User:Egm6321.f10.team4.Yoon/HW6

= Relationship Between Artanh(x) function And Legendre Functions $$\displaystyle {{Q}_{0}}(x),{{Q}_{1}}(x)$$ =

Solution for part.a
To find out the relationship between Legendre equation and artanh function, let's derivate artanh function from tanh fuction first.

let

Then, moving the denominator of the right hand side to the left hand side.

Multiplying e^y on the both side of the equation

Collecting the term of x yields

Deviding by (x-1)

Taking log on the both side of the equation,

Deviding by 2

In conclusion, artanh fuction is identical with Legendre series(Q0).

Solution for part.b
Multiplying x and substract 1 on both side of equation yield,

= Problem 2. Non-homog. L2_ODE_VC using Variable of Paramerters ==

From Meeting 33, p 33-1.

Given
Solve the Non-homog L2_ODE_VC with the condition $$\displaystyle f(x)=0 $$. K.p.28, Pb 1.1 ab '''Part. a'''
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$$\displaystyle (x-1)y''-xy'+y=f(x) $$ $$
 * $$\displaystyle (Eq. 2-1)
 * }
 * }

'''Part. b'''
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$$\displaystyle xy''+2y'+xy=f(x) $$ $$
 * $$\displaystyle (Eq. 2-2)
 * }
 * }

General trial solution is given on Lecture note p.32-1 as,
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$$ \displaystyle y={x}^{c}\cdot {e}^{rx} $$ $$
 * $$\displaystyle (Eq. 2-3)
 * }
 * }

Find
Find solutions of the equations.

'''Part. a'''
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$$\displaystyle (x-1)y''-xy'+y=f(x) $$ $$
 * $$\displaystyle (Eq. 2-1)
 * }
 * }

'''Part. b'''
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$$\displaystyle xy''+2y'+xy=f(x) $$ $$
 * $$\displaystyle (Eq. 2-2)
 * }
 * }

Solution
Taking 1st and 2nd derivatives of the trial solution is

Solution for Part. a
Plugging the derivatives of trial solutions into the equation given (eq 2.1) yields,

Collecting x terms by reduction order,

To make above equation into the format of trial solution, only one of the x term should be survived. Thus, the values for c and r by (Eq 2-7) are,

Substituting the values (Eq.2-1) into the trial solution, the homogeneous solutions are determined as

Solution for Part. b
Plugging the derivatives(from (eq.2-4) to (eq.2-6)) of trial solutions into the equation given (eq 2.2) yields,

Collecting x terms by reduction order,

To make above equation into the format of trial solution, only one of the x term should be survived. Thus, the values for c and r by (Eq.2-11) are,

Substituting the values (Eq.2-2) into the trial solution, the homogeneous solutions are determined as

Egm6321.f10.team4.Yoon 00:00, 14 Nov 2010 (UTC) - Primary Author

= Problem 1. L2_ODE_VC using Var. of Paramerters = From Meeting 33, p 33-1.

Given
Legendre Equation

Legendre Polynomials

where, $$\displaystyle m = \frac{n}{2}$$

When a particular solution P2(n=2,r=1) is given as below,

Find
Legendre Equation will be given

And then show that the second solution is as below.

Solution
When n=2, the given Legendre Equation is as

Firstly, writing the equation in standard format as

On the assumption that we have one solution, say $$\displaystyle y(x) = u_1(x)$$. We will find the other solution by differentiating $$\displaystyle y(x) = U(x)\cdot {u}_{1}(x)$$ gives

Plugging the above deriviates into the (Eq. 1-7) gives

Since u1(x) is a solution of (Eq. 1-7), the term is zero. We will use $$\displaystyle Z:=U'$$ to apply missing dependent variable method,

After deviding the equation in terms of Z by the highest coefficient of Z, now we can find intergrating factor h(x),

By the definition we set in advance, $$\displaystyle Z=U'=\frac{k2}{h(x)}$$, we can go back to U(x) by integrating Z

Since we have U(x) now, we can solve y(x)

By plugging what we have, h(x), u1(x), into above equation, we can get the desired solution Q2(x)

=hw7 p1=

Given
The general form of the Legendre Equation:

And the solutions have the following format.

Definitions of the Legendre Polynomials:


 * where $$\displaystyle m = \frac{n}{2}$$.