# University of Florida/Egm6321/f09.Team2/HW2

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## Problem #1

Problem Statement: Find the Euler Integrating Factor h(y) for a non-linear 1st order ODE for the particular case where we assume hxN=0

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

The first condition is that ${\displaystyle F(x,y,y')=0}$  must be in the form ${\displaystyle M(x,y)+N(x,y)y'=0}$

The second condition is that ${\displaystyle M_{y}=N_{x}}$ . This example will show how to satisfy the second condition.

Start with equation ${\displaystyle M(x,y)+N(x,y)y'=0}$  which already satisfies the first condition.

If we assume the equation is not exact we can multiply it by an unkown factor ${\displaystyle h(x,y)}$  (Euler Integrating Factor) to attempt to find a factor that will make the equation exact.

The equation becomes:

${\displaystyle h(x,y)[M(x,y)dx+N(x,y)dy]=0}$

${\displaystyle (hM)dx+(hN)dy=0,where\ hM={\overline {M}},and\ hN={\overline {N}}}$

${\displaystyle {\overline {M}}_{y}=(hM)_{y}=h_{y}M+hM_{y}}$

${\displaystyle {\overline {N}}_{x}=(hN)_{y}=h_{x}N+hN_{x}}$

Set ${\displaystyle {\overline {M}}_{y}={\overline {N}}_{x}}$  and rearrange,

${\displaystyle h_{x}N-h_{y}M+h(N_{x}-M_{y})=0}$

To solve for ${\displaystyle h_{y}}$  we will let ${\displaystyle h_{x}N=0}$  which implies that ${\displaystyle h_{x}=0}$  since ${\displaystyle \scriptstyle M\neq 0}$

${\displaystyle -h_{y}M+h(N_{x}-M_{y})=0}$ , rearrage terms

${\displaystyle {h_{y} \over h}dy={1 \over M}(N_{x}-M_{y})}$  take the integral of both sides,

${\displaystyle \int {h_{y} \over h}dy=\int {1 \over M}(N_{x}-M_{y})dy}$

If the right hand side of the integral is only a function of y then it can be written as ${\displaystyle g(y)}$

${\displaystyle {1 \over M}(N_{x}-M_{y})=g(y)}$

${\displaystyle ln\mid h\mid =\int g(y)dy}$

${\displaystyle h(y)=exp\int _{}^{y}g(s)ds}$

You have lost a negative sign in your expression for ${\displaystyle g(y)}$ --Egm6321.f09.TA 02:38, 28 September 2009 (UTC)

## Problem 2

Problem Statement: Given the Non-homogeneous linear 1st order ODE with VC ${\displaystyle y'+{1 \over x}y=x^{2}}$  show the steps to obtain ${\displaystyle y={x^{3} \over 4}+{C \over x}}$  where C=Const.

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

${\displaystyle P_{(x)}y'+Q_{(x)}y=R_{(x)}}$  ; Yes [pg.(8-1) Eg.(1)]

If ${\displaystyle P_{(x)}\neq 0}$  for all (x) then:

${\displaystyle y'+{Q_{(x)} \over P_{(x)}}y={R_{(x)} \over P{(x)}}}$

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

${\displaystyle N_{(x,y)}{dy \over dx}+M_{(x,y)}=0}$

Multiply by ${\displaystyle h_{(x,y)}}$

${\displaystyle h_{(x,y)}[N_{(x,y)}dy+M_{(x,y)}dx]}$

${\displaystyle (hM)dx+(hN)dy=0}$  where ${\displaystyle (hM)={\overline {M}};\ (hN)={\overline {N}}}$

Satisfy the exactness condition ${\displaystyle {\overline {M}}_{y}={\overline {N}}_{x}}$  where:

${\displaystyle {\overline {M}}_{y}=(hM)_{y}=h_{y}M+hM_{y}}$

${\displaystyle {\overline {N}}_{x}=(hN)_{x}=h_{x}N+hN_{x}}$

${\displaystyle h_{x}N-h_{y}M+h(Nx-My)=0}$

To solve for h(x,y) we must make an assumption that h is a function of (x) only.

You do not have to make this assumption. You have chosen ${\displaystyle h(x,y)=h(x)}$  instead of ${\displaystyle h(x,y)=h(y)}$ . --Egm6321.f09.TA 02:42, 28 September 2009 (UTC)

Assume hyM=0, so our equation becomes;

${\displaystyle h_{x}N+h(N_{x}-M_{y})=0}$

${\displaystyle {h_{x} \over h}=-{1 \over N_{(x,y)}}(N_{x}-M_{y})}$

Let ${\displaystyle N_{(x,y)}=1}$  and ${\displaystyle M_{(x,y)}={Q_{(x)} \over P_{(x)}}y={1 \over x}y}$

${\displaystyle N_{x}=0}$  and ${\displaystyle M_{y}={1 \over x}}$

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

${\displaystyle {1 \over 1}(0-{1 \over x})={1 \over x}=f(x)}$

Using that result we can now solve for ${\displaystyle h}$

${\displaystyle \int {h_{x} \over h}dx=\int {1 \over x}dx}$

${\displaystyle ln|h|=ln|x|}$

${\displaystyle h_{x}=x}$

Now that we have ${\displaystyle h_{x}}$  we can multiply through our original equation to obtain,

${\displaystyle (x)y'+(x){1 \over x}y=(x)x^{2}\ =xy'+y=x^{3}}$

Integrating yields:

${\displaystyle xy={x^{4} \over 4}+C}$  divide by x,

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.--Egm6321.f09.TA 02:42, 28 September 2009 (UTC)

${\displaystyle y={x^{3} \over 4}+{C \over x}}$

## Problem #3

Problem Statement: Show that ${\displaystyle {\frac {1}{2}}x^{2}y'+[x^{4}y+10]=0}$  is exact.

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.

To be exact, it must meet 2 conditions:

(1)it must meet the form ${\displaystyle M(x,y)+N(x,y)y'=0}$
(2)${\displaystyle M_{y}=N_{x}}$

If ${\displaystyle M(x,y)dy=a(x)}$  , then ${\displaystyle M(x,y)=a(x)y+k}$  where k is a constant.

Similarly:
${\displaystyle N(x,y)dx=b(x)}$  , then ${\displaystyle N(x,y)=\int _{}^{x}b(s)\,ds=c(x)={\overline {b}}(x)}$ . Replacing M and N in our initial form

${\displaystyle M(x,y)dx+N(x,y)dy=[a(x)y+k]dx+c(x)dy=0}$

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

For condition 2, ${\displaystyle M_{y}={\frac {d}{dy}}[a(x)+k]={\frac {d}{dy}}(x^{4}y+10)=x^{4}}$  and ${\displaystyle N_{x}={\frac {d}{dx}}({\frac {1}{2}}x^{2})=x}$

But these two are not equal. Therefore, using the Integrating Factors Method, we must find ${\displaystyle h(x,y)}$  such that ${\displaystyle (hM)_{y}=(hN)_{x}}$  and thereby satisfy condition 2.
Differentiating and rearranging terms gives us:
${\displaystyle h_{x}N-h_{y}M+h(N_{x}-M_{y})=0}$ . If we assume ${\displaystyle h_{y}=0}$  since ${\displaystyle M_{y}}$  and ${\displaystyle N_{x}}$  are both functions of x, and rearrange terms, we get:
${\displaystyle {\frac {h_{x}}{h}}={\frac {-1}{N}}(N_{x}-M_{y})dx}$ . Integrating both sides creates ${\displaystyle log(h)}$ . Solving for h:
${\displaystyle h=e}$ ^${\displaystyle (-2)(\int _{}^{x}N_{x}-M_{y},\ dx}$ . We know ${\displaystyle N_{x}}$  and ${\displaystyle M_{y}}$ . Therefore ${\displaystyle h=e}$ ^${\displaystyle (-2)({\frac {1}{2}}x^{2}-{\frac {1}{5}}x^{5})}$ .
Multiplying ${\displaystyle M(x,y)}$  and ${\displaystyle N(x,y)}$  by ${\displaystyle h}$  causes our function to meet the 2nd condition and therefore be exact.

${\displaystyle (e}$ ^${\displaystyle ({\frac {2}{5}}x^{5}-{\frac {1}{2}}x^{2})({\frac {1}{2}}x^{2}y'+[x^{4}y+10]=0}$  is exact.

This equation is incomplete. There are parenthesis missing. Where are the differentials?--Egm6321.f09.TA 02:47, 28 September 2009 (UTC)

You need to show that you have arrived at the correct integrating factor. Specifically: ${\displaystyle {\frac {\partial {\overline {M}}}{\partial y}}={\frac {\partial {\overline {N}}}{\partial x}}}$ .--Egm6321.f09.TA 02:47, 28 September 2009 (UTC)

## Problem #4

Problem Statement: Show that ${\displaystyle ({\frac {1}{3}}x^{3})(y^{4})y'+(5x^{3}+2)({\frac {1}{5}}y^{5})=0}$  is an exact nonlinear, first order ODE.

To prove this equation is an exact, non-linear, 1st order ODE, it must meet 2 conditions:
(1) it must fit the form ${\displaystyle M(x,y)y'+N(x,y)=0}$  and
(2) ${\displaystyle M_{y}=N_{x}}$

It is obvious from simple observation that our equation meets condition 1; however it fails to meet condition 2.
${\displaystyle M(x,y)={\frac {1}{3}}x^{3}y^{4}}$  and ${\displaystyle N(x,y)=(5x^{3}-2)({\frac {1}{5}}y^{5})}$  then
${\displaystyle M_{y}={\frac {4}{3}}x^{3}y^{3}}$  and ${\displaystyle N_{x}=5x^{3}y^{4}}$ . Clearly ${\displaystyle M_{y}\neq N_{x}}$ .

To meet condition 2, we must find h(x,y) such that ${\displaystyle (hM)_{y}=(hN)_{x}}$ . As ${\displaystyle M_{y}}$  and ${\displaystyle N_{x}}$  only differ in ${\displaystyle y}$ , let us assume that ${\displaystyle h_{x}=0}$ . Differentiating and rearranging terms gives us:
${\displaystyle {\frac {h_{y}}{h}}={\frac {1}{M}}(N_{x}-M_{y})dx}$ . Integrating both sides creates ${\displaystyle logh}$ . Solving for h:
${\displaystyle h=e}$ ^${\displaystyle ({\frac {1}{M}}\int _{}^{y}N_{x}-M_{y},\ dy)}$ . We know ${\displaystyle N_{x}}$  and ${\displaystyle M_{y}}$ . Substituting for ${\displaystyle N_{x}}$  and ${\displaystyle M_{y}}$  and simplifying:
${\displaystyle h=e}$ ^${\displaystyle (15y-4(log(y))}$ . This, multiplied by our original equation will cause it to be an exact, nonlinear, first order ODE.

The ${\displaystyle {\frac {1}{M}}}$  should be inside the integral.--Egm6321.f09.TA 02:51, 28 September 2009 (UTC)

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:

${\displaystyle {\overline {b}}(x)c(y)y'+a(x){\overline {c}}(y)=0}$

where ${\displaystyle {\overline {b}}(x)=\int _{}^{x}b(x)\,dx}$
and ${\displaystyle {\overline {c}}(y)=\int _{}^{y}c(y)\,dy}$

If we assume:
${\displaystyle a(x)=5x^{3}+2}$
${\displaystyle b(x)=x^{2}}$  and
${\displaystyle c(y)=y^{4}}$  then
${\displaystyle {\overline {b}}(x)=\int _{}^{x}x^{2}\,dx={\frac {1}{3}}x^{3}}$  and ${\displaystyle {\overline {c}}(y)=\int _{}^{y}y^{4}\,dy={\frac {1}{5}}y^{5}}$

Since our equation does fit the form for this class, it is an exact, nonlinear, first order ODE.

Need to verify that ${\displaystyle {\overline {N}}_{x}={\overline {M}}_{y}}$ --Egm6321.f09.TA 02:51, 28 September 2009 (UTC)

## Problem #5

Problem Statement: Show that the second exactness condition for ${\displaystyle xyy''+x(y')^{2}+yy'=0}$  is satisfied.

The 2nd exactness condition actually consists of 2 parts, and our equation must meet both:
(1) fxx + (2p) fxy + (p2) fyy = gxp + p(gyp) - gy

(2) fxp + p (fyp) + 2 (fy) = gpp

where p is y'

In our equation,
f(x,y,p) = ${\displaystyle xy}$  and g(x,y,p) = ${\displaystyle x(y')^{2}+yy'}$

Looking at our equation, for the first half of the condition:
fxx = ${\displaystyle {\frac {d}{dx}}}$ [${\displaystyle {\frac {d}{dx}}}$ (${\displaystyle xy}$ )] = ${\displaystyle {\frac {d}{dx}}}$ (${\displaystyle y}$ ) = 0

fxy = ${\displaystyle {\frac {d}{dy}}}$ [${\displaystyle {\frac {d}{dx}}}$ (${\displaystyle xy}$ )] = ${\displaystyle {\frac {d}{dy}}}$ (${\displaystyle y}$ ) = 1

fyy = ${\displaystyle {\frac {d}{dy}}}$ [${\displaystyle {\frac {d}{dy}}}$ (${\displaystyle xy}$ )] = ${\displaystyle {\frac {d}{dy}}}$ (${\displaystyle x}$ ) = 0

gxp = ${\displaystyle {\frac {d}{dp}}}$ [${\displaystyle {\frac {d}{dx}}}$ (${\displaystyle x(p)^{2}+yp}$ )] = ${\displaystyle {\frac {d}{dp}}}$ (${\displaystyle p^{2}}$ ) = 2p

gyp = ${\displaystyle {\frac {d}{dp}}}$ [${\displaystyle {\frac {d}{dy}}}$ (${\displaystyle x(p)^{2}+yp}$ )] = ${\displaystyle {\frac {d}{dp}}}$ (${\displaystyle p}$ ) = 1

gy = ${\displaystyle {\frac {d}{dy}}}$ (${\displaystyle x(p)^{2}+yp}$ ) = p

Plugging back into the first half of our condition,
${\displaystyle 0+2p(1)+p2(0)=2p+p(1)-p}$
${\displaystyle 2p=2p}$  and therefore the first half of the condition is satisfied.

What is sup? Check to verify that your code has rendered properly.--Egm6321.f09.TA 02:57, 28 September 2009 (UTC)

Looking at the 2nd half of the condition:
fxp = ${\displaystyle {\frac {d}{dp}}}$ [${\displaystyle {\frac {d}{dy}}}$ (${\displaystyle xy}$ )] = ${\displaystyle {\frac {d}{dp}}}$ (${\displaystyle y}$ ) = 0

fyp = ${\displaystyle {\frac {d}{dp}}}$ [${\displaystyle {\frac {d}{dx}}}$ (${\displaystyle xy}$ )] = ${\displaystyle {\frac {d}{dp}}}$ (${\displaystyle x}$ ) = 0

fy = ${\displaystyle {\frac {d}{dy}}}$ (${\displaystyle xy}$ )] = x

gpp = ${\displaystyle {\frac {d}{dp}}}$ [${\displaystyle {\frac {d}{dp}}}$ (${\displaystyle x(p)^{2}+yp}$ )] = ${\displaystyle {\frac {d}{dp}}}$ (${\displaystyle 2xp+y}$ ) = ${\displaystyle 2x(p')}$

Plugging back into the second half of our condition (remembering that p = y'),
${\displaystyle 0+p(0)+2(x)p'=2x(p')}$  thereby satisfying the second half of the condition.

What is ${\displaystyle p'}$  ? This would be ${\displaystyle y''}$  and this should not appear in this problem.--Egm6321.f09.TA 02:57, 28 September 2009 (UTC)

## Problem #6

Problem Statement: Derive ${\displaystyle f_{xp}+pf_{yp}+2f_{y}=g_{pp}}$  using the following relations:

(1) ${\displaystyle f(x,y,p):=\phi _{p}(x,y,p)}$
(2) ${\displaystyle g(x,y,p):=\phi _{x}+\phi _{y}p}$
(3) ${\displaystyle \phi _{xp}=\phi _{px}}$
(4) ${\displaystyle \phi _{yp}=\phi _{py}}$

Start by taking the partial derivative of g with respect to p twice: ${\displaystyle g_{pp}={\frac {\partial ^{2}g}{\partial p^{2}}}}$

${\displaystyle g_{p}=\phi _{xp}+\phi _{yp}p+\phi _{y}}$
${\displaystyle g_{pp}=\phi _{xpp}+\phi _{ypp}p+\phi _{yp}+\phi _{yp}}$
${\displaystyle =>g_{pp}=\phi _{xpp}+\phi _{ypp}p+2\phi _{yp}}$

Now use (1), (3), and (4) to substitute ${\displaystyle \phi _{xpp}}$ , ${\displaystyle \phi _{ypp}}$ , and ${\displaystyle \phi _{yp}}$

${\displaystyle f_{x}=\phi _{px}=\phi _{xp}}$  by (1) and (3)
${\displaystyle f_{xp}=\phi _{xpp}}$
${\displaystyle f_{y}=\phi _{py}=\phi _{yp}}$  by (1) and (4)
${\displaystyle f_{yp}=\phi _{ypp}}$

Therefore,

${\displaystyle g_{pp}=f_{xp}+f_{yp}p+2f_{y}=f_{xp}+pf_{yp}+2f_{y}}$

Clear and concise.--Egm6321.f09.TA 02:58, 28 September 2009 (UTC)

## Problem #7

Problem Statement: Derive ${\displaystyle f_{xx}+2pf_{xy}+p^{2}f_{yy}=g_{xp}+pg_{yp}-g_{y}}$  using the following relations:
(1) ${\displaystyle f(x,y,p):=\phi _{p}(x,y,p)}$
(2) ${\displaystyle g(x,y,p):=\phi _{x}+\phi _{y}p}$
(3) ${\displaystyle \phi _{px}=\phi _{xp}}$
(4) ${\displaystyle \phi _{py}=\phi _{yp}}$
(5) ${\displaystyle \phi _{xy}=\phi _{yx}}$

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

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

Noting ${\displaystyle f_{yx}=f_{xy}}$ , ${\displaystyle g_{px}=g_{xp}}$ , ${\displaystyle g_{py}=g_{yp}}$ , and moving the f terms to the left and the g terms to the right

${\displaystyle f_{xx}+2pf_{xy}+p^{2}f_{yy}=g_{xp}+pg_{yp}-g_{y}}$

You have performed much unnecessary rearranging of ${\displaystyle \Phi }$  terms to get here. You need to rearrange until the end. Equate ${\displaystyle \Phi _{xy}=\Phi _{yx}}$  using the appropriate relations and your solution will be must more straight forward.--Egm6321.f09.TA 03:01, 28 September 2009 (UTC)

## Problem #8

Problem Statement: Equations 4&5 on p.(10-2).

Given:

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

Show that the Nonlinear 2nd-Order ODE is exact

### Solution

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

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

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

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

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

The Following partial derivatives are then identified:

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

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

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

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

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

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

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

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

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

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

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

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,

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

is Exactness Second ODE is an exact, nonlinear second ODE

Nice work. One caveat: For your final statement, express the equation in terms of ${\displaystyle x,y,y'}$ . The equation involves an unknown function ${\displaystyle y(x)}$ , where ${\displaystyle p}$  was a dummy variable which helped us perform differentiations. --Egm6321.f09.TA 03:05, 28 September 2009 (UTC)

## Problem #9

Problem Statement: Verify the exactness of the ODE ${\displaystyle {\sqrt {x}}y''+2xy'+3y=0}$ .

Given

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

Prove that the equation is not exact.

### Solution

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

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

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

This satisfy satisfies the First Condition of Exactness.

The Second Exactness Condition for a second order ODE is as follows:

${\displaystyle \ F_{xx}+2pF_{xy}+p^{2}F_{yy}=G_{xp}+pG_{yp}-G_{y}\qquad Page10-2Eq(4)}$

${\displaystyle \ F_{xp}+pF_{yp}+2F_{y}=G_{pp}\qquad Page10-2Eq(5)}$

The following partial derivatives are found:

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

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

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

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

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

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

${\displaystyle F_{yp}=0\qquad \qquad }$

Using these values in Eq(4):

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

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

Page 10-2 Eq(4) is not satisfied therefore the ODE is not exact.

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.--Egm6321.f09.TA 03:09, 28 September 2009 (UTC)

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

${\displaystyle \ 0=0+p(0)+2(0)}$

${\displaystyle \ 0=0}$

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

Egm6321.f09.Team 2.walker 20:58, 20 September 2009 (UTC) (Walker, Matthew)

Joe Gaddone 14:04, 21 September 2009 (UTC)

Egm6321.f09.Team2.sungsik 19:20, 23 September 2009 (UTC)

Kumanchik 19:53, 23 September 2009 (UTC)