User:Egm6321.f12.team7/report2

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Problem 1: Verify the Homogeous Solutions
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Given: Two homogeous solutions
The Legendre differential equation is given by,

When the $$n=1$$, we have,

We also have two linearly-independent solutions, which are given by,

Find
Verify that,$$ L_0(Y^2_H(x))= L_0(Y^1_H(x))=0$$.

Solution

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For $$ L_0(Y^1_H(x))$$, we have,

Then the $$\displaystyle L_0(Y^1_H(x))$$ comes to,

For $$\displaystyle L_0(Y^2_H(x))$$, we have,

So the $$\displaystyle L_0(Y^2_H(x))$$ changes into,

So we have,

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Problem 2: Verify solution
Report problem 2.2 from.

Given: The solution of a particular L1-ODE-CC (linear first order ordinary differential equation with constant coefficients)
A linear first order ordinary differential equation

has the solution of

Find: Verify the solution
Verify the ($$) is indeed the solution of ($$).

Solution: It is indeed the solution

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Take first order derivative of ($$),

Substitute ($$) and ($$) into left hand side of ($$),

Thus,

This proved the ($$) is the solution of ($$)

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Problem 3: Demonstrate affine relationship
Report problem 2.3 from.

Given: A class of N1-ODE
A particular class of nonlinear, 1st order ordinary differential equations can be expressed as

Find: Show affineness
Show that ($$) is affine with respect to $$y'$$. Also explain why ($$) is a non-linear, 1st order, ordinary differential equation. Finally, give an example of a more general nonlinear, 1st order, ordinary differential equation.

Solution: It is affine with respect to y'

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An affine function or operator is similar to a linear operator, except it is shifted by a constant. Therefore, with respect to $$y'$$, $$M(x,y)$$ can be considered a constant. For a function to be linear, the following must be true.

The linear part of ($$) is given by

This implies that

Therefore, if we substitute $$y' = \alpha u + \beta v$$

Since $$f(\alpha u + \beta v) = \alpha f(u) + \beta f(v)$$, ($$) is linear with respect to $$y'$$ and adding in the constant (with respect to $$y'$$) $$M(x,y)$$ makes ($$) affine with respect to $$y'$$.

However, ($$) is not linear with respect to $$y$$ because the coefficient of $$y'$$ is dependent on $$y$$. Powers greater than one of the dependent variable, or dependent variables multiplied by their derivatives implies nonlinearities. The equation is an ordinary differential equation, as apposed to a partial differential equation, since the dependent variable, $$y$$, is only dependent on one independent variable, $$x$$, and all the derivatives are total derivatives as apposed to partial derivatives. Finally, the equation is 1st order because the highest derivative is the first derivative of the dependent variable.

A more general form of a nonlinear, 1st order, differential equation is given by

where $$y', y$$ and $$x$$ are related nonlinearly. An example is

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Problem 4: Show linear independence
Based on lecture notes section 9

Given: Homogeneous solutions to a Legendre differential operator
The homogeneous solutions $$\displaystyle y^1_H(x) $$ in (3) p.7-1 and $$\displaystyle y^2_H(x) $$ in (4) p.7-1 are linearly independent.

Find: Linear independence
Show that $$\displaystyle \forall \alpha \in \mathbb{R}, y^1_H(\cdot) \ne \alpha y^2_H(\cdot) $$

i.e., for any given $$\displaystyle \alpha, $$ show that $$\displaystyle \exists \hat x $$ such that $$\displaystyle y^1_H(\hat x) \ne \alpha y^2_H(\hat x) $$

Plot $$\displaystyle y^1_H(x) $$ and $$\displaystyle y^2_H(x) $$

Solution: The solutions are linearly indepenent

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This solution was prepared without referring to previous solutions. The Legendre Differential Operator is:
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When n=1, the Legendre Differential Operator becomes:

The two linearly-independent homogeneous solutions are:

Suppose that there exists an $$\displaystyle \alpha_1 $$ such that $$\displaystyle y^1_H(\hat x_1)=\alpha_1 y^2_H(\hat x_1) $$ and that there exists an $$\displaystyle \alpha_2 $$ such that $$\displaystyle y^1_H(\hat x_2)=\alpha_2 y^2_H(\hat x_2) $$ The two homogeneous solutions are linearly independent if $$\displaystyle \alpha_1 \ne \alpha_2 $$ Let $$\displaystyle \hat x_1 = 0.1 $$ Then we have: $$\displaystyle 0.1 = \alpha_1 [\frac{0.1}{2} \log \left( \frac{1+0.1}{1-0.1} \right) - 1] \rightarrow \alpha_1 = -0.1004 $$ And let $$\displaystyle \hat x_2 = 0.2 $$ Then we have: $$\displaystyle 0.2 = \alpha_2 [\frac{0.2}{2} \log \left( \frac{1+0.2}{1-0.2} \right) - 1] \rightarrow \alpha_2 = -0.2036 $$ Thus $$\displaystyle \alpha_1 \ne \alpha_2 $$ and therefore the 2 homogeneous solutions are linearly independent.



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Problem 5: Find the 1st derivative of a function and show that this function is N1-ODE
Report problem 2.5 from lecture notes section 9-2.

Given: A specified function
A specified function is shown as below:

Find: The 1st derivative of the function and show that it is a N1-ODE
Find the 1st derivative of ($$):

and show it is a N1-ODE.

Solution: the 1st derivative of this function has the same form with the N1-ODE

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Find the first derivative by applying chain-rule to $$\displaystyle\phi(x,y)$$ ($$):

on the left side, the derivative can be written as

on the right side, the derivative can be written as

In this specified problem, the equation can be calculate as below

after observation, it can be found that ($$) has the same form as the particular class of N1-ODE which is showed as below

Thus, ($$) can be rewritten as below

Then build the relationship between ($$) and (5.7)($$):

apparently, the equation

has the same form with the a N1-ODE:

Thus, it has fulfilled the requirement to be a N1-ODE

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Problem 6: Explain the Condition of Mixed Partial Derivative Equality
Report problem 2.6 from lecture notes

Given:The Second Exactness Condition
THe second exactness condition of $$N2-ODE$$ is given by,

Where $$M(x,y), N(x,y)$$ is nonlinear founctions.

Find: Find the Condition
Find the minimun degree of differentiability and proof the theorem.

Solution

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This solution was prepared without referring to previous solutions. The theorem is: Suppose there is a founction(x_1, y_1). If the partial derivatives$$\frac{\partial^2 f}{\partial x_1 \partial y_1}$$ and $$\frac{\partial^2 f}{\partial y_1 \partial x_1}$$ exist and are continuous at (x_1, y_1), then we have,
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*The Minimum Degree of Differentiability
The mixed partial derivitives$$\frac{\partial^2 f}{\partial x_1 \partial y_1}$$ and $$\frac{\partial^2 f}{\partial y_1 \partial x_1}$$ exists and is continuous, which implies that the minimum degree of differentiability is two, since from lecture note we know that a function is continuous at a point doesn't means that it is differentiable at the same point.

*Proof of the Thoerem
In general, the partial derivative of a function $$f(x_1,...,x_n)$$ in direction$$x_i$$ at a point $$f(a_1,...,a_n)$$ is defined to be ,

Suppose we have two real numbers $$a$$, $$b$$, then with $$we have,

Change the order of derivatives, we have,

Obviously we have,

The theorem has been proved.

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Problem 7: Verify the solution to a N1-ODE
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Given: A N1-ODE and its solution

 * A nonlinear, 1st order ordinary differential equation(N1-ODE):


 * The solution to the N1-ODE:

Find: Verify the solution to the equation
Verify that ($$) is indeed the solution for the N1-ODE ($$).

Solution

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Basic train of thought is to produce ($$) from ($$).

First we obtain ($$) from ($$):

Then make differential of ($$) with respect of $$x$$, we obtain:

Obviously, ($$) is indeed ($$). Therefore the solution is verified.

Alternate Solution
According to the elements in ($$), the following calculation is needed.

Substitute the counterparts in the left part of ($$) with ($$) and ($$), we get:

The left part of ($$) equals to its right part, so ($$) is indeed the solution of ($$). (Prove end)

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Problem 8: Solving equations for integrating factor
Report problem 2.8 from lecture notes

Given:Euler Integrating Factor Method
When we find a N1-ODE can not fit the first exactness condition but fail on the second one, we can use Euler Integrating Factor to make the equation exact, which is given by,

Apply the second exactness condition, we have,

Where,

$$h_{x}:=\frac{\partial h}{\partial x},$$ $$h_{y}:=\frac{\partial h}{\partial y},$$ $$N_{x}:=\frac{\partial N}{\partial x},$$ $$M_{y}:=\frac{\partial M}{\partial y}$$

Solution

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This solution was prepared without referring to previous solutions. In ($$), since $$h$$ is nonlinear and we treat x and y as two independent variables, so the equation turns into a $$N1-PDE$$, so we have to solve this problem in two dimensional domain. What's more, the coefficients of $$h$$, $$h_{x}$$ and $$h_{y}$$ have varying coefficients ($$N_{x}, $$$$M_{y}, $$$$M, $$$$N$$). Hence it is not easy to solve this equation. And that's also the reason to make assumpetion that $$h$$ is only a founction of $$x$$ or $$y$$.
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Problem 9: Euler Integrating Factor
Based on lecture notes Pages 3 - 4

Given: Integrating factor as a function of y only
Suppose $$\displaystyle h_x(x,y) = 0, $$ thus $$\displaystyle h $$ is a function of y only; then the integrating factor becomes

Find: Integrating factor
Find $$\displaystyle h $$

Solution: h(y)

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This solution was prepared without referring to previous solutions. To find $$\displaystyle h(y)$$ we can start by integrating equation 9.1 to get: $$\displaystyle \int^y \frac{h_y}{h} = \int^y \frac{1}{M}(N_x - M_y) $$ $$\displaystyle log[h(y)] = \int^y m(s)ds + k $$ Finally, raising both sides to the exponent, you get: $$\displaystyle h(y) = exp[\int^y m(s)ds + k] $$ with the term in the exponent being known as the primitive.
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Problem 10: Find the integrating factor and solution of certain non-homogenous L1-ODE-VC
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Given: A non-homogenous L1-ODE-VC
The non-homogenous linear 1st-order ordinary differential equation with varying coefficients(L1-ODE-VC):

Find: Get the integrating factor and solution to the equation
Using Euler Integrating Factor Method(IFM), obtain the integrating factor and solution to ($$).

Solution: Using IFM

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This solution was produced without referring to previous solutions. Basic train of thought: To solve non-homogenous L1-ODE-VC like ($$), the ultimate goal is to reduce the order of the equation. However, first we need to check whether or not the equation satisfies the Two Exactness Conditions, that is:
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 * 1st Condition:


 * 2nd Condition:

If the 1st condition isn't satisfied, then we can transform the equation into the form of ($$) using algebraic method. Then if the 2nd condition isn't satisfied, we could adopt Euler Integrating Factor Method(IFM).

Based on the methodology above, the equation will be solved below: First put the right-hand side of ($$) to the left-hand side, we obtain:

Thus ($$) is transformed into form of ($$).

Then introduce $$h(x)$$ to ($$) such that:(Here we neglect $$h(y)$$ situation which will leads to contradictory result.)

Apply 2nd exactness condition ($$):

We obtain:

Thus:

Make integration on both sides of the equation ($$), we obtain:

thus:

where $$k, K$$ are integration constants and $$K\neq 0$$.

Institute ($$) into ($$), we obtain:

Reduce $$K$$, we get:

Again, make integration on($$), thus:

where $$C$$ is an integration constant.

To put in a nutshell, the solution to the problem is ($$)&($$).

Discussion: how come there are TWO constants in the process but finally there is only ONE
It should be noted that $$h(x)$$ we just produced is actually a function group $$h(x)=K\cdot x$$ which varies with $$K(K\neq 0)$$. However, $$k$$ will be reduced when instituted into the equation($$). Therefore, $$K$$ will not affect the result of $$y(x)$$ and the fact that there is only ONE integration constant $$C$$ in the result, which is corresponding to FIRST ORDER.

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Problem 11: Solve The General L1-ODE-VC
Report problem 2.7 from Pea1.f12.sec11.

Given: General L1-ODE-VC
As to a general L1-ODE-VC:

Find: Solution and Proof
1.Solve $$, when

1.1.

1.2.

2.Find $$y(x)$$ in terms of$$ a_1(x), a_0(x), b(x)$$.

3.Solve $$, when

3.1.

3.2.

Solution

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Subproblem 1
For general non-homogenous L1-ODE-VC

If $$P(x)\neq 0$$ then $$\forall x$$

Check condition

If the condition is accepted, then

Condition 1
According to the condition $$, $$yields to

We have

Check condition,

The condition is accepted. So we get

Condition 2
According to the condition $$, $$yields to

We have

Check the condition,

The condition is accepted. So we get

Subproblem 2
For $$,

Then we get:

Subproblem 3
General non-homogenous L1-ODE-VC is defined as below:

When $$P(x)\neq 0$$ then ($$) can be rewritten as below

In this specific problem, the equation is given as

First, I rearrange this equation as below

As shown in lecture

and

To make the Integration more feasible, I can rearrange the new $$a_0(x)$$ as below

the new $$b(x)$$ as below

Thus

Then

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