University of Florida/Egm6321/f09.Team2/HW2

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Problem #1Edit

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   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 2Edit

Problem Statement: Given the Non-homogeneous linear 1st 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 hyM=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 #3Edit

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 #4Edit

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 #5Edit

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) 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) =   and g(x,y,p) =  

Looking at our equation, for the first half of the condition:
fxx =  [ ( )] =  ( ) = 0

fxy =  [ ( )] =  ( ) = 1

fyy =  [ ( )] =  ( ) = 0

gxp =  [ ( )] =  ( ) = 2p

gyp =  [ ( )] =  ( ) = 1

gy =  ( ) = 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:
fxp =  [ ( )] =  ( ) = 0

fyp =  [ ( )] =  ( ) = 0

fy =  ( )] = x

gpp =  [ ( )] =  ( ) =  


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 #6Edit

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 #7Edit

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 #8Edit

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



Given:


 


Show that the Nonlinear 2nd-Order ODE is exact

SolutionEdit

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 #9Edit

Problem Statement: Verify the exactness of the ODE  .


Given


 


Prove that the equation is not exact.


SolutionEdit

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 MembersEdit

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)