N2-ODE (See Problem 7 Eq. 1),


 * {| style="width:100%" border="0" align="left"



f(x,y,p)=\sqrt{x} $$ $$
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 5)
 * }
 * }


 * {| style="width:100%" border="0" align="left"

g(x,y,p)=2xp+3y $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 6)
 * }
 * }

Taking the partial derivatives of (5) and (6) respectively yields


 * {| style="width:100%" border="0" align="left"

\begin{array}{*{20}l} {f_x = \frac{x^{-0.5}}{2}} & {f_{xx}  = -\frac{x^{-1.5}}{4}} & {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


 * }
 * }

Substituting the elements of Eq. 2 results in


 * {| style="width:100%" border="0" align="left"

(-\frac{x^{-1.5}}{4}) + 2 p (0) + p^2 (0)  = (2) + p (0) - (3) $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }


 * {| style="width:100%" border="0" align="left"


 * style="width:30%; padding:10px; border:2px solid #8888aa" |
 * $$ \displaystyle
 * $$ \displaystyle

(-\frac{x^{-1.5}}{4}) = -1 $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 7)
 * }
 * }

Obviously, these equations are not equal, therefore given N2-ODE is NOT EXACT.

=Contributing Team Members= Egm6321.f09.Team1.vasquez 13:52, 23 September 2009 (UTC)

Egm6321.f09.Team1.AH 18:19, 23 September 2009 (UTC)

Egm6321.f09.Team1.sallstrom 18:29, 23 September 2009 (UTC)

Egm6321.f09.Team1.andy 18:48, 23 September 2009 (UTC)