User:Egm6321.f11.team4/HW5
Contents
 1 Problem R*5.1  Equivalence of twos forms of 2nd exactness condition
 2 Problem R*5.2  Verification of the exactness of Legendre and Hermite equations
 3 References
 4 Problem R*5.3 Method of undetermined coefficients
 5 Problem R*5.4  Find in terms of the derivatives of y with respect to t
 6 Problem R*5.5  Solve y and plot the solution
 7 Problem R5.6: Equivalence of Two methods to solve Euler LnODEVC
 8 Problem R*5.7Euler L2ODEs
 9 Problem R*5.8  Variation of parameters
 10 Problem R*5.9 Special IFM Solution
 11 Contributing Members
Problem R*5.1  Equivalence of twos forms of 2nd exactness condition[edit]
From Mtg 226
Given[edit]

(1.1) 



Find[edit]
Equivalence of the form of 2nd exactness condition:

Solution[edit]
 Solved on our own
Let's assume that p=y' and q=y.

(1.1) 


(1.2) 


(1.3) 
As a result,

Author[edit]
Problem R*5.2  Verification of the exactness of Legendre and Hermite equations[edit]
From Mtg 271
Given[edit]
Legendre equation:

(2.1) 
Hermite equation:

(2.2) 
Find[edit]
1. Verify the exactness of the designated L2ODEVC (Eq. 2.1, 2.2).
2. If Hermite equation is not exact, check whether it is in power form, and see whether it can be made exact using IFM with
3. The first few Hermite polynomials are given as below.

(2.3) 
Verify that the equations in (2.3) are homogeneous solutions of the Hermite differential equation (Eq. 2.2).
Solution[edit]
 Solved on our own
Part 1[edit]
Legendre equation:
The first exactness condition is satisfied since the equation has a form of Eq. (2.4).

(2.4) 
Method 1 : the 2nd exactness condition
In order to verify the second exactness condition, the following terms are computed.

(2.5) 

(2.6) 

(2.7) 
 The second exactness condition is satisfied when or .
Method 2 : the 2nd exactness condition
Following is another condition to prove the second exactness.

(2.8) 
Using the definition of , the followings are computed.

(2.9) 
After substituting the terms in Eq. (2.9), Eq. (2.8) becomes as following.

(2.10) 
 The second exactness condition is satisfied when or .
Hermite equation:
The first exactness condition is satisfied since the equation has a form of Eq. (2.8).

(2.11) 
Method 1 : the 2nd exactness condition
In order to verify the second exactness condition, the following terms are computed.

(2.12) 

(2.13) 

(2.14) 
 The second exactness condition is satisfied only when , but the condition is not satisfied if .
Method 2 : the 2nd exactness condition
Following is another condition to prove the second exactness.

(2.15) 
Using the definition of , the followings are computed.

(2.16) 
After substituting the terms in Eq. (2.16), Eq. (2.15) becomes as following.

(2.17) 
 The second exactness condition is satisfied only when .
Part 2[edit]
As shown in the Eq. (2.10), the Hermite equation does not satisfy the second exactness condition if n is a positive value. To see if the equation can be made exact using IFM with , multiply the Eq. (2.2) by . Also, the term n in the Eq. (2.2) is replaced with to avoid confusion.

(2.18) 

(2.19) 
It is assumed that the Eq. (2.13) is exact. Then, the integrating factor can be obtained by utilizing the second exactness condition as following.

(2.20) 
With the terms computed above, the two equations below which represent the second exactness condition are exploited.

(2.21) 

(2.22) 
First, substitute the terms in the Eq. (2.14) into the Eq. (2.16).

(2.23) 
The only way the Eq. (2.17) is satisfied is to set n equal to zero.

(2.24) 
With knowing n=0, substitute the terms in the Eq. (2.14) into the Eq. (2.15).

(2.25) 
The both sides of the Eq. (2.19) are equal only when the coefficients of are zero. Then the following is concluded.

(2.26) 
The first equation in the Eq. (2.20) implies that or . Then is determined using the second one in the Eq. (2.20).

(2.27) 
Since n is determined to be zero already, m can not be zero. Hence, m is chosen to be one and is 2.

(2.28) 

(2.29) 
Here, the exactness of the Hermite equation with is justified.

(2.30) 
In order to prove the second exactness of the Hermite equation, the Method 2 is used.

(2.31) 
Using the definition of , the followings are computed.

(2.32) 
After substituting the terms in Eq. (2.32), Eq. (2.31) becomes as following.

(2.33) 
 Hermite equation with IFM is exact.
Part 3[edit]
Hermite differential equation is known to have a polynomial solution, which has following form.

(2.34) 
Compute the first and the second derivative of Eq. (2.34) with respect to x,

(2.35) 
Substitute Eq. (2.34), (2.35) into Eq. (2.2),

(2.36) 
Shift the first summation up by two units,

(2.37) 
The second summation starts at n=1, while the other summations start at n=0. Following justifies the second summation starting at n=0.

(2.38) 
Combine all summations in Eq. (2.37),

(2.39) 
Hence, the following recurrence relation is obtained from Eq. (2. 39).

(2.40) 
Simplify Eq. (2.40),

(2.41) 
Using the recurrence relation in Eq. (2.41), the Hermite polynomials given in the problem are obtained.
Let the initial conditions and be as follows.

(2.42) 
Then,

(2.43) 
 Hence, Hermite polynomial is in this case.
Let the initial conditions and be as follows.

(2.44) 
Then,

(2.45) 
 Hence, Hermite polynomial is in this case.
Let the initial conditions and be as follows.

(2.46) 
Then,

(2.47) 
 Hence, Hermite polynomial is in this case.
References [edit]
 Hermite Polynomial, WolframMathworld [1]
Author[edit]
Problem R*5.3 Method of undetermined coefficients[edit]
From Mtg 303
Given[edit]
Given L4ODECC,
(3.1)
Assuming,
(3.2)
and substituting this in Eq. (3.1), we get 4 solutions for r,
(3.3)
Find[edit]
Expressions for X(x) in terms of
Solution[edit]
 Solved on my own
We can write final solution X(x) as,
(3.4)
Substituting all the r_{i} values from Eq. (3.3),
(3.5)
Expanding the exponential terms into sin, cos, sinh and cosh terms, we get,
(3.6)
for X(x) to be real, imaginary part in above Eq. (3.6) must be real, i.e. (C_{3}  C_{4}) must be imaginary
and, (C_{3} + C_{4}) must be real,
Therefore, C3 and C4 will be given by,
This means C3 and C4 given above are complex conjugates.
Therefore, we get final solution to be,
(3.7)
Author[edit]
Problem R*5.4  Find in terms of the derivatives of y with respect to t[edit]
From Mtg 311
Given[edit]
In Euler L2ODEVC, suppose: .
Find[edit]
Solution[edit]
 Solved on my own
Since:
(4.1) 
So:
(4.2) 
(4.3) 
(4.4) 
(4.5) 
(4.6) 
Author[edit]
Problem R*5.5  Solve y and plot the solution[edit]
From Mtg 313
Given[edit]
Euler L2ODEVC:

(5.1) 
and trial solution:

(5.2) 
with boundary conditions:

(5.3) 
Find[edit]
The solution and plot it.
Solution[edit]
 Solved on my own
Substitute for in equation (5.1):

(5.4) 
So the characteristic equation:

(5.5) 
So the solution is:

(5.6) 
Use the boundary conditions:

(5.7) 
So the solution is:

(5.8) 
Plot .
Matlab Code:

clear all close all clc %homogeneous solutions y=@(x) (15/2)*(x^2)(23/2)*x; %plotting hold on fplot(y,[7 7],'r'); title('Plotting'); xlabel('x'); ylabel('y(x)'); legend('y=(15/2)*(x^2)(23/2)*x')
Author[edit]
Problem R5.6: Equivalence of Two methods to solve Euler LnODEVC[edit]
From Mtg 315
Given[edit]
Given Euler LnODEVC,
(6.1)
where, is i^{th} derivative of y w.r.t. x,
Two methods of solving above equation,
Method 1:
Stage 1: Transformation of variables
(6.2)
Stage 2: Trial Solution
(6.3)
Method 2: Trial Solution
(6.4)
Find[edit]
Show equivalence of methods 1 and 2 shown above in Eq. (6.2), (6.3) and Eq. (6.4) respectively
Solution[edit]
 Solved on our own
Let's take method  2 shown in Eq. (6.4), first. Differentiating Eq. (6.4) w.r.t. x, we get
(6.5)
For, i^{th} derivative of y w.r.t. x, i.e. y^{(i)}, we have,
(6.6)
Substituting Eq. (6.6), i.e. definition of y^{(i)} in Eq. (6.1), we get,
(6.7a)
(6.7b)
Eliminating x^{r}, which is a common factor in the all the terms of above summation, we get,
(6.7c)
So, we solve the above equation for r, and that provides us solution for Eq. (6.1).
Now, for method  1, we will try to get the same form as in Eq. (6.7c)
Stage 1: Substitute x = e^{t} in Eq. (6.1), we write y_{x} as,
(6.8)
Stage 2: Substitute y = e^{rt},
Using above we can write y_{t} as, r*e^{rt}. Substitute this into Eq. (6.8), and we get,
(6.9)
Performing another differentiation of above equation to get y_{xx}, we get,
(6.10)
We can see that as done above for y^{(2)}, we can write in general for y^{(i)}, we can write,
(6.10)
Substituting definitions of both x and y as seen in Eq. (6.2) and Eq. (6.10) respectively, into Eq. (6.1) we get,
(6.11a)
Above can be simplified to following form very easily (combining both the exponential terms in above expression),
(6.11b)
Since, e^{(rt)} is not dependent on i, we can take this term out of the summation as a common factor, and write rest of the terms as,
(6.11c)
Above Eq. (6.11c) is exactly same as Eq. (6.7c), and therefore it proves that by both methods 1 & 2 we will get same values for r, and hence both methods are equivalent.
Author[edit]
Problem R*5.7Euler L2ODEs [edit]
From Mtg 321
Given[edit]
Characteristic equation

(7.1) 

(7.2) 

(7.3) 
Find[edit]
1.1) Find such that (7.1) is characteristic equation of (7.2)
1.2) 1st homogeneous solution :
1.3) Complete solution : Find c(x) such that
1.4) Find the 2nd homogeneous solution
2.1) Find such that (7.1) is characteristic equation of (7.3)
2.2) 1st homogeneous solution
2.3) Complete solution : Find c(x) such that
2.4) Find the 2nd homogeneous solution
Solution[edit]
 Solved on our own
Part 1.1[edit]
Assume that the trial solution is following.

(7.4) 
Then, compute the first and the second derivatives of the trial solution,

(7.5) 

(7.6) 
Substitute the equation (7.4), (7.5), (7.6) into (7.2) and simplify the equation,

(7.7) 
Then, we get

(7.8) 
Compare (7.8) to (7.1),

(7.9) 
Comparing the coefficients of LHS and RHS,

(7.10) 
Part 1.2[edit]
From the characteristic equation in Eq. (7.1), the solution of the characteristic equation is determined as follows.

(7.11) 
Hence, the 1st homogeneous solution can be

(7.12) 
Part 1.3[edit]

(7.13) 
Compute the first and the second derivatives of Eq. (7.13),

(7.14) 
Substitute (7.13) and (7.14) into (7.2) and simplify the equation,

(7.15) 
The first term in Eq. (7.15) cancels out using Eq. (7.2). Then,

(7.16) 
Substitute coefficients and homogeneous solution,



(7.17) 
Let ,

(7.18) 

(7.19) 

(7.20) 
Then, compute the function ,

(7.21) 
As a result,

(7.22) 
Part 1.4[edit]
From 1.3, the general homogeneous solution is

From (7.12), is .
So, is

Part 2.1[edit]
The trial solution for the Euler equation with constant coefficient:

(7.23) 
Compute the first and the second derivatives

(7.24) 

(7.25) 
Substitute (7.23),(7.24),(7.25) into (7.3) and simplify it,

Then,


Compare LHS and RHS,

(7.26) 

(7.27) 

(7.28) 
Part 2.2[edit]
From (7.1), , As a result,

(7.29) 
Part 2.3[edit]

(7.30) 
Compute the first and the second derivatives

(7.31) 

(7.32) 
Substituting (7.25),(7.26) and (7.27) into (7.3)

From (7.3),

Then,

Substituting coefficients and homogeneous solution,

Then,

(7.33) 
Solve (7.33),

(7.34) 
As a result,

(7.35) 
Part 2.4[edit]
From 2.3, the general homogeneous solution is

From (7.29), is .
So, is

Author[edit]
Problem R*5.8  Variation of parameters[edit]
From Mtg 322
Given[edit]
A General nonhomogenous L1ODEVC (Linear, 1^{st} order, Ordinary Differential Equation with Variable Coefficient),

(8.1) 
which can be written in the form given below,

(8.2) 
The form can be written as,

(8.3) 
Find[edit]
Find the particular solution, by the use of the method of variation of parameters after knowing the homogeneous solution , i.e., let , with being the unknown to be found.
Solution[edit]
 Solved on our own
For particular solution, , the use variation of parameters requires to be following,

(8.4) 
for order nonlinear homogenous ODE. For our case, n = 1, therefore we get,

(8.5) 
It is important to note for method of variation of parameters that above, is a function of , and is determined by substituting the particular solution. Thus,

(8.6) 
Substitute Eq. 8.5 into Eq. 8.6 to get,

(8.7) 
and, rearrange the terms,

(8.8) 
Therefore is given by,

(8.9) 
Integrate both sides w.r.t. x,

(8.10) 
Substituting for the expression of from the homework problem 2.17 into above, we will get,

(8.11) 
Therefore particular solution can finally be written as,

(8.12) 
Author[edit]
Problem R*5.9 Special IFM Solution[edit]
From Mtg 327
Given[edit]

(9.1) 
Find[edit]
Special IFM to solve for general
Solution[edit]
 Solved on our own
First, find integrating factor
(Eq 9.2) 
Suppose is a function of , thus
Rearrange Eq 9.2, we have
(Eq 9.3) 
(Eq 9.4) 
Thus
(Eq 9.5) 
Where
According to the 2nd Exactness Condition
(Eq 9.6) 
(Eq 9.7) 
We have
(Eq 9.8) 
(Eq 9.9) 
(Eq 9.10) 
(Eq 9.11) 
(Eq 9.12) 
(Eq 9.13) 
Substituting Eq 9.8~9.13 in Eq 9.6, we get
(Eq 9.14) 
Now we assume , where is a constant
So Eq 9.14 can be expressed as
(Eq 9.15) 
Then we try to reduce order for Eq 9.2 and express Eq 9.2 as
(Eq 9.16) 
We can get
(Eq 9.17) 
Therefore, we have
(Eq 9.18) 
Rearrange Eq 9.16, we have
(Eq 9.19) 
Using , where
Thus, we can get
(Eq 9.20) 
Therefore, we have

(Eq 9.21)
Where and are constants.
Author[edit]
Contributing Members[edit]


Problem Number  Lecture(Mtg)  Assigned To  Solved By  Typed By  Signature(Author)  
R*5.1  Mtg 226  Chung  Chung  Chung  Chung 17 Oct. 2011 at 1:30 (UTC)  
R*5.2  Mtg 271  Shin  Shin  Shin  Shin 31 Oct. 2011 at 14:38 (UTC)  
R*5.3  Mtg 303  Ankush  Ankush  Ankush  Ankush 1 Nov. 2011 at 18:00 (UTC)  
R*5.4  Mtg 311  YuChen  YuChen  YuChen  YuChen 30 Oct. 2011 at 08:47 (UTC)  
R*5.5  Mtg 313  YuChen  YuChen  YuChen  YuChen 30 Oct. 2011 at 10:01 (UTC)  
R5.6  Mtg 315  Ankush  Ankush  Ankush  Ankush 1 Nov. 2011 at 20:00 (UTC)  
R*5.7  Mtg 321  Chung  Chung  Chung  Chung 29 Oct. 2011 at 0:19 (UTC)  
R*5.8  Mtg 322  Shin  Shin  Shin  Shin 2 Nov. 2011 at 15:43 (UTC)  
R*5.9  Mtg 327  Ren  Ren  Ren  Ren 1 Nov. 2011 at 23:50 (UTC) 