User:Egm6321.f11.team4/HW5

= Problem R*5.1 - Equivalence of twos forms of 2nd exactness condition = From [[media:Pea1.f11.mtg22.23.djvu|Mtg 22-6]]

Find
Equivalence of the form of 2nd exactness condition:

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

As a result,

Author
Contributed by Chung

= 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. = References =
 * Hermite Polynomial, WolframMathworld

Author
Contributed by Shin

= Problem R*5.3 Method of undetermined coefficients = From [[media:Pea1.f11.mtg30.djvu|Mtg 30-3]]

Given
Given L4-ODE-CC,

$$\displaystyle X^{(4)} - K^4 X = 0 $$  (3.1)

Assuming,

$$\displaystyle X(x) = e^{(rx)} $$  (3.2)

and substituting this in Eq. (3.1), we get 4 solutions for r,

$$\displaystyle r_{1,2} = \pm K $$

$$\displaystyle r_{3,4} = \pm i \, K $$   (3.3)

Find
Expressions for X(x) in terms of $$ \displaystyle \cos K x, \ \sin K x, \ \cosh K x, \ \sinh K x $$

Solution
- Solved on my own

We can write final solution X(x) as,

$$\displaystyle X(x) = \sum_{i=1}^4 C_i e^{(r_i x)} $$  (3.4)

Substituting all the ri values from Eq. (3.3),

$$\displaystyle X(x) = C_1 e^{K x} + C_2 e^{-K x} + C_3 e^{iK x} + C_4 e^{-iK x} $$  (3.5)

Expanding the exponential terms into sin, cos, sinh and cosh terms, we get,

$$\displaystyle X(x) = C_1 \cosh K x \ + C_1 \sinh K x \ + C_2 \cosh K x \ - C_2 \sinh K x \ + C_3 \cos K x \ + i C_3 \sin K x \ + C_4 \cos k x - \ i C_4 \sin k x $$  (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

$$\displaystyle C_3 - C_4 = ib_4 $$

and, (C3 + C4) must be real,

$$\displaystyle C_3 + C_4 = b_3 $$

Therefore, C3 and C4 will be given by,

$$\displaystyle C_3 = \frac{1}{2}(b_3 + ib_4) $$

$$\displaystyle C_4 = \frac{1}{2}(b_3 - ib_4) $$

This means C3 and C4 given above are complex conjugates.

Therefore, we get final solution to be,

$$\displaystyle X(x) = (C_1 + C_2) \cosh K x + (C_1 - C_2) \sinh K x + b_3 \cos K x - b_4 \sin K x $$  (3.7)

Author
Contributed by Ankush

= Problem R*5.4  - Find $$\displaystyle y_{xxxxx} $$ in terms of the derivatives of y with respect to t= From [[media:Pea1.f11.mtg31.djvu|Mtg 31-1]]

Given
In Euler L2-ODE-VC, suppose: $$\displaystyle x=e^t $$.

Find
$$\displaystyle y_{xxxxx} $$

Solution
- Solved on my own Since:

So:

Author
Contributed by YuChen

= Problem R*5.5  - Solve y and plot the solution= From [[media:Pea1.f11.mtg31.djvu|Mtg 31-3]]

Given
Euler L2-ODE-VC:

and trial solution:

with boundary conditions:

Find
The solution $$\displaystyle y(x) $$ and plot it.

Solution
- Solved on my own Substitute $$\displaystyle x^r $$ for $$\displaystyle y $$ in equation (5.1):

So the characteristic equation:

So the solution is:

Use the boundary conditions:

So the solution is:

Plot $$\displaystyle y(x) $$.

Matlab Code:


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Author
Contributed by YuChen

= Problem R5.6: Equivalence of Two methods to solve Euler Ln-ODE-VC = From [[media:Pea1.f11.mtg31.djvu|Mtg 31-5]]

Given
Given Euler Ln-ODE-VC,

$$ \displaystyle \sum_{i=0}^n a_i x^i y^{(i)} = 0 $$  (6.1)

where, $$ \displaystyle y^{(i)} $$ is ith derivative of y w.r.t. x,

Two methods of solving above equation,

Method 1:

Stage 1: Transformation of variables

$$ \displaystyle x = e^t $$  (6.2)

Stage 2: Trial Solution

$$ \displaystyle y = e^{rt} ,\ {\color{red} r = \text{constant}} $$  (6.3)

Method 2: Trial Solution

$$ \displaystyle y = x^{r} ,\ {\color{red} r = \text{constant}} $$  (6.4)

Find
Show equivalence of methods 1 and 2 shown above in Eq. (6.2), (6.3) and Eq. (6.4) respectively

Solution
- 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

$$ \displaystyle y_x = y' = r x^{r-1} $$  (6.5)

For, ith derivative of y w.r.t. x, i.e. y(i), we have,

$$ \displaystyle y^{(i)} = r(r-1)...(r-i+1)x^{(r-1)} $$  (6.6)

Substituting Eq. (6.6), i.e. definition of y(i) in Eq. (6.1), we get,

$$ \displaystyle \sum_{i=0}^n a_i x^i r(r-1)...(r-i+1) x^{(r-i)} = 0 $$  (6.7a)

$$ \displaystyle \sum_{i=0}^n a_i x^r r(r-1)...(r-i+1) = 0 $$  (6.7b)

Eliminating xr, which is a common factor in the all the terms of above summation, we get,

$$ \displaystyle \sum_{i=0}^n a_i r(r-1)...(r-i+1) = 0 $$  (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,

$$ \displaystyle y_x = y_t \frac {dt}{dx} = e^{-t} y_t $$ <p style="text-align:right"> (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,

$$ \displaystyle y^{(1)} = y_x = re^{rt}e^-t = r e^{(r-1)t} $$ <p style="text-align:right"> (6.9)

Performing another differentiation of above equation to get yxx, we get,

$$ \displaystyle y^{(2)} = y_{xx} = \frac {d}{dx} (y_x) = \frac {d}{dx} (r e^{(r-1)t}) = \frac {d}{dt} (r e^{(r-1)t}) \frac {dt}{dx} = r(r-1) e^{(r-1)t} e^{-t} = r(r-1)e^{(r-2)t} $$ <p style="text-align:right"> (6.10)

We can see that as done above for y(2), we can write in general for y(i), we can write,

$$ \displaystyle y^{(i)} = r(r-1)...(r-i+1) e^{(r-i)t} $$ <p style="text-align:right"> (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,

$$ \displaystyle \sum_{i=0}^n a_i e^{it} r(r-1)...(r-i+1) e^{(r-i)t} = 0 $$ <p style="text-align:right"> (6.11a)

Above can be simplified to following form very easily (combining both the exponential terms in above expression),

$$ \displaystyle \sum_{i=0}^n a_i e^{rt} r(r-1)...(r-i+1) = 0 $$ <p style="text-align:right"> (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,

$$ \displaystyle \sum_{i=0}^n a_i r(r-1)...(r-i+1) = 0 $$ <p style="text-align:right"> (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
Contributed by Ankush

= Problem R*5.7-Euler L2-ODEs = From [[media:Pea1.f11.mtg32.djvu|Mtg 32-1]]

Given
Characteristic equation

Find
1.1) Find $$a_2, a_1, a_0$$ such that (7.1) is characteristic equation of (7.2)

1.2) 1st homogeneous solution : $$y_1(x)=x^\lambda$$

1.3) Complete solution : Find c(x) such that $$y(x)=c(x)y_1(x)$$

1.4) Find the 2nd homogeneous solution $$y_2(x)$$

2.1) Find $$b_2, b_1, b_0$$ such that (7.1) is characteristic equation of (7.3)

2.2) 1st homogeneous solution

2.3) Complete solution : Find c(x) such that $$y(x)=c(x)y_1(x)$$

2.4) Find the 2nd homogeneous solution $$y_2(x)$$

Solution
- Solved on our own

Part 1.1
Assume that the trial solution is following. Then, compute the first and the second derivatives of the trial solution,

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

Then, we get

Compare (7.8) to (7.1),

Comparing the coefficients of LHS and RHS,

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

Hence, the 1st homogeneous solution can be

Part 1.3
Compute the first and the second derivatives of Eq. (7.13),

Substitute (7.13) and (7.14) into (7.2) and simplify the equation, The first term in Eq. (7.15) cancels out using Eq. (7.2). Then, Substitute coefficients and homogeneous solution,

Let $$\displaystyle c^{\prime}(x)=z(x) $$,

Then, compute the function $$\displaystyle c(x) $$,

As a result,

Part 1.4
From 1.3, the general homogeneous solution is From (7.12), $$\displaystyle y_1(x)$$ is $$\displaystyle x^\lambda$$.

So, $$\displaystyle y_2(x)$$ is

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

Compute the first and the second derivatives

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

Then,

Compare LHS and RHS,

Part 2.2
From (7.1), $$\displaystyle r=\lambda$$, As a result,

Part 2.3
Compute the first and the second derivatives

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

From (7.3),

Then, Substituting coefficients and homogeneous solution, Then, Solve (7.33),

As a result,

Part 2.4
From 2.3, the general homogeneous solution is From (7.29), $$\displaystyle y_1(x)$$ is $$\displaystyle e^{\lambda x}$$.

So, $$\displaystyle y_2(x)$$ is

Author
Contributed by chung

= 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

= Problem R*5.9 Special IFM Solution= From [[media:Pea1.f11.mtg32.djvu|Mtg 32-7]]

Find
Special IFM to solve for general $$\displaystyle f(t)$$

Solution
- Solved on our own First, find integrating factor $$\displaystyle h(t,y)$$ Suppose $$\displaystyle h(t,y)$$ is a function of $$\displaystyle t$$, thus $$\displaystyle h(t,y)=h(t)$$

Rearrange Eq 9.2, we have Thus Where $$\displaystyle p:=y'$$

According to the 2nd Exactness Condition We have Substituting Eq 9.8~9.13 in Eq 9.6, we get Now we assume $$\displaystyle h(t)=e^{\alpha t}$$, where $$\displaystyle {\alpha}$$ is a constant

So Eq 9.14 can be expressed as Then we try to reduce order for Eq 9.2 and express Eq 9.2 as We can get

Therefore, we have Rearrange Eq 9.16, we have Using $$\displaystyle h(t)=e^{{\beta}t}$$, where $$\displaystyle {\beta}=\frac{\bar{a_0}}{\bar{a_1}}$$

Thus, we can get

Therefore, we have


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$$ \displaystyle y(t)={{C}_{1}}{{e}^{-\alpha t}}+{{C}_{2}}{{e}^{-\beta t}}+\frac\int{{{e}^{(\beta -\alpha )t}}\left( \int{{{e}^{\alpha t}}f(t)dt} \right)dt}$$ (Eq 9.21) Where $$\displaystyle {C_1}$$ and $$\displaystyle {C_2}$$ are constants.
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Author
Contributed by Kexin Ren

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