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

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**Problem #1**

[edit | edit source]__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 1^{st} order ODE to be exact.

The first condition is that must be in the form

The second condition is that . This example will show how to satisfy the second condition.

Start with equation which already satisfies the first condition.

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

The equation becomes:

Set and rearrange,

To solve for we will let which implies that since

, rearrage terms

take the integral of both sides,

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

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

**Problem 2**

[edit | edit source]__Problem Statement:__ Given the Non-homogeneous linear 1^{st} order ODE with VC show the steps to obtain where C=Const.

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

; Yes [pg.(8-1) Eg.(1)]

If for all (x) then:

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

Multiply by

where

Satisfy the exactness condition where:

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

Assume *h _{y}M*=0, so our equation becomes;

Let and

and

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

Using that result we can now solve for

Now that we have we can multiply through our original equation to obtain,

Integrating yields:

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)

**Problem #3**

[edit | edit source]__Problem Statement:__ Show that 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

(2)

If , then where k is a constant.

Similarly:

, then . Replacing M and N in our initial form

Dividing by and re-arranging:

. Our initial problem meets condition 1, where

where

For condition 2,
and

But these two are not equal. Therefore, using the Integrating Factors Method, we must find such that and thereby satisfy condition 2.

Differentiating and rearranging terms gives us:

. If we assume since and are both functions of x, and rearrange terms, we get:

. Integrating both sides creates . Solving for h:

^. We know and . Therefore ^.

Multiplying and by causes our function to meet the 2nd condition and therefore be exact.

^ 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: .--Egm6321.f09.TA 02:47, 28 September 2009 (UTC)

**Problem #4**

[edit | edit source]__Problem Statement:__ Show that 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 and

(2)

It is obvious from simple observation that our equation meets condition 1; however it fails to meet condition 2.

and then

and . Clearly .

To meet condition 2, we must find h(x,y) such that . As and only differ in , let us assume that . Differentiating and rearranging terms gives us:

. Integrating both sides creates . Solving for h:

^. We know and . Substituting for and and simplifying:

^. This, multiplied by our original equation will cause it to be an exact, nonlinear, first order ODE.

The 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:

where

and

If we assume:

and

then

and

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

Need to verify that --Egm6321.f09.TA 02:51, 28 September 2009 (UTC)

**Problem #5**

[edit | edit source]__Problem Statement:__ Show that the second exactness condition for 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) = and g(x,y,p) =

Looking at our equation, for the first half of the condition:

f_{xx} = [()] = () = 0

f_{xy} = [()] = () = 1

f_{yy} = [()] = () = 0

g_{xp} = [()] = () = 2p

g_{yp} = [()] = () = 1

g_{y} = () = p

Plugging back into the first half of our condition,

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} = [()] = () = 0

f_{yp} = [()] = () = 0

f_{y} = ()] = x

g_{pp} = [()] = () =

Plugging back into the second half of our condition (remembering that p = y'),

thereby satisfying the second half of the condition.

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

**Problem #6**

[edit | edit source]__Problem Statement:__ Derive using the following relations:

(1)

(2)

(3)

(4)

Start by taking the partial derivative of g with respect to p twice:

Now use (1), (3), and (4) to substitute , , and

by (1) and (3)

by (1) and (4)

Therefore,

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

**Problem #7**

[edit | edit source]__Problem Statement:__ Derive using the following relations:

(1)

(2)

(3)

(4)

(5)

Solving (1) for and taking the partial with respect to yields

Then solving (2) for and taking the partial with respect to yields

Equating these two by (3) and solving for yields

Performing a similar task, take (1), solve for and take the partial with respect to

Then solving (2) for and taking the partial with respect to yields

Equating these two by (4) and solving for yields

Finally, and as previously shown and so the two equations are

(a)

(b)

Taking the partial with respect to for (a) and the partial with respect to for (b) and equating based on (5) yields

Noting , , , and moving the f terms to the left and the g terms to the right

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

**Problem #8**

[edit | edit source]__Problem Statement:__ Equations 4&5 on p.(10-2).

Given:

Show that the Nonlinear 2nd-Order ODE is exact

### Solution

[edit | edit source]Second Condition of Exactness: 10-2 Eq (4),(5)

The Following partial derivatives are then identified:

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

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

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,

is Exactness Second ODE is an exact, nonlinear second ODE

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

**Problem #9**

[edit | edit source]__Problem Statement:__ Verify the exactness of the ODE .

Given

Prove that the equation is not exact.

### Solution

[edit | edit source]First Condition of Exactness Page10-1 Eq(1):

This satisfy satisfies the First Condition of Exactness.

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

The following partial derivatives are found:

Using these values in Eq(4):

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:

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**

[edit | edit source]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)