User:Egm6321.f11.team4/HW5

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Problem R*5.1 - Equivalence of twos forms of 2nd exactness condition[edit]

From Mtg 22-6

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]

Contributed by Chung

Problem R*5.2 - Verification of the exactness of Legendre and Hermite equations[edit]

From Mtg 27-1

Given[edit]

Legendre equation:

(2.1)

Hermite equation:

(2.2)

Find[edit]

1. Verify the exactness of the designated L2-ODE-VC (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]

Contributed by Shin

Problem R*5.3 Method of undetermined coefficients[edit]

From Mtg 30-3

Given[edit]

Given L4-ODE-CC,

(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 ri 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. (C3 - C4) must be imaginary

and, (C3 + C4) 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]

Contributed by Ankush

Problem R*5.4 - Find in terms of the derivatives of y with respect to t[edit]

From Mtg 31-1

Given[edit]

In Euler L2-ODE-VC, 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]

Contributed by YuChen

Problem R*5.5 - Solve y and plot the solution[edit]

From Mtg 31-3

Given[edit]

Euler L2-ODE-VC:

(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')

Plot of y(x).jpg

Author[edit]

Contributed by YuChen

Problem R5.6: Equivalence of Two methods to solve Euler Ln-ODE-VC[edit]

From Mtg 31-5

Given[edit]

Given Euler Ln-ODE-VC,

(6.1)

where, is ith 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, ith 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 xr, 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 = et in Eq. (6.1), we write yx as,

(6.8)

Stage 2: Substitute y = ert,

Using above we can write yt as, r*ert. Substitute this into Eq. (6.8), and we get,

(6.9)

Performing another differentiation of above equation to get yxx, 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]

Contributed by Ankush

Problem R*5.7-Euler L2-ODEs [edit]

From Mtg 32-1

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]

Contributed by chung

Problem R*5.8 - Variation of parameters[edit]

From Mtg 32-2

Given[edit]

A General non-homogenous L1-ODE-VC (Linear, 1st 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 non-linear 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]

Contributed by Shin

Problem R*5.9 Special IFM Solution[edit]

From Mtg 32-7

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]

Contributed by Kexin Ren

Contributing Members[edit]

Team Contribution Table
Problem Number Lecture(Mtg) Assigned To Solved By Typed By Signature(Author)
R*5.1 Mtg 22-6 Chung Chung Chung Chung 17 Oct. 2011 at 1:30 (UTC)
R*5.2 Mtg 27-1 Shin Shin Shin Shin 31 Oct. 2011 at 14:38 (UTC)
R*5.3 Mtg 30-3 Ankush Ankush Ankush Ankush 1 Nov. 2011 at 18:00 (UTC)
R*5.4 Mtg 31-1 YuChen YuChen YuChen YuChen 30 Oct. 2011 at 08:47 (UTC)
R*5.5 Mtg 31-3 YuChen YuChen YuChen YuChen 30 Oct. 2011 at 10:01 (UTC)
R5.6 Mtg 31-5 Ankush Ankush Ankush Ankush 1 Nov. 2011 at 20:00 (UTC)
R*5.7 Mtg 32-1 Chung Chung Chung Chung 29 Oct. 2011 at 0:19 (UTC)
R*5.8 Mtg 32-2 Shin Shin Shin Shin 2 Nov. 2011 at 15:43 (UTC)
R*5.9 Mtg 32-7 Ren Ren Ren Ren 1 Nov. 2011 at 23:50 (UTC)