User:Egm6321.f11.team4.shin.js/HW5

= Problem R*5.2 - Verification of the exactness of Legendre and Hermite equations = From [[media:Pea1.f11.mtg27.28.29.djvu|Mtg 27-1]]

Given
Legendre equation: Hermite equation:

Find
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 $$\displaystyle h(x,y)=x^m y^n $$ 3. The first few Hermite polynomials are given as below. Verify that the equations in (2.3) are homogeneous solutions of the Hermite differential equation (Eq. 2.2).

Solution
- Solved on our own

Part 1
Legendre equation: The first exactness condition is satisfied since the equation has a form of Eq. (2.4). Method 1 : the 2nd exactness condition In order to verify the second exactness condition, the following terms are computed.

- The second exactness condition is satisfied when $$\displaystyle n=0 $$ or $$\displaystyle n=-1 $$. Method 2 : the 2nd exactness condition Following is another condition to prove the second exactness. Using the definition of $$\displaystyle g_i := \frac{\partial G}{\partial y ^{(i)}}$$, the followings are computed.

After substituting the terms in Eq. (2.9), Eq. (2.8) becomes as following. - The second exactness condition is satisfied when $$\displaystyle n=0 $$ or $$\displaystyle n=-1 $$. Hermite equation: The first exactness condition is satisfied since the equation has a form of Eq. (2.8). Method 1 : the 2nd exactness condition In order to verify the second exactness condition, the following terms are computed.

- The second exactness condition is satisfied only when $$\displaystyle n=-1 $$, but the condition is not satisfied if $$\displaystyle n \neq -1 $$. Method 2 : the 2nd exactness condition Following is another condition to prove the second exactness. Using the definition of $$\displaystyle g_i := \frac{\partial G}{\partial y ^{(i)}}$$, the followings are computed.

After substituting the terms in Eq. (2.16), Eq. (2.15) becomes as following. - The second exactness condition is satisfied only when $$\displaystyle n=-1 $$.

Part 2
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 $$\displaystyle h(x,y) = x^m y^n $$, multiply the Eq. (2.2) by $$\displaystyle h(x,y) $$. Also, the term n in the Eq. (2.2) is replaced with $$\displaystyle \lambda $$ to avoid confusion.

It is assumed that the Eq. (2.13) is exact. Then, the integrating factor $$\displaystyle h(x,y) $$ can be obtained by utilizing the second exactness condition as following. With the terms computed above, the two equations below which represent the second exactness condition are exploited. First, substitute the terms in the Eq. (2.14) into the Eq. (2.16). The only way the Eq. (2.17) is satisfied is to set n equal to zero. With knowing n=0, substitute the terms in the Eq. (2.14) into the Eq. (2.15). The both sides of the Eq. (2.19) are equal only when the coefficients of $$\displaystyle x^{m-1}, x^m $$ are zero. Then the following is concluded. The first equation in the Eq. (2.20) implies that $$\displaystyle m=0 $$ or $$\displaystyle m=1 $$. Then $$\displaystyle \lambda $$ is determined using the second one in the Eq. (2.20). Since n is determined to be zero already, m can not be zero. Hence, m is chosen to be one and $$\displaystyle \lambda $$ is 2. Here, the exactness of the Hermite equation with $$\displaystyle h(x,y) $$ is justified. In order to prove the second exactness of the Hermite equation, the Method 2 is used. Using the definition of $$\displaystyle g_i := \frac{\partial G}{\partial y ^{(i)}}$$, the followings are computed. After substituting the terms in Eq. (2.32), Eq. (2.31) becomes as following. - Hermite equation with IFM $$\displaystyle h(x,y) = x $$ is exact.

Part 3
Hermite differential equation is known to have a polynomial solution, which has following form. Compute the first and the second derivative of Eq. (2.34) with respect to x, Substitute Eq. (2.34), (2.35) into Eq. (2.2), Shift the first summation up by two units, The second summation starts at n=1, while the other summations start at n=0. Following justifies the second summation starting at n=0. Combine all summations in Eq. (2.37), Hence, the following recurrence relation is obtained from Eq. (2. 39). Simplify Eq. (2.40), Using the recurrence relation in Eq. (2.41), the Hermite polynomials given in the problem are obtained. Let the initial conditions and $$\displaystyle \lambda $$ be as follows. Then, - Hence, Hermite polynomial is $$\displaystyle H_0(x) = 1 $$ in this case.

Let the initial conditions and $$\displaystyle \lambda $$ be as follows. Then, - Hence, Hermite polynomial is $$\displaystyle H_1(x) = 2x $$ in this case.

Let the initial conditions and $$\displaystyle \lambda $$ be as follows. Then, - Hence, Hermite polynomial is $$\displaystyle H_2(x) = 4x^2-2 $$ in this case.

Author
Contributed by Shin

= Problem R*5.8 - Variation of parameters = From [[media:Pea1.f11.mtg32.djvu|Mtg 32-2]]

Given
A General non-homogenous L1-ODE-VC (Linear, 1st order, Ordinary Differential Equation with Variable Coefficient), which can be written in the form given below, The form can be written as,

Find
Find the particular solution, $$\displaystyle y_P(x) $$ by the use of the method of variation of parameters after knowing the homogeneous solution $$\displaystyle y_H(x)$$, i.e., let $$\displaystyle y(x) = A(x)y_H(x)$$, with $$\displaystyle A(x)$$ being the unknown to be found.

Solution
- Solved on our own For particular solution, $$\displaystyle y_P (x) $$, the use variation of parameters requires $$\displaystyle y_P(x)$$ to be following, for $$ n^{th} $$ order non-linear homogenous ODE. For our case, n = 1, therefore we get, It is important to note for method of variation of parameters that above, $$\displaystyle c_1 $$ is a function of $$\displaystyle x $$, and is determined by substituting the particular solution. Thus, Substitute Eq. 8.5 into Eq. 8.6 to get, and, rearrange the terms, Therefore $$\displaystyle A(x) $$ is given by, Integrate both sides w.r.t. x, Substituting for the expression of $$\displaystyle y_H(x) $$ from the homework problem 2.17 into above, we will get,

Therefore particular solution can finally be written as,

Author
Contributed by Shin