User:Egm6321.f11.team1/report4

= R4.1 Direct Derivation of Alternate Version of RTT =

Given
Another version of the Reynolds Transport Theorem:


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

$$
 * $$\frac{D}{Dt}\int_{\mathcal B_t}f(x,t)\,d\mathcal B_t=\int_{\mathcal B_t}\frac{\partial f}{\partial t}\,d\mathcal B_t+\int_{\partial\mathcal B_t}\mathbf n\cdot(f\mathbf u)\,d(\partial\mathcal B_t)$$
 * $$\displaystyle (Equation\;4.1.1)
 * $$\displaystyle (Equation\;4.1.1)
 * }
 * }

== Find ==

Provide a different and direct derivation of Equation 4.1.1.

Solution
Solution is adapted from Malvern. If $$\displaystyle f(x,t)$$ denotes any property of the material volume $$\displaystyle\mathcal B_t$$, for a spatial volume bounded by a control surface $$\displaystyle\partial\mathcal B_t$$, the following is true:

$$ \begin{bmatrix} \rm Total\,rate\,of\,change\,of\\ \rm{an\,arbitrary\,function} \\ f(x,t)\,\rm in\,region\,\mathcal B_t \end{bmatrix}=\begin{bmatrix} \rm Time\,rate\,of\,change\\ \rm of\,\mathit f(x,t)\,\rm within\,\mathcal B_t\\ \end{bmatrix}+\begin{bmatrix} \rm Flux\,of\,\mathit{f(x,t)}\\ \rm thru\,surface\,\partial\mathcal B_t \end{bmatrix} $$

Now expressing mathematically what is stated above:

= R*4.2 Exactness Verification=

Given
Non-exact L2-ODE-VC

Example: Consider the following ODE:

$$\sqrt{x} \, y'' + 2 x y' +3y=0$$ $$\displaystyle (Equation\;4.2.1) $$

Find
Verify the exactness of Equation 4.2.1

Solution

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[[media:pea1.f11.mtg21.djvu|Mtg 21 (c,x)]]

To be exact, the equation 4.2.1 must satisfy two conditions.

1st Condition of Exactness

Fist, the equation must be presented in the following format

$$\displaystyle F(x,y,y',y)=g(x,y,y')+f(x,y,y')y$$$$\displaystyle (Equation\;4.2.2) $$

By letting y'=p, the first condition of exactness is satisfied as shown in the following equation:

$$\displaystyle F(x,y,p,y)=\underbrace{2xp+3y}_{g(x,y,p)}+\underbrace{\sqrt{x}}_{f(x,y,p)}y$$$$\displaystyle (Equation\;4.2.3) $$

2nd Condition of Exactness

In order to meet the second condition of exactness, the following equations must be satisfied:

$$\displaystyle f_{xx}+2pf_{xy}+p^2f_{yy}=g_{xp}+pg_{yp}-g_y$$ $$\displaystyle (Equation\;4.2.4) $$

$$\displaystyle f_{xp}+pf_{yp}+2f_y=g_{pp}$$ $$\displaystyle (Equation\;4.2.5) $$

Each element of equationgs 4.2.4 and 4.2.5 are calculated to be the following:

$$\displaystyle f_{xx}=-\frac{1}{4}x^{-\frac{3}{2}}$$

$$\displaystyle f_y=f_{xy}=f_{yy}=f_{xp}=f_{yp}=0$$

$$\displaystyle g_{xp}=g_{pp}=2$$

$$\displaystyle g_y=3$$

$$\displaystyle g_{yp}=0$$

Plugging each relevant element into equation 4.2.4 yields the following:

$$\displaystyle -\frac{1}{4}x^{-\frac{3}{2}}+2p \cdot 0+ p^2 \cdot 0 \neq 2+ p\cdot 0 -3$$

Plugging each relevant element into equation 4.2.5 yields the following:

$$\displaystyle 0+p\cdot 0+2 \cdot 0 \neq 2$$

Equations 4.2.4 and 4.2.5 have not been satisfied, therefore proved equation 4.2.1 is not exact.


 * }

=Problem R*4.3: Find $$\displaystyle m$$ and $$\displaystyle n$$ of the integrating factor $$\displaystyle H(x,y) = x^{m}y^{n} $$ such that the given N2-ODE is exact then solve for $$\displaystyle y(x)$$.=

Given
A function is given by


 * $$\displaystyle x^{m}y^{n} \lbrack \sqrt{x}y''+2xy'+3y \rbrack = 0$$

$$\displaystyle (1) p.21-3 $$

Per Lecture notes: [[media:pea1.f11.mtg21.djvu|Mtg 21 (c)]] where


 * $$\displaystyle h(x,y) = x^{m}y^{n}$$

$$\displaystyle (4.3.1) $$

Find
Part A: Find $$\displaystyle (m,n) $$ such that $$\displaystyle (1) p.21-3 $$ is exact. 

Part B: Show that the first integral is a L1-ODE-VC and solve


 * $$\displaystyle \phi(x,y,p) = xp+(2x^{\frac{3}{2}}-1)y = k$$ $$\displaystyle (2)p.21-3

$$

for $$\displaystyle y(x)$$.

Solution

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Part A:

The first exactness criteria for second order non-liner ODE is
 * $$\displaystyle F(x,y,p) = f(x,y,p)y'' + g(x,y,p) $$

$$\displaystyle (2) p.16-4 $$

where


 * $$\displaystyle p = y'$$

Comparing (1) p.21-3 with the form of (2) p.16-4, the functions $$\displaystyle f $$ and $$\displaystyle g $$ can be defined as


 * $$\displaystyle f(x,y,p)=x^{0.5+m}y^{n}$$$$\displaystyle (4.3.2) $$


 * $$\displaystyle g(x,y,p)=2x^{m+1}y^{n}p+3x^{m}y^{n+1}$$$$\displaystyle (4.3.3) $$

So the form of the 1st exactness condition is met.

For a second order non-liner ODE, the first part of the second exactness condition is
 * $$\displaystyle f_{xx} + 2pf_{xy} + p^2f_{yy} = g_{xp} + pg_{yp} - g_{y} $$$$\displaystyle (1) p.16-5 $$

For a second order non-liner ODE, the second part of the second exactness condition is
 * $$\displaystyle f_{xp} + pf_{yp} + 2f_{y} = g_{pp}$$$$\displaystyle (2) p.16-5 $$

Using the second part of the second exactness condition given above in $$\displaystyle (2) p.16-5 $$ and recognizing the given$$\displaystyle f(x,y,p) $$ from equation $$\displaystyle (4.3.2) $$  is not a function of $$\displaystyle p$$, so the first two terms are zero. Additionally $$\displaystyle g_p$$ is constant wrt. $$\displaystyle p$$ and therefore $$\displaystyle g_{pp}$$ is zero. $$\displaystyle (2) p.16-5 $$ can be simplified as shown below


 * $$\displaystyle 2f_y=g_{pp}=0$$<p style="text-align:right;">$$\displaystyle (4.3.4)$$

Taking the derivative .wrt $$\displaystyle y $$ of $$\displaystyle f(x,y,p) = f_y $$ in equation $$\displaystyle (4.3.2) $$ and substituting into $$\displaystyle (4.3.4)$$ yields


 * $$\displaystyle 2\left[x^{0.5+m}ny^{n-1}\right]=0$$<p style="text-align:right;">$$\displaystyle (4.3.5) $$

In order for equation $$\displaystyle (4.3.5) $$ to be true $$\displaystyle m=0$$


 * $$\displaystyle n=0$$<p style="text-align:right;">$$\displaystyle (4.3.6) $$

For clarity we can now re-write equations $$\displaystyle (4.3.2) $$$$\displaystyle (4.3.3) $$ with $$\displaystyle n=0$$ as shown below:


 * $$\displaystyle f=x^{0.5+m}$$ <p style="text-align:right;">$$\displaystyle (4.3.7) $$


 * $$\displaystyle g=2x^{m+1}p+3x^{m}y $$ <p style="text-align:right;">$$\displaystyle (4.3.8) $$

Next, use the first part of the the second exactness condition given in equation $$\displaystyle (1) p.16-5  $$ with the following derivatives of $$\displaystyle f $$ and $$\displaystyle g $$ plugged in.


 * $$\displaystyle {f_{xx} = (m+0.5)(m-0.5)x^{-1.5+n}}$$


 * $$\displaystyle{f_{xy} = 0}$$


 * $$\displaystyle{f_{yy} = 0}$$


 * $$\displaystyle{g_{xp} = 2(m+1)x^{m}}$$


 * $$\displaystyle{g_{yp} = 0}$$


 * $$\displaystyle{g_y=3x^{m}}$$

Which yeilds:
 * $$\displaystyle(m+0.5)(m-0.5)x^{-1.5+m} = 2(m+1)x^{m}-3x^{m}$$ <p style="text-align:right;">$$\displaystyle (4.3.9) $$

Solving this equation for $$ \displaystyle m $$ yields


 * $$ \displaystyle m = \frac{1}{2}$$ <p style="text-align:right;">$$\displaystyle (4.3.10)$$

So the integrating factor to make $$\displaystyle (1) p.21-3 $$ exact is


 * $$\displaystyle h(x,y) = x^{0.5} $$ <p style="text-align:right;">$$\displaystyle (4.3.11) $$

Part B:

So using the integrating factor found above and multiplying through to equation $$\displaystyle (1) p.21-3 $$ to get and exact N2-ODE one can use equation $$\displaystyle (3) p.16-5 $$ to obtain the first integral given in $$\displaystyle (2)p.21-3 $$

Rearranging equation $$\displaystyle (2)p.21-3 $$ and solving for $$\displaystyle y(x)$$ by dividing through by x to put the ODE into the general form:


 * $$ \displaystyle y' + a_0(x)y = b(x) $$<p style="text-align:right;">$$\displaystyle (3)p.11-3 $$

Yields:


 * $$ p + (2x^{\frac{1}{2}}-{\frac{1}{x}} )y=k $$

Where:
 * $$ a_o(x)=\frac{(2x^{\frac{3}{2}}-1)}{x}=2x^{\frac{1}{2}}-x^{-1}$$ <p style="text-align:right;">$$\displaystyle (4.3.12) $$
 * $$ b(x)=\frac{k}{x}$$ <p style="text-align:right;">$$\displaystyle (4.3.13) $$

The ODE can be solved by finding an integrating factor $$h(x,y)$$ per


 * $$h(x)=exp(\int a_o(x)dx)$$ <p style="text-align:right;">$$\displaystyle (3)p.11-4 $$


 * $$y(x)= \frac{1}{h(x)}\int h(x)b(x)dx+C_1 $$ <p style="text-align:right;">$$\displaystyle (1)p.11-5 $$

Substituting in $$h(x)=exp(\int a_o(x)dx)$$ gives:


 * $$y(x)=\frac{\int exp(\int a_0(x)dx)b(x)+C_1}{exp(\int a_0(x)dx)}$$ <p style="text-align:right;">$$\displaystyle (4.3.14) $$

Integrating $$a_0(x)$$ which is given in $$\displaystyle (4.3.12) $$


 * $$\int a_0(x)dx=\int 2x^{\frac{1}{2}}-x^{-1}=\frac{4}{3}x^{\frac{3}{2}}-ln(x)+C_2$$ <p style="text-align:right;">$$\displaystyle (4.3.15) $$

Plugging in $$\displaystyle (4.3.15) $$ and $$\displaystyle (4.3.13) $$ into $$\displaystyle (4.3.14) $$ yields:
 * $$y(x)=\frac{\int C_3x^{-1}exp(\frac{4}{3}x^{\frac{3}{2}})\frac{k}{x}+C_1}{ C_3x^{-1}exp(\frac{4}{3}x^{\frac{3}{2}})}$$ <p style="text-align:right;">$$\displaystyle (4.3.16) $$

where $$C_3 = exp(C_2) $$

Equation $$\displaystyle (4.3.16) $$ is the solution for $$\displaystyle y(x) $$

=R*4.4 Generate a class of exact L2-ODE-VC = Question from meeting 21, pages 5-6: Given (2)-(3) from p. 21-4 $$G = \underbrace{\underbrace{R(x)}_{\displaystyle a_0(x)} y}_{\displaystyle \ne \phi_x} + \underbrace{\underbrace{Q(x)}_{\displaystyle a_1(x)} y'}_{\displaystyle \ne \phi_y} + \underbrace{\underbrace{P(x)}_{\displaystyle a_2(x)} y''}_{\displaystyle = \phi_p}$$ <p style="text-align:right;">$$\displaystyle (Equation\;4.4.1)$$ <p">$$\phi_x(x,y,p) + \phi_y(x,y,p) \,y' = R(x)y + Q(x)y'$$ <p style="text-align:right;">$$\displaystyle (Equation\;4.4.2)$$  Find that the solution gives (1) from p. 21-5 <p style="font-size:125%">$$\phi(x,y,p) = P(x)p + T(x)y + k$$ <p style="text-align:right;">$$\displaystyle (Equation\;4.4.3)$$

From lecture notes [[media:pea1.f11.mtg21.djvu|Mtg 21]]

Solution
The equations (2) and (3) are within the category of L2-ODE-VC. The key is to use the substitution p = y' for the derivations and p' = y''.The first step is to prove the exactness of the equations below. There are two exactness conditions for the N2-ODE. Note that the functions R and Q are of x but not y. This affects the 2nd exactness condition and subsequent derivations and integrations.

$$G = \underbrace{\underbrace{R(x)}_{\displaystyle a_0(x)} y}_{\displaystyle \ne \phi_x} + \underbrace{\underbrace{Q(x)}_{\displaystyle a_1(x)} y'}_{\displaystyle \ne \phi_y} + \underbrace{\underbrace{P(x)}_{\displaystyle a_2(x)} y''}_{\displaystyle = \phi_p}$$ <p style="text-align:right;">$$\displaystyle (Equation\;4.4.1)$$

$$g(x,y,p) = \phi_x(x,y,p) + \phi_y(x,y,p) \,y' = R(x)y + Q(x)y'$$ <p style="text-align:right;">$$\displaystyle (Equation\;4.4.2$$  The function with p=y' can be rearranged.  $$g(x,y,p) = \phi_x(x,p) + \phi_y(x,p) \,y' = R(x)y + Q(x)p$$ <p style="text-align:right;">$$\displaystyle (Equation\;4.4.3)$$

The equations satisfy the following form of the 1st exactness condition. $$G = \frac{d \phi}{dx} = \phi_x + \phi_y \,y'+ \phi_p \,y'' $$ <p style="text-align:right;">$$\displaystyle (Equation\;4.4.4)$$ The 2nd exactness condition is derived below. First, assume that <p style="font-size:125%" align="center">$$f(x,p) := \phi_p(x,p)= P(x)$$ Then the 2nd exactness condition can be applied satisfactorily. <p style="font-size:125%" align="center">$$f_{xx} + 2pf_{xy} + p^2f_{yy} = g_{xp} + pg_{yp} - g_y$$ <p style="font-size:125%" align="center">$$f_{xp} + pf_{yp} + 2f_y = g_{pp}$$ <p style="text-align:right;">$$\displaystyle (Equation\;4.4.5)$$ This can be shown by placing the values of R and Q into the equation for 2nd exactness. <p style="font-size:125%" align="center">$$P_{xx} + 0 + 0 = g_{xp} + pg_{yp} - g_y = Q_{x} - R$$ The next step is to find the integration factor h(x) from the equation below. <p style="font-size:125%" align="center">$$g(x,yp) := \phi_x + \phi_y \,y' = \phi_x(x,p) + \phi_y(x,p) \,p = R(x)y + Q(x)p$$ Recall that the integration factor in this case is a function of x, so the following equation from meeting 11 can be used. <p style="font-size:125%" align="center">$$\frac{h_x}{h} = -\frac{1}{N}(N_x - M_y) =: \color{blue}{n(x)}$$ The integration is simple. <p style="font-size:125%" align="center">$$h(x) = \int \frac{h_x}{h} = -\int \frac{1}{N}(N_x - M_y) = -\int\frac{1}{Q(x)}(Q_x - R)= T(x)y + k$$ Subsequently, the final solution is equated below. <p style="font-size:125%" align="center">$$\phi(x,y,p) = h(x) + \int(f,x,y,p) \,dp = P(x)p + T(x)y + k$$ References for R4.4 Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 11) [[media:pea1.f11.mtg11.djvu|Mtg 11]] 13 Sep 2011. Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 16) [[media:pea1.f11.mtg16.djvu|Mtg 16]] 22 Sep 2011. Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 21) [[media:pea1.f11.mtg21.djvu|Mtg 21]] 04 Oct 2011.

= R*4.5 Solve a L2-ODE-VC = == Given ==
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$$
 * $$\displaystyle G=(cosx)y''+(x^2-sinx)y'+2xy=0$$
 * <p style="text-align:right;">$$\displaystyle (Equation\;4.5.1)
 * <p style="text-align:right;">$$\displaystyle (Equation\;4.5.1)
 * }
 * }

Find
1）Show Equation(4.5.1) is exact. 2）Find $$\displaystyle\, \phi$$ 3）Solve for $$\displaystyle y(x)$$.

Solution
1) The particular form of N2-ODE-VC is $$\displaystyle G(x,y,y',y)=g(x,y,p)+f(x,y,p)y$$ where $$\displaystyle g(x,y,p)=(x^2-sinx)y'+2xy$$ $$\displaystyle f(x,y,p)=cosx$$ This is shown that Equation(4,5,1) satisfies the 1st exactness condition. To verify its exactness, we calculate following terms: $$\displaystyle f_x=-sinx;f_{xx}=-cosx;$$ $$\displaystyle f_{xy}=f_{xp}=0;$$ $$\displaystyle f_y=0;f_{yy}=f_{yp}=0;$$ $$\displaystyle g_x=2xy'-cosx\cdot p+2y;$$ $$\displaystyle g_{xp}=2x-cosx;$$ $$\displaystyle g_y=2x;g_{yp}=0;$$ $$\displaystyle g_p=x^2-sinx;g_{pp}=0.$$ Recall the 2nd exactness condition for L2-ODE-VC:
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$$
 * $$\displaystyle f_{xx}=2pf_{xy}+p^2f_{yy}=g_{xp}+pg_{yp}-g_y$$
 * <p style="text-align:right;">$$\displaystyle (Equation\;4.5.2)
 * <p style="text-align:right;">$$\displaystyle (Equation\;4.5.2)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$ Substituting those calculated terms into Equation(4,5,2) and Equation(4.5.3). As for Equation(4,5,2) , $$\displaystyle LHS=-cosx+0+0=-cosx$$ $$\displaystyle RHS=2x-cosx+0-2x=-cosx$$ which means $$\displaystyle LHS=RHS$$ As for Equation(4,5,3) , $$\displaystyle LHS=0+0+0=0$$ $$\displaystyle RHS=0$$ which means $$\displaystyle LHS=RHS$$ Therefore, Equation(4,5,1) is exact.
 * $$\displaystyle f_{xp}+pf_{yp}+2f_y=g_{pp}$$
 * <p style="text-align:right;">$$\displaystyle (Equation\;4.5.3)
 * <p style="text-align:right;">$$\displaystyle (Equation\;4.5.3)
 * }
 * }

2) We find the first integral $$\displaystyle \phi$$ by integrating $$\displaystyle f=\phi_p=cosx$$. $$\displaystyle \phi=h(x,y)+\int fdp$$ $$\displaystyle \phi=h(x,y)+cosx\cdot p$$ Also, $$\displaystyle g=(x^2-sinx)p+2xy=\phi_x+\phi_y p$$ where, $$\displaystyle \phi_x=h_x-sinx\cdot p$$ $$\displaystyle \phi_y=h_y$$. Substituting those partial derivatives of $$\phi$$ into function g. $$\displaystyle (x^2-sinx)p+2xy=(h_y-sinx)p+h_x$$. To solving h(x,y), we assume that $$\displaystyle h_x=2xy $$ Therefore, $$\displaystyle h(x,y)=x^2 y+k_1(y)$$ Partial derivate respect to y, $$\displaystyle h_y=x^2+k_1'(y)-sinx=x^2-sinx$$ We can find that $$\displaystyle k_1'(y)=0$$ which means $$\displaystyle k_1(y)=k_1=constant$$. Therefore, the first integral is $$\displaystyle \phi=x^2y+cosx\cdot p+k_!$$

3) Recall that $$\displaystyle \phi=k_2$$, we obtain a L1-ODE-VC that $$\displaystyle x^2y+cosx\cdot p=k_2 -k_1:=k_3$$. Rewrite it into $$\displaystyle y'+\underbrace{\frac{x^2}{cosx}}_{\displaystyle a_0(x)}y=\underbrace{\frac{k_3}{cosx}}_{\displaystyle b(x)}$$

Therefore, the integration factor is $$\displaystyle h(x)=exp[\int^x a_0'(s)ds]$$ $$\displaystyle =exp[\int^x \frac{s^2}{cos s}ds]$$. The solution is

= R*4.6 Show the equivalence of two forms of 2nd exactness condition of N2-ODE =

Given
$$ \displaystyle g_0 - \frac{d}{dx}g_1+\frac{d^2}{dx^2}g_2=0 $$

$$ \displaystyle g_i:=\frac{\partial G}{\partial y^{(i)}} \, \quad for \ i=0,1,2 $$

Find
It's equal to the 2nd exactness condition in form:

$$ \displaystyle \phi_{xy}=\phi_{yx}\ ,\quad \phi_{px}=\phi_{xp}\ ,\quad \phi_{py}=\phi_{yp} $$

Solution
As defined:

$$ \displaystyle G=\frac{d \phi}{dx} (x,y,p)=0 $$

$$ \displaystyle G=\phi_x+\phi_y \cdot p +\phi_p \cdot q $$

$$ \displaystyle for \quad p=y' \, \ q=y'' $$

$$ \displaystyle g_0 = \frac{\partial G}{\partial y} = \frac{\partial}{\partial y}(\phi_x+\phi_yp+\phi_pq) = \frac{\partial}{\partial y}(\frac{d \phi}{dx}) $$

$$ \displaystyle \frac{d}{dx}g_1 = \frac{d}{dx}(\frac{\partial G}{\partial p}) = \frac{d}{dx}(\phi_{xp}+\phi_{yp}p+\phi_y+\phi_{pp}q) $$

$$ \displaystyle \begin{align} \frac{d^2}{dx^2} g_2& = \frac{d^2}{dx^2}(\frac{\partial G}{\partial q})\\ &=\frac{d^2}{dx^2} [\frac{\partial}{\partial q}(\phi_x + \phi_yp +\phi_pq)] \\ &= \frac{d^2}{dx^2} \phi_p\\ &= \frac{d}{dx}(\phi_{px}+\phi_{py}p+\phi_{pp}q) \end{align} $$

For we have:

$$ \displaystyle g_0-\frac{d}{dx}g_1+\frac{d^2}{dx^2}g_2=0 $$

$$ \displaystyle \frac{\partial}{\partial y}(\frac{d \phi}{dx}) - \frac{d}{dx}(\phi_{xp}+\phi_{yp}p+\phi_y+\phi_{pp}q) + \frac{d}{dx}(\phi_{px}+\phi_{py}p+\phi_{pp}q) = 0 $$

Rearranging:
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\displaystyle [\frac{\partial}{\partial y} (\frac{d\phi}{dx})-\frac{d}{dx}(\frac{\partial \phi}{\partial y})] + \frac{d}{dx} [(\phi_{px}-\phi_{xp})+(\phi_{py}-\phi_{yp})p+(\phi_{pp}-\phi_{pp})q] = 0 $$ $$
 * <p style="text-align:right;">$$\displaystyle (Equation\;4.6.1 )
 * }
 * }

So, both two parts in the left side of equation 4.6.1 are equal to 0 separately:

For first part:

$$ \displaystyle \frac{\partial}{\partial y} (\frac{d\phi}{dx})-\frac{d}{dx}(\frac{\partial \phi}{\partial y})=0 $$

$$ \displaystyle (\phi_{xy}-\phi_{yx})+(\phi_{yy}-\phi_{yy})p+(\phi_{py}-\phi_{yp}q)=0 $$

so:

$$ \displaystyle \phi_{xy}=\phi_{yx} $$

$$ \displaystyle \phi_{py}=\phi_{yp} $$

And, second part:

$$ \displaystyle \frac{d}{dx} [(\phi_{px}-\phi_{xp})+(\phi_{py}-\phi_{yp})p+(\phi_{pp}-\phi_{pp})q] = 0 $$

$$ \displaystyle (\phi_{px}-\phi_{xp})+(\phi_{py}-\phi_{yp})p+(\phi_{pp}-\phi_{pp})q = 0 $$

thus we get:

$$ \displaystyle \phi_{px}=\phi_{xp} $$

= References =

= Team Work Distribution =

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