User:Egm6321.f11.team2/HW2
Contents
 1 R*2.1  Homogenous Solutions of the SecondOrder Legendre Equation
 2 R*2.2 ReducedOrder Solution of an ODE with Missing Dependent Variable
 3 R*2.3 Linearity and Nonlinearity in General FirstOrder ODEs
 4 R*2.4 Verification of Nonlinearity in a Particular FirstOrder ODE
 5 R*2.5 Converting a General FirstOrder ODE into Particular Form
 6 R*2.6  Linear Independence of Solutions to the SecondOrder Legendre ODE
 7 R*2.7  Deriving Exact First Order Differential Equations from Level Sets
 8 R*2.8  First Exactness Condition
 9 R*2.9 Commutative Partial Derivatives
 10 R*2.10Verification of solution for N1ODE
 11 R 2.11Solving PDE for Integrating Factor
 12 R*2.12  Case 2: Finding as a Function of y
 13 R*2.13  Using an Integrating Factor to Render an Inexact L1ODEVC Exact
 14 R*2.14  General L1ODEVC
 15 R*2.15 Integration constant
 16 R*2.16Solution of L1ODEVC
 17 R*2.17Solving for homogenous counterpart
 18 R*2.18  Integrating Factor Method
 19 References
 20 Contributing Members
R*2.1  Homogenous Solutions of the SecondOrder Legendre Equation[edit]
Given[edit]
The Legendre Differential Equation^{[1]} is a secondorder ordinary differential equation given as follows:
2.1.1 
For this problem, we take the specific case n = 1, in which the Legendre Differential Equation takes the form
2.1.2 
It is asserted that the following functions satisfy 2.1.2:
2.1.3 

2.1.4 
Find[edit]
Show that and satisfy 2.1.2.
Solution[edit]
is by far the simpler of the two functions, so we prove that it satisfies 2.1.2 first in order to make the process clear.
First, we find the first and second derivatives of :
2.1.5 

2.1.6 
Substituting 2.1.5 and 2.1.6 into 2.1.2 yields
which proves that satisfies equation 2.1.2.
The process for showing that satisfies 2.1.2 is exactly the same as above.
We begin by carefully taking the first two derivatives of :
Simplifying the last term yields
2.1.7 
For the second derivative, we can easily differentiate the first term by using the results from taking the first derivative.
simplifying yields
2.1.8 
Finally, substituting 2.1.7 and 2.1.8 into 2.1.2 yields
which proves that satisfies 2.1.2.
Since the Legendre Differential Equation is secondorder, and we now possess two linearly independent solutions to the homogenous equation (see R2.6), every solution of the homogenous form of the equation can be expressed as a linear combination of and .
R*2.2 ReducedOrder Solution of an ODE with Missing Dependent Variable[edit]
Given[edit]
(2.2.1)
Solution is that
(2.2.2)
Find[edit]
Verify that eq (2.2.2) is indeed solution for eq (2.2.1).
Solution[edit]
Equation 2.2.1 is that
(2.2.1)
And now Let
Plug in eq (2.2.1)
The equation 2.2.1 is following that
(2.2.2)
It shows that eq (2.2.2) is indeed solution for eq (2.2.1).
If you want double check,
Thus,
R*2.3 Linearity and Nonlinearity in General FirstOrder ODEs[edit]
Given[edit]
An ODE of 1st order is read as
(2.3.1)
Find[edit]
show that the eqn (2.3.1) is linear in , and that it is in general an N1ODE. But it is not the most general N1ODE as represented by
(2.3.2)
Give an example of a more general N1ODE.
Solution[edit]
Definition of Linear[edit]
If is a linear operator, then it must satisfy
Where are the independent variables, and
is linear in .
And it is easy to proof that
So, is in general an N1ODE.
An example of the general N1ODE[edit]
.
(2.3.3)
where is an arbitrary function of .
R*2.4 Verification of Nonlinearity in a Particular FirstOrder ODE[edit]
Given[edit]
(2.4.1) 
Find[edit]
Verify that (2.4.1) is a N1ODE.
Solution[edit]
General nonlinear ODE of order 1 (N1ODE) is
(2.4.2) 
(2.4.1) is a statement of (2.4.2) showing that the highest order is one. Thus, (2.4.1) is a fist order differential equation.
Linearty has to satisfy the following condition:
(2.4.3) 
Nonlinearty is the condition not satisfing the above statement as follows:
(2.4.4) 
(2.4.1) can be rewritten as follows:
(2.4.5) 
And then, the nonlinearity can be checked with the following two functions:
(2.4.6) 
(2.4.7) 
From (2.4.6) and (2.4.7), is found so that (2.4.1) is verified to be a nonlinear equation.
Finally it is concluded that (2.4.1) is a N1ODE (Nonlinear First Oder Differential Equation).
R*2.5 Converting a General FirstOrder ODE into Particular Form[edit]
Given[edit]
2.5.1 
Equation 2.5.1 satisfies the most general form , but can be converted into particular form .
Find[edit]
Using the hint:
2.5.2 
and
2.5.3 
Illustrate a way to convert equation 2.5.1 into particular form .
Solution[edit]
2.5.4 
Let
And
Equation 2.5.4 becomes which is a particular form of Equation 2.5.1 .
R*2.6  Linear Independence of Solutions to the SecondOrder Legendre ODE[edit]
Given[edit]
The Legendre Differential Operator/Equation takes the form:
(1)
(2)
Two homogeneous solutions for this equation are:
(3)
(4)
Find[edit]
Show that equations (3) and (4) are linearly independent, i.e., show that
i.e., for any given , show that
Plot and .
Solution[edit]
Since we know there exists an and only one such that for
We also know that for any other point there exists an such that
The two homogeneous solutions are linearly independent if
For
Solving for we determine = 0.5677175.
For
Solving for we determine = 0.2571305.
Since the homogeneous solutions must be linearly independent. This can be seen by looking at their graph.
R*2.7  Deriving Exact First Order Differential Equations from Level Sets[edit]
Given[edit]
The family of level sets
2.7.1 
Find[edit]
and show that it is an N1ODE.
Solution[edit]
We know from the chainrule^{[2]} that
Applying this directly yields
Thus
2.7.2 
which, as can be seen, is already in the form .
Due to the way that it was derived, it is expected that 2.7.2 is exact, and indeed
It is obvious that G is an ordinary differential equation and that it is firstorder.
To prove that it is nonlinear, suppose that and are functions of x. Then, slightly changing our notation to G(y) for clarity, we see that
The powers of y in G make it nonlinear.
R*2.8  First Exactness Condition[edit]
Given[edit]
First exactness condition is that an ODE can be written in the form
Find[edit]
Does the following N1ODE satisfy the first exactness condition?
(2.8.1)
Solution[edit]
In order to satisfy the first exactness condition, it must be in the form of:
(2.8.2)
So rearrange eq (2.8.1) to fit eq (2.8.2)
is any function of y'. If the function has no explicit inverse
then it can not satisfy in the form of eq (2.8.2)
In this case, does not have exact inverse
Thus, This does not satisfy the first exactness condition.
R*2.9 Commutative Partial Derivatives[edit]
Given[edit]
Review calculus, and differentiation
(2.9.1)
Find[edit]
Find the minimum degree of differentiability of the function such that the above equation is satisfied, State the full theorem and provide a proof.
Solution[edit]
Clairaut's theorem[edit]
According to the Schwarz's theorem,^{[3]}, if
has continuous second partial derivatives at any given point in , then for
And we then know that the minimum degree of differentiability of the function must be three when it satisfies eqn (2.9.1)
Proof of Clairaut's theorem[edit]
The result was first discovered by L.Euler around 1734 in connection with problems in hydrodynamics ^{[4]}. In order to prove this theorem, we shall first introduce the Mean Value theorem as follows:
Let be a continuous function on the closed interval , and derivative on the open interval , where , Then such that
(2.9.2)
Suppose Q is a closed rectangle with sides parallel to the coordinate axes, put
(2.9.3)
(2.9.4)
So, we put

 ,
Chose . if and are sufficiently small, we have
(2.9.5)
Fix , and let , then
(2.9.6)
Since was is arbitrary, and (2.9.6) holds for all sufficiently small
(2.9.7)
Since is an arbitrary point in , so
R*2.10Verification of solution for N1ODE[edit]
Given[edit]
(2.10.1) 
(2.10.2) 
where and
Find[edit]
Verify that (2.10.1) is the solution for the (2.10.2).
Solution[edit]
(2.10.1) would be rewritten as follows:
(2.10.3) 
Differentiating the both sides of (2.10.3), (2.10.3) results in
(2.10.4) 
Thus, (2.10.1) is verified to be the solution for (2.10.2.)
R 2.11Solving PDE for Integrating Factor[edit]
Given[edit]
For
2.11.1 
Find[edit]
Why solving equation 2.11.1 for the integrating factor is usually not easy ?
Solution[edit]
Solving equation 2.11.1 for the integrating factor is usually not easy because in general equation 2.11.1 is a nonlinear partial differential equation for unknown function and it is having terms which makes it very difficult to solve for .
R*2.12  Case 2: Finding as a Function of y[edit]
Given[edit]
A N1ODE satisfies the first exactness condition if it takes the particular form:
(1)
The second exactness condition states that
(2)
If a N1ODE satisfies the first exactness condition but not the second exactness condition the Euler Integrating factor method (IFM) can be used.
This requires an integrating factor be found such that the following N1ODE is exact:
(3)
This can be written as:
(4)
To find we must apply the second exactness condition to obtain:
(5)
Find[edit]
Suppose , thus is a function of y only, then equation (5) becomes:
(6)
Find using equation (6).
Solution[edit]
To find we can integrate equation (6) to obtain:
(7)
R*2.13  Using an Integrating Factor to Render an Inexact L1ODEVC Exact[edit]
Given[edit]
2.13.1 
Find[edit]
Show that h(x) = x is an integrating factor for 2.13.1, and that
is its solution.
Solution[edit]
We first show that 2.13.1 is not already exact, or else its integrating factor would be !
Rearranging 2.13.1, we arrive at the standard form
Now,
so 2.13.1 is not exact.
We thus proceed to multiply 2.13.2 by h(x) = x which yields
2.13.2 
Now,
so 2.13.2 is exact and h(x) = x is the integrating factor that makes 2.13.1 exact.
Since every exact differential equation may be produced from a first integral such that
and
we may integrate and to find as follows:
2.13.3 
2.13.4 
Combining 2.13.3 and 2.13.4 yields
which upon solving for yields
While this already proves the assertion, we can also show that this is the correct expression for by plugging it back in to equation 2.13.1:
R*2.14  General L1ODEVC[edit]
Given[edit]
(2.14.1)
(2.14.2)
(2.14.3)
Find[edit]
Solve the general L1ODEVC
1.
2. Find in term of
3.
Solution[edit]
an integrating factor is a function that is chosen to facilitate the solving of a given equation involving differentials. It is commonly used to solve ordinary differential equations, but is also used within multivariable calculus, in this case often multiplying through by an integrating factor allows an inexact differential to be made into an exact differential.
From lecture note,
And applying second exactness condition, the equation is following that
If we assume that , then h is function of x only and the equation is following that
And we obtain some equations through rearranging equation above
And we can obtain h(x) through dividing by h
(2.14.1)
And in order to obtain y(x), we can apply the integrating factor to obtain the following equations
and
Thus,
(2.14.2)
part 1[edit]
1. In order to solve this problems, plug given information in eq (2.14.3)
the equation is following that
Also, eq (2.14.1) is following that
And eq (2.14.2) is following that
Let
Finally, the equation is that
In order to solve more, numerical method must be used.
part 2[edit]
From eq 2.14.3
and plug in eq 2.14.1
part 3[edit]
At this point,
Let
(2.14.4)
Then plug in eq 2.14.2
Let
Final solution is that
R*2.15 Integration constant[edit]
Given[edit]
Since
(2.15.1)
(2.15.2)
is an L1ODE_VC, there should be only one integration constant, not two.
Find[edit]
Show that the integration constant in
(2.15.3)
is not necessary, i.e., only in
(2.15.4)
is necessary.
Solution[edit]
Substituting the eqn (2.15.3) to eqn (2.15.4), we obtain the following expression
Then, we can define a new constant such that
hence, this new constant contains . In other words, is unnecessary.
R*2.16Solution of L1ODEVC[edit]
Given[edit]
(2.16.1) 
(2.16.2) 
The derivated result in King 2003 p.512 is
(2.16.3) 
Find[edit]
Show that (2.16.1) agrees with (2.16.3).
Solution[edit]
(2.16.4) 
In King 2003 p.512, the L1ODEVC is presented as follows:
(2.16.5) 
From (2.16.3) and (2.164), the following relations can be derived.
(2.16.6) 
(2.16.7) 
(2.16.8) 
From the relations, in King 2003 P512, the integrating factor is shwon as follows:
(2.16.9) 
When (2.16.1) is substituted by (2.16.9), (2.16.1) becomes
(2.16.10) 
Substituting the relation (2.16.8) for (2.16.10), (2.16.10) can be rewritten as follows:
(2.16.11) 
(2.16.10) is the solution presented in King 2003 p.512 identifying A, and as follows:
(2.16.12) 
(2.16.13) 
(2.16.14) 
Thus, it is verified that (2.16.1) agrees with (2.16.3).
R*2.17Solving for homogenous counterpart[edit]
Given[edit]
Instead of identifying from .
Find[edit]
Solve the homogenous counterpart of .
Solution[edit]
R*2.18  Integrating Factor Method[edit]
Given[edit]
A L1ODEVC satisfies the first exactness condition if it takes the particular form:
(1)
The second exactness condition states that
(2)
If a L1ODEVC satisfies the first exactness condition but not the second exactness condition the Euler Integrating factor method (IFM) can be used.
This requires an integrating factor be found such that the following N1ODE is exact:
(3)
Find[edit]
(4)
Find out whether equation (4) is exact. If it is not exact find the integrating factor to make it exact.
Solution[edit]
Since equation (4) satisfies the first exactness condition we must test the second exactness condition.
Therefore, equation (4) is not exact. To make it exact an integration factor must be found such that:
Solving for h(x) produces:
References[edit]
Contributing Members[edit]


Problem Number  Assigned To  Solved By  Typed By  Proofread By  Signature 
2.1  Jonathan  Jonathan  Jonathan  all  Egm6321.f11.team2.epps 13:22, 21 September 2011 (UTC) 
2.2  kim  kim  kim  all  Egm6321.f11.team2.kim 19:51, 21 September 2011 (UTC) 
2.3  Langshah  Langshah  Langshah  all  
2.4  Seounghyun  Seounghyun  Seounghyun  all  Egm6321.f11.team2.rho 19:09, 21 September 2011 (UTC) 
2.5  Yuvraj  Yuvraj  Yuvraj  all  Egm6321.f11.team2.johri.y 17:36, 21 September 2011 (UTC) 
2.6  Gary  Gary  Gary  all  
2.7  Jonathan  Jonathan  Jonathan  all  Egm6321.f11.team2.epps 13:22, 21 September 2011 (UTC) 
2.8  Kim  Kim  Kim  all  Egm6321.f11.team2.kim 17:08, 21 September 2011 (UTC) 
2.9  Langshah  Langshah  Langshah  all  
2.10  Seounghyun  Seounghyun  Seounghyun  all  Egm6321.f11.team2.rho 19:09, 21 September 2011 (UTC) 
2.11  Yuvraj  Yuvraj  Yuvraj  all  Egm6321.f11.team2.johri.y 17:36, 21 September 2011 (UTC) 
2.12  Gary  Gary  Gary  all  
2.13  Jonathan  Jonathan  Jonathan  all  Egm6321.f11.team2.epps 13:22, 21 September 2011 (UTC) 
2.14  Kim  Kim  Kim  all  Egm6321.f11.team2.kim 17:08, 21 September 2011 (UTC) 
2.15  Langshah  Langshah  Langshah  all  
2.16  Seounghyun  Seounghyun  Seounghyun  all  Egm6321.f11.team2.rho 19:09, 21 September 2011 (UTC) 
2.17  Yuvraj  Yuvraj  Yuvraj  all  Egm6321.f11.team2.johri.y 17:36, 21 September 2011 (UTC) 
2.18  Gary  Gary  Gary  all 
Egm6321.f11.team2.rho 20:41, 21 September 2011 (UTC)
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