User:Egm6321.f10.team4.Yoon/HW22009

= 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 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} $$ $$ =Problem 3=
 * <p style="text-align:right;">$$\displaystyle (Eq. 9)
 * }
 * }



Given
A non-homogenous L1_ODE_VC


 * $$ \frac{1}{2}x^{2}y'+[x^{4}y+10]= 0 $$

Find
Show that the above non-homogenous L1_ODE_VC is exact

Solution
For the non-homogenous L1_ODE_VC to be exact it must staisfy satisfy the 2 conditions of exactness



\begin{align} N(x,y)y' + M(x,y) &= 0 \\ M_y(x,y) &= N_x(x,y) \end{align} $$

So the first condition is met by:


 * $$ \underbrace{\frac{1}{2}x^{2}}_{N(x,y)}y'+\underbrace{[x^{4}y+10]}_{M(x,y)}= 0 $$

Let's take a look at the 2nd condition:



\begin{align} M_y(x,y) & = x^4 \\ N_x(x,y) & = x \end{align} $$

Clearly $$M_y(x,y) \neq N_x(x,y)$$, so we make use of the integrating factor to see if we can make it exactly integrable.

So we re-write the ODE


 * $$ \underbrace{\frac{1}{2}x^{2}}_{N(x,y)}y'+\underbrace{[x^{4}y]}_{M(x,y)}= -10 $$

Looking at case #1 for the integrating factor we obtain:



\begin{align} \frac{1}{N}\left(N_x - M_y\right) &= -f(x) \qquad \text{replacing values found above}\\ \frac{2}{x^2}\left(x - x^4\right) &= -f(x) \qquad \text{applying knowledge from integrating factor we know}\\ h(x)&=\exp \int^{x}f(s)ds \qquad \text{replacing f(x) into the integral and evaluating it, obtain}\\ h(x) &= \frac{1}{x^2}\exp \left(\frac{2x^3}{3}\right) \end{align} $$

Now we can try the 2nd condition of exactness to see if by using integrating factor we can make this ODE exact



\begin{align} \bar{M}_y &= \bar{N}_x\\ \underbrace{(hM)}_{\bar{M}}dx&=\underbrace{(hN)}_{\bar{N}}dy\qquad \text{calculating the respective values we obtain}\\ \bar{N}_x &=x^2 \exp\left(\frac{2x^3}{3}\right)\\ \bar{M}_y &=x^2 \exp\left(\frac{2x^3}{3}\right) \end{align} $$

Hence the non-homogenous L1_ODE_VC is exact.

<div style="width: 80%; margin-left: auto; margin-right: auto; padding: 4px; border: 2px solid #FF0000; background-color: #FFDDDD; text-align: left;"> Nice work. --Egm6321.f09.TA 03:14, 28 September 2009 (UTC) =Problem 4=
 * }

Given
An aribitrary 1st order ODE:


 * $$ \left(\frac{1}{3}x^{3}\right)(y^{4})y'+(5x^{3}+2)\left(\frac{1}{5}y^{5}\right)=0$$

Find
Show that the above is non-linear and exact

Solution
So let's determine that the above ODE is non-linear by making use of the $$D(.)\!$$ operator

To be a linear ODE the following must be true:


 * $$D\left(\alpha u + \beta v \right) = D\left(\alpha u \right) + D\left(\beta v \right)$$

Now we will define the $$D(.)\!$$ operator for the above ODE:


 * $$D(.)= \left(\frac{1}{3}x^{3}\right)(\cdot)^{4}\frac{d(\cdot)}{dx}+(5x^{3}+2)\left(\frac{1}{5}(\cdot)^{5}\right)$$

Then let's evaluate the linearity condition to see if it's true:



\begin{align} D\left(\alpha u + \beta v \right) &= \left(\frac{1}{3}x^{3}\right)(\alpha u + \beta v)^{4}\frac{d}{dx}(\alpha u + \beta v)+(5x^{3}+2)\left(\frac{1}{5}(\alpha u + \beta v)^{5}\right)\\ D\left(\alpha u \right) + D\left(\beta v \right) &= \left(\frac{1}{3}x^{3}\right)(\alpha u)^{4}\frac{d(\alpha u)}{dx}+(5x^{3}+2)\left(\frac{1}{5}(\alpha u)^{5}\right) +  \left(\frac{1}{3}x^{3}\right)(\beta v)^{4}\frac{d(\beta v)}{dx}+(5x^{3}+2)\left(\frac{1}{5}(\beta v)^{5}\right) \end{align} $$

Clearly $$D\left(\alpha u + \beta v \right) \neq D\left(\alpha u \right) + D\left(\beta v \right)$$ so the ODE is non-linear

<div style="width: 80%; margin-left: auto; margin-right: auto; padding: 4px; border: 2px solid #FF0000; background-color: #FFDDDD; text-align: left;"> Nonlinearity can be determined by inspection. This would have sufficed for this problem, although your explanation is very complete. --Egm6321.f09.TA 03:18, 28 September 2009 (UTC) To prove that the ODE is exact, let's first look at the 1st condition of exactness:


 * $$ \underbrace{\left(\frac{1}{3}x^{3}\right)(y^{4})}_{N(x,y)}y'+\underbrace{(5x^{3}+2)\left(\frac{1}{5}y^{5}\right)}_{M(x,y)}=0$$

hence $$N(x,y)y' + M(x,y) = 0\!$$ so 1st condition is met.

Looking at the 2nd condition where $$M_y = N_x\!$$. However, from above we can clearly see that:


 * $$M_y \neq N_x$$

So we will try to find an integrating factor to find if our ODE can be exactly integrable



\begin{align} \left(\frac{1}{3}x^{3}\right)(y^{4})y'+(5x^{3}+2)\left(\frac{1}{5}y^{5}\right) &=0 \qquad \text{let us multiply througout  by} \qquad \frac{3}{x^3y^4}\\ y' +\frac{3}{x^3y^4}(5x^{3}+2)\left(\frac{1}{5}y^{5}\right) &=0\qquad \text{re-writing the equation obtain}\\ \underbrace{1}_{N}\cdot y' + \underbrace{\left(3 + \frac{6}{5x^3}\right)}_{a_0}y &=0 \end{align} $$ throughout

Now we will make use of integrating factor knowledge:



\begin{align} h(x,y) &= \exp\int^{x}a_0(s)ds\qquad \text{replacing the value of} \quad a_0\\ h(x,y) &= \exp\int^{x}\left(3 + \frac{6}{5s^3}\right)ds\qquad \text{evaluating the integral obtain}\\ h(x,y) &= \exp\left(3x - \frac{3}{5x^2}\right) \end{align} $$

Now we will try to evaluate the 2nd condition of exactness which is:



\begin{align} (hM)_{y}&=(hN)_x \qquad \text{the condition says:} \\ \bar{M}_y &= \bar{N}_x \qquad \text{evaluating the derivatives we obtain}\\ \dfrac{6\exp(3x)}{\left(\exp^{\frac{1}{x^2}}\right)^{\frac{3}{5}} x^3} + \dfrac{3\exp(3x)}{\left(\exp^{\frac{1}{x^2}}\right)^{\frac{3}{5}} } & = \dfrac{6\exp(3x)}{\left(\exp^{\frac{1}{x^2}}\right)^{\frac{3}{5}} x^3} + \dfrac{3\exp(3x)}{\left(\exp^{\frac{1}{x^2}}\right)^{\frac{3}{5}}} \end{align} $$

Hence the ODE given is then, exact N1-ODE-VC
 * }

= 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"

$$\displaystyle f_{xx} + 2 p f_{xy} + p^2 f_{yy}  = g_{xp} + p g_{yp} - g_y $$ $$ = Problem 9: Exactness of N2-ODE=
 * style="width:45%; padding:10px; border:2px solid #8888aa" |
 * style="width:45%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right;">$$\displaystyle (Eq. 11)
 * }
 * }

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.