# 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 h$_x$N=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 $\scriptstyle M\neq 0$
$-h_{y}M+h(N_{x}-M_{y})=0$, 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 h$_y$M=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) f$_{xx}$ + (2p) f$_{xy}$ + (p$^2$) f$_{yy}$ = g$_{xp}$ + p(g$_{yp}$) - g$_y$
(2) f$_{xp}$ + p (f$_{yp}$) + 2 (f$_y$) = g$_{pp}$
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: f$_{xx}$ = ${\frac {d}{dx}}$[${\frac {d}{dx}}$($xy$)] = ${\frac {d}{dx}}$($y$) = 0
f$_{xy}$ = ${\frac {d}{dy}}$[${\frac {d}{dx}}$($xy$)] = ${\frac {d}{dy}}$($y$) = 1
f$_{yy}$ = ${\frac {d}{dy}}$[${\frac {d}{dy}}$($xy$)] = ${\frac {d}{dy}}$($x$) = 0
g$_{xp}$ = ${\frac {d}{dp}}$[${\frac {d}{dx}}$($x(p)^{2}+yp$)] = ${\frac {d}{dp}}$($p^{2}$) = 2p
g$_{yp}$ = ${\frac {d}{dp}}$[${\frac {d}{dy}}$($x(p)^{2}+yp$)] = ${\frac {d}{dp}}$($p$) = 1
g$_y$ = ${\frac {d}{dy}}$($x(p)^{2}+yp$) = p
Plugging back into the first half of our condition,
$0+2p(1)+p<sup>2</sup>(0)=2p+p(1)-p$ $2p=2p$ 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: f$_{xp}$ = ${\frac {d}{dp}}$[${\frac {d}{dy}}$($xy$)] = ${\frac {d}{dp}}$($y$) = 0
f$_{yp}$ = ${\frac {d}{dp}}$[${\frac {d}{dx}}$($xy$)] = ${\frac {d}{dp}}$($x$) = 0
f$_y$ = ${\frac {d}{dy}}$($xy$)] = x
g$_{pp}$ = ${\frac {d}{dp}}$[${\frac {d}{dp}}$($x(p)^{2}+yp$)] = ${\frac {d}{dp}}$($2xp+y$) = $2x(p')$
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)