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=='''Problem #1'''==
<u>Problem Statement:</u> Find the Euler Integrating Factor h(y) for a non-linear 1st order ODE for the particular case where we assume h<sub>x</sub>N=0

----

Two conditions must be satisfied in order for a nonlinear 1<sup>st</sup> order ODE to be exact.

The first condition is that <math>F(x,y,y')=0</math> must be in the form <math>M(x,y)+N(x,y)y'=0</math>

The second condition is that <math>M_y=N_x</math>.  This example will show how to satisfy the second condition.

Start with equation <math>M(x,y)+N(x,y)y'=0</math> which already satisfies the first condition.

If we assume the equation is not exact we can multiply it by an unkown factor <math>h(x,y)</math> (Euler Integrating Factor) to attempt to find a factor that will make the equation exact.

The equation becomes:

<math>h(x,y)[M(x,y)dx+N(x,y)dy]=0</math>


<math>(hM)dx+(hN)dy=0, where\ hM=\overline M, and\ hN=\overline N</math>


<math>\overline M_y=(hM)_y=h_yM+hM_y</math>


<math>\overline N_x=(hN)_y=h_xN+hN_x</math>


Set <math>\overline M_y=\overline N_x</math> and rearrange,


<math>h_xN-h_yM+h(N_x-M_y)=0</math>


To solve for <math>h_y</math> we will let <math>h_xN=0</math> which implies that <math>h_x=0</math> since <math>\scriptstyle M\ne 0</math>


<math>-h_yM+h(N_x-M_y)=0</math>, rearrage terms


<math>{h_y\over h}dy={1\over M}(N_x-M_y)</math>
take the integral of both sides,


<math>\int {h_y\over h}dy=\int {1\over M}(N_x-M_y)dy</math>


If the right hand side of the integral is only a function of y then it can be written as <math>g(y)</math>

<math>{1\over M}(N_x-M_y)=g(y)</math>


<math>ln\mid h\mid=\int g(y)dy</math>

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<math>h(y)=exp\int_{}^{y}g(s)ds</math>
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You have lost a negative sign in your expression for <math>g(y)</math>--[[User:Egm6321.f09.TA|Egm6321.f09.TA]] 02:38, 28 September 2009 (UTC)
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=='''Problem 2'''==
<u>Problem Statement:</u> Given the Non-homogeneous linear 1<sup>st</sup> order ODE with VC <math>y'+{1\over x}y=x^2</math> show the steps to obtain <math>y={x^3\over 4}+{C\over x}</math> where C=Const.

----


The first step is to check if the given equation is in the proper form:

<math>P_{(x)}y'+Q_{(x)}y=R_{(x)}</math> ; Yes  [pg.(8-1) Eg.(1)]


If <math>P_{(x)}\ne 0</math> for all (x) then:


<math>y'+{Q_{(x)}\over P_{(x)}}y={R_{(x)}\over P{(x)}}</math>

In this form we can use the Integration Factor Method to solve for y

<math>N_{(x,y)}{dy\over dx}+M_{(x,y)}=0</math>

Multiply by <math>h_{(x,y)}</math>


<math>h_{(x,y)}[N_{(x,y)}dy+M_{(x,y)}dx]</math>


<math>(hM)dx+(hN)dy=0</math> where <math>(hM)=\overline M;\ (hN)=\overline N</math>

Satisfy the exactness condition <math>\overline M_y=\overline N_x</math> where:

<math>\overline M_y=(hM)_y=h_yM+hM_y</math>

<math>\overline N_x=(hN)_x=h_xN+hN_x</math>


<math>h_xN-h_yM+h(Nx-My)=0</math>

To solve for h(x,y) we must make an assumption that h is a function of (x) only.
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You do not <b> have</b> to make this assumption.  You have chosen <math>h(x,y)=h(x)</math> instead of <math>h(x,y)=h(y)</math>.  --[[User:Egm6321.f09.TA|Egm6321.f09.TA]] 02:42, 28 September 2009 (UTC)
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Assume ''h<sub>y</sub>M''=0, so our equation becomes;

<math>h_xN+h(N_x-M_y)=0</math>


<math>{h_x\over h}=-{1\over N_{(x,y)}}(N_x-M_y)</math>

Let <math>N_{(x,y)}=1</math> and <math>M_{(x,y)}={Q_{(x)}\over P_{(x)}}y={1\over x}y</math>

<math>N_x=0</math> and <math>M_y={1\over x}</math>

Substitute back into the right hand side of the equationto verify h is a function of (x) only,

<math>{1\over 1}(0-{1\over x})={1\over x}=f(x)</math>

Using that result we can now solve for <math>h</math>

<math>\int {h_x\over h}dx=\int {1\over x}dx</math>


<math>ln|h|=ln|x|</math>


<math>h_x=x</math>

Now that we have <math>h_x</math> we can multiply through our original equation to obtain,

<math>(x)y'+(x){1\over x}y=(x)x^2\ =xy'+y=x^3</math>

Integrating yields:

<math>xy={x^4\over 4}+C</math> divide by x,
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Show the steps - the terms must be grouped together as the derivative of a product before you can integrate this.  It is not directly integrable in this form.--[[User:Egm6321.f09.TA|Egm6321.f09.TA]] 02:42, 28 September 2009 (UTC)
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<math>y={x^3\over 4}+{C\over x}</math>
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=='''Problem #3'''==
<u>Problem Statement:</u> Show that  <math> \frac{1}{2}x^{2}y'+[x^{4}y+10]= 0 </math>  is exact.<br>

----

In this problem, we are asking if: (1) this equation is only a function of x, and (2) it is "exact", meaning exact or exactly integrable by the Integrating Factors Method.<br>

To be exact, it must meet 2 conditions:<br>

(1)it must meet the form <math>M(x,y) + N(x,y)y' = 0</math><br>
(2)<math>M_y = N_x</math><br>


If <math>M(x,y)dy = a(x)</math> , then <math>M(x,y) = a(x)y + k</math> where k is a constant.<br>

Similarly:<br>
<math>N(x,y)dx = b(x)</math> , then <math>N(x,y) = \int_{}^{x}b(s)\, ds = c(x) = \overline{b}(x)</math>. Replacing M and N in our initial form<br>

<math>M(x,y)dx + N(x,y)dy = [a(x)y + k]dx + c(x)dy = 0</math><br>


Dividing by <math>dx</math> and re-arranging:<br>
<math>[a(x)y +k] + \overline{b}(x) y' = 0</math>. Our initial problem meets condition 1, where<br>
<math>\overline{b}(x) = \frac{1}{2}x^2 = N(x,y)</math><br>
<math>a(x) + k = x^4y + 10 = M(x,y)</math> where <math>k = 10</math><br>


For condition 2,
<math>M_y = \frac{d}{dy}[a(x) + k] = \frac{d}{dy}(x^4y + 10) = x^4</math> and <math>N_x = \frac{d}{dx}(\frac{1}{2}x^2) = x</math><br>

But these two are not equal.  Therefore, using the Integrating Factors Method, we must find <math>h(x,y)</math> such that <math> (hM)_y = (hN)_x</math> and thereby satisfy condition 2. <br>
Differentiating and rearranging terms gives us:<br>
<math>h_xN - h_yM + h(N_x - M_y) = 0</math>.  If we assume <math>h_y=0</math> since <math>M_y</math> and <math>N_x</math> are both functions of x, and rearrange terms, we get:<br>
<math>\frac{h_x}{h} = \frac{-1}{N}(N_x - M_y)dx</math>.  Integrating both sides creates <math>log  (h)</math>. Solving for h:<br>
<math> h = e</math>^<math>(-2)(\int_{}^{x}N_x - M_y,\ dx</math>.  We know <math>N_x</math> and <math>M_y</math>.  Therefore <math>h = e</math>^<math>(-2)(\frac{1}{2}x^2 - \frac{1}{5}x^5)</math>.<br>
Multiplying <math>M(x,y)</math> and <math>N(x,y)</math> by <math>h</math> causes our function to meet the 2nd condition and therefore be exact.<br>

<math>(e</math>^<math>(\frac{2}{5}x^5 - \frac{1}{2}x^2) (\frac{1}{2}x^{2}y'+[x^{4}y+10]= 0 </math>  is exact.
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This equation is incomplete.  There are parenthesis missing.  Where are the differentials?--[[User:Egm6321.f09.TA|Egm6321.f09.TA]] 02:47, 28 September 2009 (UTC)
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You need to show that you have arrived at the correct integrating factor.  Specifically: <math>\frac{\partial\overline{M}}{\partial y}=\frac{\partial\overline{N}}{\partial x}</math>.--[[User:Egm6321.f09.TA|Egm6321.f09.TA]] 02:47, 28 September 2009 (UTC)
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=='''Problem #4'''==
<u>Problem Statement:</u> Show that  <math> (\frac{1}{3}x^{3})(y^{4})y'+(5x^{3}+2)(\frac{1}{5}y^{5})=0</math> is an exact nonlinear, first order ODE.<br>

----

To prove this equation is an exact, non-linear, 1st order ODE, it must meet 2 conditions:<br>
(1) it must fit the form <math>M(x,y)y' + N(x,y) = 0</math>  and<br>
(2) <math>M_y = N_x</math><br>

It is obvious from simple observation that our equation meets condition 1; however it fails to meet condition 2.<br>
<math>M(x,y) = \frac{1}{3}x^3y^4</math>   and   <math>N(x,y) = (5x^3 - 2)(\frac{1}{5}y^5)</math>  then<br>
<math>M_y = \frac{4}{3}x^3y^3</math>  and <math>N_x = 5x^3y^4</math>.  Clearly <math>M_y \ne N_x</math>.

To meet condition 2, we must find h(x,y) such that <math>(hM)_y = (hN)_x</math>. As <math>M_y</math> and <math>N_x</math> only differ in <math>y</math>, let us assume that <math>h_x = 0</math>. Differentiating and rearranging terms gives us:<br>
<math>\frac{h_y}{h} = \frac{1}{M}(N_x - M_y)dx</math>.  Integrating both sides creates <math>log h</math>. Solving for h:<br>
<math> h = e</math>^<math>(\frac{1}{M}\int_{}^{y}N_x - M_y,\ dy)</math>.  We know <math>N_x</math> and <math>M_y</math>.  Substituting for <math>N_x</math> and <math>M_y</math> and simplifying: <br>
<math>h = e</math>^<math>(15y-4(log(y))</math>.  This, multiplied by our original equation will cause it to be an exact, nonlinear, first order ODE.<br>
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The <math>\frac{1}{M}</math> should be inside the integral.--[[User:Egm6321.f09.TA|Egm6321.f09.TA]] 02:51, 28 September 2009 (UTC)
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The simpliest method to prove this equation as an exact, nonlinear, first order ODE is to show that (1) it fits the form for a class of exact, non-linear 1st order ODE's.  Specifically:<br>
<p style="text-align:center;"><math>\overline{b}(x)c(y)y' + a(x)\overline{c}(y) = 0</math><br></p>
where <math>\overline{b}(x) = \int_{}^{x} b(x)\, dx</math><br>
and <math>\overline{c}(y) = \int_{}^{y} c(y)\, dy</math><br>

If we assume:<br>
<math>a(x) = 5x^{3}+2</math><br>
<math>b(x) = x^{2}</math>  and <br>
<math>c(y) = y^{4}</math>  then<br>
<math>\overline{b}(x) = \int_{}^{x} x^{2}\, dx = \frac{1}{3}x^{3}</math>  and  
<math>\overline{c}(y) = \int_{}^{y} y^{4}\, dy = \frac{1}{5}y^{5}</math><br>

Since our equation does fit the form for this class, it is an exact, nonlinear, first order ODE.
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Need to verify that <math>\overline{N}_x=\overline{M}_y</math>--[[User:Egm6321.f09.TA|Egm6321.f09.TA]] 02:51, 28 September 2009 (UTC)
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=='''Problem #5'''==
<u>Problem Statement:</u> Show that the second exactness condition for <math>xyy''+x(y')^{2}+yy'=0</math> is satisfied. <br>

----

The 2nd exactness condition actually consists of 2 parts, and our equation must meet both:<br>
(1)  f<sub>xx</sub> + (2p) f<sub>xy</sub> + (p<sup>2</sup>) f<sub>yy</sub> = g<sub>xp</sub> + p(g<sub>yp</sub>) - g<sub>y</sub><br>

(2)  f<sub>xp</sub> + p (f<sub>yp</sub>) + 2 (f<sub>y</sub>) = g<sub>pp</sub>

where p is y<sup>'</sup>


In our equation, <br>
f(x,y,p) = <math>xy</math>  and  g(x,y,p) = <math>x(y')^{2}+yy'</math>

Looking at our equation, for the first half of the condition:<br>
f<sub>xx</sub> = <math>\frac{d}{dx}</math>[<math>\frac{d}{dx}</math>(<math>xy</math>)] = <math>\frac{d}{dx}</math>(<math>y</math>) = 0 <br>

f<sub>xy</sub> = <math>\frac{d}{dy}</math>[<math>\frac{d}{dx}</math>(<math>xy</math>)] = <math>\frac{d}{dy}</math>(<math>y</math>) = 1 <br>

f<sub>yy</sub> = <math>\frac{d}{dy}</math>[<math>\frac{d}{dy}</math>(<math>xy</math>)] = <math>\frac{d}{dy}</math>(<math>x</math>) = 0 <br>

g<sub>xp</sub> = <math>\frac{d}{dp}</math>[<math>\frac{d}{dx}</math>(<math>x(p)^{2}+yp</math>)] = <math>\frac{d}{dp}</math>(<math>p^{2}</math>) = 2p <br>

g<sub>yp</sub> = <math>\frac{d}{dp}</math>[<math>\frac{d}{dy}</math>(<math>x(p)^{2}+yp</math>)] = <math>\frac{d}{dp}</math>(<math>p</math>) = 1 <br>

g<sub>y</sub> = <math>\frac{d}{dy}</math>(<math>x(p)^{2}+yp</math>) = p <br>

Plugging back into the first half of our condition, <br>
<math>0 + 2p(1) + p<sup>2</sup>(0) = 2p + p(1) - p</math><br>
<math>2p = 2p</math>   and therefore the first half of the condition is satisfied.<br>

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What is <b>sup</b>?  Check to verify that your code has rendered properly.--[[User:Egm6321.f09.TA|Egm6321.f09.TA]] 02:57, 28 September 2009 (UTC)
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Looking at the 2nd half of the condition:<br>
f<sub>xp</sub> = <math>\frac{d}{dp}</math>[<math>\frac{d}{dy}</math>(<math>xy</math>)] = <math>\frac{d}{dp}</math>(<math>y</math>) = 0 <br>

f<sub>yp</sub> = <math>\frac{d}{dp}</math>[<math>\frac{d}{dx}</math>(<math>xy</math>)] = <math>\frac{d}{dp}</math>(<math>x</math>) = 0 <br>

f<sub>y</sub> = <math>\frac{d}{dy}</math>(<math>xy</math>)] = x <br>

g<sub>pp</sub> = <math>\frac{d}{dp}</math>[<math>\frac{d}{dp}</math>(<math>x(p)^{2}+yp</math>)] = <math>\frac{d}{dp}</math>(<math>2xp + y</math>) = <math>2x(p')</math> <br>


Plugging back into the second half of our condition (remembering that p = y'),<br>
<math>0 + p(0) + 2(x)p' = 2x(p')</math>  thereby satisfying the second half of the condition.
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What is <math>p'</math> ? This would be <math> y''</math> and this should not appear in this problem.--[[User:Egm6321.f09.TA|Egm6321.f09.TA]] 02:57, 28 September 2009 (UTC)
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=='''Problem #6'''==
<u>Problem Statement:</u> Derive <math>f_{xp}+pf_{yp}+2f_{y}=g_{pp}</math> using the following relations:

(1)   <math>f(x,y,p):=\phi_{p}(x,y,p)</math><br>
(2)   <math>g(x,y,p):=\phi_{x}+\phi_{y}p</math><br>
(3)   <math>\phi_{xp}=\phi_{px}</math><br>
(4)   <math>\phi_{yp}=\phi_{py}</math><br>
----
Start by taking the partial derivative of g with respect to p twice: <math>g_{pp}=\frac{\partial^2 g}{\partial p^2}</math>

<math>g_{p}=\phi_{xp}+\phi_{yp}p+\phi_{y}</math><br>
<math>g_{pp}=\phi_{xpp}+\phi_{ypp}p+\phi_{yp}+\phi_{yp}</math><br>
<math>=>g_{pp}=\phi_{xpp}+\phi_{ypp}p+2\phi_{yp}</math>

Now use (1), (3), and (4) to substitute <math>\phi_{xpp}</math>, <math>\phi_{ypp}</math>, and <math>\phi_{yp}</math>

<math>f_{x}=\phi_{px}=\phi_{xp}</math> by (1) and (3)<br>
<math>f_{xp}=\phi_{xpp}</math><br>
<math>f_{y}=\phi_{py}=\phi_{yp}</math> by (1) and (4)<br>
<math>f_{yp}=\phi_{ypp}</math><br>

Therefore,<br>
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<math>g_{pp}=f_{xp}+f_{yp}p+2f_{y}=f_{xp}+pf_{yp}+2f_{y}</math>
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Clear and concise.--[[User:Egm6321.f09.TA|Egm6321.f09.TA]] 02:58, 28 September 2009 (UTC)
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=='''Problem #7'''==
<u>Problem Statement:</u> Derive <math>f_{xx}+2pf_{xy}+p^2f_{yy}=g_{xp}+pg_{yp}-g_{y}</math> using the following relations:<br>
(1) <math>f(x,y,p):=\phi_{p}(x,y,p)</math><br>
(2) <math>g(x,y,p):=\phi_{x}+\phi_{y}p</math><br>
(3) <math>\phi_{px}=\phi_{xp}</math><br>
(4) <math>\phi_{py}=\phi_{yp}</math><br>
(5) <math>\phi_{xy}=\phi_{yx}</math><br>

----
Solving (1) for <math>\phi_{p}</math> and taking the partial with respect to <math>x</math> yields<br>
<math>\phi_{px}=f_{x}</math><br>
Then solving (2) for <math>\phi_{x}</math> and taking the partial with respect to <math>p</math> yields<br>
<math>\phi_{xp}=g_{p}-\phi_{yp}p-\phi_{y}</math><br>
Equating these two by (3) and solving for <math>\phi_{y}</math> yields<br>
<math>\phi_{y}=-f_{x}+g_{p}-\phi_{yp}p</math><br>
Performing a similar task, take (1), solve for <math>\phi_{p}</math> and take the partial with respect to <math>y</math><br>
<math>\phi_{py}=f_{y}</math><br>
Then solving (2) for <math>\phi_{y}</math> and taking the partial with respect to <math>p</math> yields<br>
<math>\phi_{yp}=\frac{g_{p}}{p}-\frac{g}{p^2}-\frac{\phi_{xp}}{p}+\frac{\phi_{x}}{p^2}</math><br>
Equating these two by (4) and solving for <math>\phi_{x}</math> yields<br>
<math>\phi_{x}=f_{y}p^2-g_{p}p+g+\phi_{xp}p</math><br>

Finally, <math>fx=\phi_{px}=\phi_{xp}</math> and <math>fy=\phi_{py}=\phi_{yp}</math> as previously shown and so the two equations are<br>
(a) <math>\phi_{y}=-f_{x}+g_{p}-f_{y}p</math><br>
(b) <math>\phi_{x}=f_{y}p^2-g_{p}p+g+f_{x}p</math><br>
Taking the partial with respect to <math>x</math> for (a) and the partial with respect to <math>y</math> for (b) and equating based on (5) yields<br>
<math>-f_{xx}+g_{px}-f_{yx}p=f_{yy}p^2-g_{py}p+g_{y}+f_{xy}p</math><br>

Noting <math>f_{yx}=f_{xy}</math>, <math>g_{px}=g_{xp}</math>, <math>g_{py}=g_{yp}</math>, and moving the f terms to the left and the g terms to the right<br>

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<math>f_{xx}+2pf_{xy}+p^2f_{yy}=g_{xp}+pg_{yp}-g_{y}</math>
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You have performed  much unnecessary rearranging of <math>\Phi</math> terms to get here.  You need to rearrange until the end. Equate <math>\Phi_{xy}=\Phi_{yx}</math> using the appropriate relations and your solution will be must more straight forward.--[[User:Egm6321.f09.TA|Egm6321.f09.TA]] 03:01, 28 September 2009 (UTC)
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=='''Problem #8'''==
<u>Problem Statement:</u> Equations 4&5 on p.([[media:Egm6321.f09.p10-2.png|10-2]]).<br>

----


Given: 


<math>

(8x^{5}y^{'})y^{''} + 2x^{2}y^{'} + 20x^{4}(y^{'})^{2} + 4xy = 0

</math> 
 


Show that the Nonlinear 2nd-Order ODE is exact

===Solution=== 


Second Condition of Exactness: 10-2 Eq (4),(5)

<math>

\ F_{xx} + 2pF_{xy} + p^{2}F_{yy} = G_{xp} + pG_{yp} - G_{y}  \   

</math> 
 

<math>

\ F_{xp} + pF_{yp} + 2F_{y} = G_{pp}  \  

</math>


<math>

\ F(x,y,p)y^{''} + G(x,y,p) = 0 

</math> 
 

<math>

\ F(x,y,p) = 8x^{5}p  \qquad \qquad G(x,y,p) = 2x^{2}p + 20x^{4}p^{2} + 4xy

</math> 


The Following partial derivatives are then identified: 


<math>

F_{x}=40x^{4}p   \qquad \qquad  G_{x}= 4xp + 80x^{3}p^{2} + 4y

</math> 


<math>

F_{xx}=160x^{3}p  \qquad \qquad  G_{xp}= 4x + 160x^{3}p 

</math> 


<math>

F_{xy}=0  \qquad \qquad  G_{y}= 4x  

</math> 


<math>

F_{xp}=40x^{4}   \qquad \qquad  G_{yp}= 0

</math> 


<math>

F_{y}=0  \qquad \qquad  G_{p}= 2x^{2} + 40x^{4}p

</math> 


<math>

F_{yy}=0  \qquad \qquad  G_{pp}= 40x^{4}

</math> 


<math>

\ 160x^{3}p +2p(0) +p^{2}(0) = 4x + 160x^{3}p + (0) -4x  

\ 160x^{3}p = 160x^{3}p
 
</math> 


Page 10-2 Eq(4) is then satisfied. 
 

Applying the results in Page 10-2 Eq(5): 


<math>

\ 40x^{4} + p(0) + 2(0) = 40x^{4}

</math> 


<math>

\ 40x^{4} = 40x^{4}

</math> 


Page 10-2 Eq(5) is also satisfied. 
 
Because Eq(4) and Eq(5) are both satisfied, then the Secound Exactness Condition is satisfied.

Therefore,
 
<math>

\ F(x,y,p) = 8x^{5}p  \qquad \qquad G(x,y,p) = 2x^{2}p + 20x^{4}p^{2} + 4xy

</math> 

is Exactness Second ODE {{font|color=ff0000|is an exact, nonlinear second ODE}}<br>
<div style="width: 80%; margin-left: auto; margin-right: auto; padding: 4px; border: 2px solid #FF0000; background-color: #FFDDDD; text-align: left;">
Nice work.  One caveat: For your final statement, express the equation in terms of <math>x,y,y'</math>.  The equation involves an unknown function <math>y(x)</math>, where <math>p</math> was a dummy variable which helped us perform differentiations.  --[[User:Egm6321.f09.TA|Egm6321.f09.TA]] 03:05, 28 September 2009 (UTC)
</div>

=='''Problem #9'''==
<u>Problem Statement:</u> Verify the exactness of the ODE <math>\sqrt{x}y''+2xy'+3y=0</math>.<br>

----

Given 


<math>

\sqrt[]{x}y^{''} + 2xy^{'} + 3y = 0

</math> 


Prove that the equation is not exact. 


===Solution=== 
 
First Condition of Exactness Page10-1 Eq(1): 

<math>

\ F(x,y,p)y^{''} + G(x,y,p) = 0 

</math> 
 

<math>

\ F(x,y,p) = \sqrt[]{x}  \qquad \qquad G(x,y,p) = 2xp + 3y

</math> 

This satisfy {{font|color=ff0000| satisfies}} the First Condition of Exactness. 
 
The Second Exactness Condition for a second order ODE is as follows: 


<math>

\ F_{xx} + 2pF_{xy} + p^{2}F_{yy} = G_{xp} + pG_{yp} - G_{y}  \qquad Page 10-2 Eq(4)

</math> 
 


<math>

\ F_{xp} + pF_{yp} + 2F_{y} = G_{pp}  \qquad Page 10-2 Eq(5)

</math> 


The following partial derivatives are found: 
 

<math>

F_{x}= -\frac{1}{2}x^{\frac{-1}{2}}   \qquad \qquad   G_{x}= 2p

</math> 


<math>

F_{xx}= -\frac{1}{4}x^{\frac{-3}{2}}   \qquad \qquad  G_{xp}= 2

</math> 


<math>

F_{xy}= 0  \qquad \qquad \qquad  G_{y}= 3  

</math> 


<math>

F_{xp}= 0   \qquad  \qquad \qquad  G_{yp}= 0

</math> 


<math>

F_{y}= 0  \qquad \qquad \qquad  G_{p}= 2x

</math> 


<math>

F_{yy}=0  \qquad \qquad \qquad  G_{pp}= 0

</math> 


<math>

F_{yp}=0  \qquad \qquad  

</math> 
 


Using these values in Eq(4): 


<math>

\frac{-1}{4}x^{\frac{-3}{2}} + 2p(0) + p^{2}(0) = 2 + p(0) -3

</math> 


<math>

\frac{-1}{4}x^{\frac{-3}{2}} = -1

</math> 


Page 10-2 Eq(4) is not satisfied therefore the ODE is not exact. <br>
 <div style="width: 80%; margin-left: auto; margin-right: auto; padding: 4px; border: 2px solid #FF0000; background-color: #FFDDDD; text-align: left;">
Add equation numbers to your solution here so that you can reference the equations in your solution, rather than a transparency that is located somewhere else on the web.  This will make your solution self contained and complete in itself.--[[User:Egm6321.f09.TA|Egm6321.f09.TA]] 03:09, 28 September 2009 (UTC)
</div>


Page 10-2 Eq(5) can be populated to find: 


<math>

\ 0 = 0 + p(0) + 2(0) 

</math> 


<math>

\ 0 = 0  

</math> 


Eq (5) is satisfied. 
 

In order for the ODE to be exact it must satisfy both Eq(4)and Eq(5), since it fails to do this it is concluded that '''the ODE is not exact.'''

=='''Contributing Team Members'''==
[[User:Egm6321.f09.Team 2.walker|Egm6321.f09.Team 2.walker]] 20:58, 20 September 2009 (UTC) (Walker, Matthew)

[[User:EGM6321.f09.Team2.gaddone|Joe Gaddone]] 14:04, 21 September 2009 (UTC)

[[User:Egm6321.f09.Team2.sungsik|Egm6321.f09.Team2.sungsik]] 19:20, 23 September 2009 (UTC)

[[User:EGM6321.f09.Team2.Kumanchik|Kumanchik]] 19:53, 23 September 2009 (UTC)
[[Category:University of Florida]]