# 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 $F(x,y,y')=0$  must be in the form $M(x,y)+N(x,y)y'=0$

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

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

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

The equation becomes:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

## Problem 2

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

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

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

If $P_{(x)}\neq 0$  for all (x) then:

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

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

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

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

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

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

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

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

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

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

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 $h(x,y)=h(x)$  instead of $h(x,y)=h(y)$ . --Egm6321.f09.TA 02:42, 28 September 2009 (UTC)

Assume hyM=0, so our equation becomes;

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

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

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

$N_{x}=0$  and $M_{y}={1 \over x}$

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

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

Using that result we can now solve for $h$

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

$ln|h|=ln|x|$

$h_{x}=x$

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

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

Integrating yields:

$xy={x^{4} \over 4}+C$  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)

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

## Problem #3

Problem Statement: Show that ${\frac {1}{2}}x^{2}y'+[x^{4}y+10]=0$  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 $M(x,y)+N(x,y)y'=0$
(2)$M_{y}=N_{x}$

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

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

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

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

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

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

$(e$ ^$({\frac {2}{5}}x^{5}-{\frac {1}{2}}x^{2})({\frac {1}{2}}x^{2}y'+[x^{4}y+10]=0$  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: ${\frac {\partial {\overline {M}}}{\partial y}}={\frac {\partial {\overline {N}}}{\partial x}}$ .--Egm6321.f09.TA 02:47, 28 September 2009 (UTC)

## Problem #4

Problem Statement: Show that $({\frac {1}{3}}x^{3})(y^{4})y'+(5x^{3}+2)({\frac {1}{5}}y^{5})=0$  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 $M(x,y)y'+N(x,y)=0$  and
(2) $M_{y}=N_{x}$

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

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

The ${\frac {1}{M}}$  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:

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

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

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

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

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

## Problem #5

Problem Statement: Show that the second exactness condition for $xyy''+x(y')^{2}+yy'=0$  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) = $xy$  and g(x,y,p) = $x(y')^{2}+yy'$

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

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

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

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

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

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

Plugging back into the first half of our condition,
$0+2p(1)+p2(0)=2p+p(1)-p$
$2p=2p$  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 = ${\frac {d}{dp}}$ [${\frac {d}{dy}}$ ($xy$ )] = ${\frac {d}{dp}}$ ($y$ ) = 0

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

fy = ${\frac {d}{dy}}$ ($xy$ )] = x

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

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

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

## Problem #6

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

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

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

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

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

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

Therefore,

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

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

## Problem #7

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

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

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

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

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

You have performed much unnecessary rearranging of $\Phi$  terms to get here. You need to rearrange until the end. Equate $\Phi _{xy}=\Phi _{yx}$  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:

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

Show that the Nonlinear 2nd-Order ODE is exact

### Solution

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

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

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

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

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

The Following partial derivatives are then identified:

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

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

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

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

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

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

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

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

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

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

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

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,

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

is Exactness Second ODE is an exact, nonlinear second ODE

Nice work. One caveat: For your final statement, express the equation in terms of $x,y,y'$ . The equation involves an unknown function $y(x)$ , where $p$  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 ${\sqrt {x}}y''+2xy'+3y=0$ .

Given

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

Prove that the equation is not exact.

### Solution

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

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

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

This satisfy satisfies the First Condition of Exactness.

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

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

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

The following partial derivatives are found:

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

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

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

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

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

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

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

Using these values in Eq(4):

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

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

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:

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

$\ 0=0$

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)