User:Egm6321.f12.team7/report3

=R*3.1=

Problem: Show Redundancy of Integration Constants
Problem R*3.1 from lecture notes section 12.

Given: L1-ODE-VC
Given a general non-homogenous, linear, first order differential equation with variable coefficients of the form:

Find: Redundancy of Integration Constants
Show how only one integration constant is needed.

Solution: Solve using Euler IFM with only one constant

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($$) can be written as:

where

Hence, the first exactness condition of nonlinear, first order ODEs is satisfied. Thus, to satisfy the second exactness condition, the integrating factor, $$h$$, must be such that:

Assuming $$h_y (x,y) = 0$$ (i.e. $$h$$ is a function of $$x$$ only), the following condition must be satisfied, where $$n(x)$$ is a function of only $$x$$:

Substituting into ($$) from ($$) and ($$), produces

which indeed is a function of $$x$$ only. Solving for $$h$$ results in

where $$s$$ is a dummy variable for integration. From the original ODE in ($$)

From ($$)

Substituting into ($$)

By definition of the product rule of differentiation

where q is a dummy variable for integration. Substituting ($$) into the solution yields

Since $$k_2$$ and $$C$$ are constants, the term $$k_2 / C$$ can be replaced with one constant, $$C_2$$. Hence, only one integration constant is necessary in the solution of the linear, first order, ordinary differential equation with variable coefficients of the form given by ($$). Intuitively, this makes sense since the differential equation is first order, so there should only be one integration constant.

=R*3.2=

Problem 2: Compare the result
Report problem 3.2 from

Given: The result in lecture note and King's Book
The result of general homogeous L1-ODE-VC is given by ,

The solution of King's book is given by,

Find
Show that the solution of lecture agrees with the King's solution

Solution

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The homogeous solution of King 's result is given by,

The particular solution of King's result is given by,

Follow the notation used in lecture, the homogeous solution of of equation is given by,

The particular solution of equation is given by,

Obviouly, $$y_p ,y_h $$are similar to $$y_p'. y_h'$$.

=R*3.3=

Problem 3: Calculate the homogeous result
Report problem 3.2 from

Find:The homogenous solution
Calculate the homogenous solution.

Solution

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Follow the notation used in lecture,the solution of $$a_o(x)y+y'=0 $$ is given by,

Where $$h(x)$$ is integration factor, $$b(x)$$ is equal to zero.

So, the homogeous solution is given by,

=R*3.4=

Problem 4: Finding the Integrating Factor h to make a L1-ODE-VC exact
Based on lecture notes section 12

Given: Exactness Conditions and the Integrating Factor
A General Nonlinear 1st Order Ordinary Differential Equation is exact if it satisfies the following two exactness conditions (from section 8 and section 9) : 1st Condition: An N1-ODE has to take the form:

2nd Condition: The partial derivatives of the N1-ODE:

If the N1-ODE satisfies the 1st condition ($$) but not the 2nd condition ($$), there exists an integrating factor $$\displaystyle h(x,y) $$ such that the following N1-ODE is exact (from section 10) :

Furthermore $$\displaystyle h(x,y) $$ is only a function of x if the following 3 conditions are true (from section 12) :

Given all of those conditions, the integrating factor can be written as:

Find: Integrating Factor h to make the L1-ODE-VC exact
For the following L1-ODE-VC, if it is not exact, find the integrating factor h to make it exact:

Solution: Integrating Factor h(x)

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Though this is a L1-ODE-VC, we can use the Integrating Factor Method for general N1-ODEs to make it exact. To begin, we must first test the two exactness conditions ($$) and ($$)

For the first condition, let

From here we can see that

For the second condition,

We can see that

Therefore ($$) is satisfied but ($$) is not.

Furthermore, we can see that ($$), ($$), and ($$) are all functions of x, thus the integrating factor is going to be:

Thus the integrating factor is

Where k is the integration constant.

=R*3.5=

Problem 5: Show that a Nonlinear 1st Order ODE meets certain conditions
Based on lecture notes section 13

Given: An N1-ODE
The following N1-ODE

where

has an integrating factor h(x) that can be found to render it exact using:

only if

$$\displaystyle k_1(y) = d_1 $$ (constant)

Find: Support that for the N1-ODE k(y) must be constant
Prove that the above statement is true. Furthermore, show that ($$) includes

as a particular case.

Solution: k(y) is constant and 5.5 is a particular case of 5.1

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Since

$$\displaystyle N(x,y) = \bar b(x,y)c(y) $$

$$\displaystyle M(x,y) = a(x)\bar c(x,y) $$

then, using ($$) and ($$),

Equation ($$) becomes:

Since $$\displaystyle n(x) $$ is defined as only a function of x, $$\displaystyle k_1(y) $$ must be a constant for that to be true.

Now to prove ($$) is a particular case, let us suppose that

$$\displaystyle c(y) = 1 $$

$$\displaystyle k_1(y) = 0 $$

We would then have,

$$\displaystyle \bar b(x,y) := \int^x b(s)ds + 0 = \bar b(x) $$

$$\displaystyle \bar c(x,y) := \int^y 1ds + k_2(x) = y + k_2(x) $$

$$\displaystyle c(y) $$ then becomes

$$\displaystyle c(y) = \frac{\partial (\bar c(x,y))}{\partial y} = 1 $$

Since a(x) is arbitrary, ($$) then reduces down to ($$) or:

$$\displaystyle [a(x)y+k_2(x)] + \bar b(x)y' = 0 $$

Therefore it is a particular case of ($$)

=R*3.6=

Problem 6:show the exactness and find the integrating factor
Report problem 3.6 from lecture notes section 13-4.

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

Find: show the exactness and find the integrating factor
Show whether the function is exact, or can be made exact by the IFM, then find the integrating factor h.

Solution: Test the function with 1st and 2nd exactness condition and find the integrating factor h

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To examine whether the function is exact is to find out that whether the function satisfies the 1st and 2nd Exactness Condition.

1. 1st exactness condition satisfies

with

2. 2nd exactness condition could be found as below

Apparently,

Thus, This function satisfying the 1st exactness condition, however, not satisfying the 2nd exactness condition.

To make the function exact by the IFM, first find the integrating factor h(x,y) such that the following function can be exact

It is given that

Then find the n(x) as below

The integrating factor h(x) can be calculate as below

=R*3.7=

Problem 7:Find the exactness and Find the first integral
Report problem 3.7 from lecture notes section 13-4.

Given: A specified function
Composites of specific function is shown as below

Find: show the exactness and find the first integral
Set up a function in terms of given composites, and make it exact, or show that it could be made exact by the IFM, then find the first integral

Solution: examine the function with 1st and 2nd condition and find the first integral

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Calculate other composites of this function

Set up a function in terms of given composites as below

when

Thus, function can be written as

To examine whether the function is exact is to find out that whether the function is satisfied the 1st and 2nd Exactness Condition.

1. 1st exactness condition satisfies

with

2. 2nd exactness condition could be found as below

Apparently,

Thus, This function satisfying the 1st exactness condition, however, not satisfying the 2nd exactness condition.

To make the function exact by the IFM, first find the integrating factor h(x,y) such that the following function can be exact

Then find the n(x) as below, assuming h is a function of x only

h(x) can be calculate as below

So the Eqn. 7.11 can be changed into

with

This function is exact by IFM

By definition

with

and

Calculation shown as below

Thus

=R*3.8=

Problem 3: Find the conunterpart
Report problem 3.2 from

Given:the general class of N1-ODE-VC
A general class of N!-ODE-VC that are either exact of can be exact by IFM method is given by,

Where $$\bar b(x,y)$$ and $$\bar c(x,y)$$ are given by,

Find:Find the counterpart of the general class of N1-ODE-VC
Find the counterpart of general class of N1-ODE-VC which is either exact or can be exact by IFM method.

Solution

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The equation is based on an assumption that intergration factor is only a founction of x, which is given by,

Where,

So ,the counterpart of the intergration factor is given by,

To satisfy the condition that $$h$$ is function of y, we can set the below forms,

So,for$$ M(x,y)$$, we have,

the $$k_1$$ should be a constant since the h is only a founction of y. The N(x,y) is given by,

Let's set the $$\int^y b(t)dt + k_1$$ equal to $$\bar b(y)$$, $$ \int^x c(t)dt + k_2{y}$$ equal to $$\bar c(x,y)$$, then we have the conterpart of general class of N!-ODE-VC, which is given by,

=R*3.9=

Problem 9: IFM application in the motion of a particle in the air
Report problem 9 from.

Given: the motion of a particle in the air
The schematic of the motion of a particle in the air:

Find: Verify the solution to the equation
1.  Derive the equations of motion:

2.  Particular case $$k=0$$:Verify that $$y(x)$$ is parabola

3.  Consider the case $$k\neq 0$$ and $$v_{x0}=0$$: ($$) turns into

(1). Is ($$) for $$n=0,1,2$$ either exact or can be made exact using IFM? Find $$v_y(t)$$ and $$y(t)$$ for $$m$$ constant.

(2). Find $$v_y(t)$$ and $$y(t)$$ for $$m=m(t)$$ varying with $$t$$ as following:

Solution

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1. Derive the equations of motion

According to ($$):

First, we divide $$v$$ into $$v_x$$ and $$v_y$$ components, and we have: $$v^2=(v_x)^2+(v_y)^2$$ and $$tan\alpha=\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\frac{\mathrm{d} y}{\mathrm{d} t}}{\frac{\mathrm{d} x}{\mathrm{d} t}}=\frac{v_y}{v_x}$$.

Then, we could use Newton's Second Law: $$F=ma$$, respectively, in X direction and Y direction:


 * X direction: $$-kv^n*cos\alpha=m\frac{dv_x}{dt}$$
 * Y direction: $$-kv^n*sin\alpha-mg=m\frac{dv_y}{dt}$$

Thus, ($$), ($$), ($$) and ($$) are derived.

2. Particular case $$k=0:$$ Verify that $$y(x)$$ is parabola.

When $$k=0$$, ($$) is turned into

Thus, through integration, we obtain:

3.  Consider the case $$k\neq 0$$ and $$v_{x0}=0$$:

When $$k\neq 0$$, ($$) is turned into

(1). Is ($$) for $$n=0,1,2$$ either exact or can be made exact using IFM? Find $$v_y(t)$$ and $$y(t)$$ for $$m$$ constant.

Rewrite ($$):

FIRST, suppose $$\frac{dM}{v_y}=\frac{dN}{dt}$$, under such condition, ($$) is exact:

we get, $$n=0$$. thus, when $$n=0$$, ($$) is exact.

THEN, suppose ($$) is not exact but could be made exact. We introduce Euler Integrating Factor $$h(t)$$ and dot it with ($$):

then, we have:

moreover,

we find $$n=1$$ satisfies ($$), but $$n=2$$ doesn't.

THUS: When $$n=0$$ ($$) is exact.

When $$n=1$$ ($$) is not exact but could be made exact.

When $$n=2$$ ($$) is neither exact nor could be made exact.

NEXT, we will find $$v_y(t)$$ and $$y(t)$$ for $$m$$ constant:

When $$n=0$$,($$) is reduced to:

through integration, we obtain:

When $$n\neq 0$$, we start from ($$):

through integration, we obtain:

where $$C_1$$ is the integration constant.

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

through integration:

(2). Find $$v_y(t)$$ and $$y(t)$$ for $$m=m(t)$$ varying with $$t$$ as ($$):

Under such condition:

we can verify that when $$n=0,1,2$$, $$\frac{dM'}{dv_y}\neq \frac{dN'}{dt}$$ i.e. ($$) is not exact.

Thus we need to introduce a Euler Integrating Factor $$H(t)$$ to make it exact:

dot ($$) with $$H(t)$$:

then we have:

make it explicit:

According to the graphic ($$): {{NumBlk|:|$$m=\left\{\begin{matrix} K_0t+m_0\;(0\leqslant t\leqslant t_1)\\m_1\;\;\;\;(t_1< t) \end{matrix}\right.$$|$$}} where: $$K_0=-\frac{m_0-m_1}{t_1}$$.

When $$n=0$$, we have:

thus:

institute ($$) into ($$) and take integration, we obtain:

{{NumBlk|:|$$v_y=\left\{\begin{matrix} -gt-\frac{k}{K_0}\log(\frac{K_0}{m_0}t+1)+v_{y_0}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(0\leqslant t\leqslant t_1)\\\\-gt-\frac{k}{K_0}\log(\frac{K_0}{m_0}t_1+1)+v_{y_0}+\frac{k}{m_1}(t-t_1)\;\;\;\;(t\geqslant t_1) \end{matrix}\right.$$|$$}}

{{NumBlk|:|$$y(t)=\left\{\begin{matrix}-\frac{1}{2}gt^2-\frac{k}{K_0}\int_{0}^{t}\log (\frac{K_0}{m_0}t+1)dt+v_{y_0}t\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(0\leqslant t\leqslant t_1)\\\\-\frac{1}{2}(\frac{k}{m_1}-g)^2-(\frac{kt_1}{m_1}+\frac{k}{K_0}\log \frac{m_1}{m_0}-v_{y_0})t\;\;\;\;(t\geqslant t_1) \end{matrix}\right.$$|$$}}

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

thus: {{NumBlk|:|$$H=\left\{\begin{matrix} m^{\frac{k}{K_0}}\;\;\;\;\;\;\;\;\;\;(0\leqslant t\leqslant t_1)\\\frac{e^{\frac{kt}{m_1}}}{m}\;\;\;\;(t\geqslant t_1)\end{matrix}\right.$$|$$}}

institute ($$) into ($$) and take integration, we obtain:

{{NumBlk|:|$$v_y=\left\{\begin{matrix}   -\frac{mg}{k+K_0}+\frac{m_0^{k/k_0} m^{(K_0-k)/K_0}g}{k+K_0}+v_{y_0}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(0\leqslant t\leqslant t_1)\\\\  \frac{mg(m_0^{k/K_0}-m_1^{k/K_0})}{e^{kt/m_1}(k+K_0)}+\frac{m_1(e^{kt/m_1}-e^{kt_1/m_1})}{k}+v_{y_0}\;\;\;\;(t\geqslant t_1) \end{matrix}\right.$$|$$}}

{{NumBlk|:|$$y(t)=\left\{\begin{matrix}  -\frac{(\frac{1}{2}K_0t^2+m_0t)g}{k+K_0}+\frac{K_0m_0^{k/K_0}m^{2-k/K_0}g}{(2K_0-k)(k+K_0)}+v_{y_0t}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(0\leqslant t\leqslant t_1)\\\frac{(\frac{1}{2}K_0t^2+m_0t)g(m_0^{k/K_0}-m_1^{k/K_0})}{e^{kt/m_1}(k+K_0)}+(m_1/k)^2e^{kt/m_1}-\frac{m_1t}{k}e^{kt_1/m_1}+v_{y_0}t   \;\;\;\;(t\geqslant t_1) \end{matrix}\right.$$|$$}}

When $$n=2$$, we have:

Obviously, ($$) could not be satisfied. Thus as for $$n=2$$, we still can't make it exact.

=R*3.10=

Problem: Solve a L1-ODE-CC
Problem R*3.10 from lecture notes section 15.

Given: A L1-ODE-CC
Given a linear, first order, ordinary differential equation with constant coefficients of the form:

Find: The solution to the L1-ODE-CC
Show that the solution is given by:

using Euler's integrating factor method. Identify the integrating factor, homogeneous solution and particular solution. Also show that when the coefficients $$a$$ and $$b$$ vary with time, the solution becomes

Solution: Based on Euler's IFM

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($$) can be rearranged into the following form

where

Hence, the first exactness condition of N1-ODEs is satisfied. Thus, to satisfy the second exactness condition, the integrating factor, $$h$$, must be such that:

Assuming $$h_x (t,x) = 0$$ (i.e. $$h$$ is a function of $$t$$ only), the following condition must be satisfied, where $$n(t)$$ is a function of only $$t$$:

undefined Substituting into ($$) from ($$) and ($$), produces

which indeed is a function of $$t$$ only. Solving for $$h$$ results in

where $$s$$ was used as a dummy integration variable. This $$h$$ is the integrating factor used to make the original ODE exact. Multiplying into both sides of the original ODE produces

From the product rule of differentiation, and observing that $$h_t = -ah$$, the left hand side of ($$) can be simplified into:

Integrating both sides from $$t_0$$ to $$t$$ and solving for x(t) produces

where $$\tau$$ is a dummy integrating variable and $$h(t_0) = 1$$. Substituting in ($$) for $$h$$ yields the solution given in ($$)

The first term on the right hand side is the homogeneous solution, as it is independent of the forcing function. The second term is the particular solution, as it depends on the nonhomogeneous term from the original ODE. When the coefficients $$a$$ and $$b$$ vary with time, the solution process is identical, except the integral on the right hand side of ($$) cannot be evaluated. Therefore, the integrating factor becomes:

Substituting into ($$) and noting that $$s$$ and $$\tau$$ are dummy integrating variables, the solution becomes

=R*3.11=

Problem 11: Free Vibration of Coupled Pendulums
Report problem 3.11 from Pea1.f12.sec15-5.

Given: Variables
Pendulums:

No applied forces:

Initial conditions:

Find: Show the Process of Verification

 * 1) Use MATLAB's ode45 command to integrate the system (1)(2)p.14-5 for $$t \in [0,7]$$. Can also use equivalent Octave.
 * 2) Use (2)p.15-2to find the solution at the same time stations as in Q1.
 * 3) Plot $$\theta_1(t)$$ from Q1 and from Q2.
 * Plot $$\theta_2(t)$$ from Q1 and from Q2.

Solution

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Integration
If we arrange the ($$) in the preparing form of matrices, we get:

Define:

$$ \mathbf{x}=\begin{Bmatrix} x_1\\x_2\\x_3\\x_4 \end{Bmatrix}=\begin{Bmatrix} \theta_1\\ \dot\theta_1\\\theta_2\\ \dot \theta_2 \end{Bmatrix} \mbox{ and } \mathbf{\dot x}=\begin{Bmatrix} x_5\\x_6\\x_7\\x_8 \end{Bmatrix}=\begin{Bmatrix} \dot\theta_1\\\ddot\theta_1\\\dot\theta_2\\ \ddot \theta_2 \end{Bmatrix} $$

Obviously

Conbining $$, we can easily write $$into a matrix form:

detailed with

in which

As we know $$g=9.8$$. Substitute $$with $$and$$, when $$t\in[0,7]$$then we get

With $$,coding on MATLAB, and get the codes.

With the MATLAB programs above, we get the plot of $$ \mathbf{x}=\begin{Bmatrix} x_1\\x_2\\x_3\\x_4 \end{Bmatrix}=\begin{Bmatrix} \theta_1\\ \dot\theta_1\\\theta_2\\ \dot \theta_2 \end{Bmatrix} $$ below.

Find Solution
According to (2)p.15-2, we get:

On the condition of $$, $$turns to be

in which $$\mathbf{A}$$ as obtained above. Substitute the relevent coefficiants with $$. When $$t\in[0,7]$$ and $$\mathbf{A}$$ as obtained above, and

we get the solution of this model:

Then we get the MATLAB codes below:

So we get the figure of $$ \mathbf{x}=\begin{Bmatrix} x_1\\x_2\\x_3\\x_4 \end{Bmatrix}=\begin{Bmatrix} \theta_1\\ \dot\theta_1\\\theta_2\\ \dot \theta_2 \end{Bmatrix} $$ shows their changes in given time space.

Notice
 By examing the previous reports, I notice that some people got incorrect plots as show below.The reason of the error comes from the misusing of MATLAB command. The command "exp(n)" is used for figure,while"expm(A)" is used for matrix.

by Hao Lin

Draw the plot
We plot $$\theta_1(t)\mbox{ and }\theta_2$$from Q1 and Q2 as below accoring to the former data.

And it is obvious that the two plots are identical because the two methords can all get the correct answer.

=R*3.12=

Problem 12: Derive 2nd exactness condition of general N2-ODE and verify the 2nd exactness condition of a particular N2-ODE
Report problem 3.12 from.

Given: 2nd exactness condition of N2-ODE
A N2-ODE

is exact if a function

exists to satisfy

($$) has the form of

where

The 2nd exactness condition for N2-ODE ($$) is

A particular N2-ODE is given as

Find: Derive and verify
1.Derive ($$)

2.Derive ($$)

3.Verify ($$) satisfies the 2nd exactness condition.

Solution

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Since

from ($$),

take partial derivative wrt p,

from ($$) and ($$)

thus,

from ($$), take partial derivative wrt x,

substitute ($$) and ($$) into ($$),

rearrange ($$),

take partial derivative wrt p,

substitute into ($$) to find

This proved the 2nd relation in the 2nd exactness condition.

substitute ($$) into ($$),

take partial derivative wrt y,

take partial derivative of ($$) wrt y,

substitute ($$) and ($$) into ($$) to find

This proved the 1st relation in the 2nd exactness condition.

In ($$),

To verify the 1st relation in the 2nd exactness condition:

take partial derivative of ($$) wrt x,

take partial derivative of ($$) wrt x,

take partial derivative of ($$) wrt y,

take partial derivative of ($$) wrt y,

take partial derivative of ($$) wrt y,

substitute ($$), ($$) and ($$) into LHS of ($$),

take partial derivative of ($$) wrt x,

take partial derivative of ($$) wrt p,

take partial derivative of ($$) wrt y,

take partial derivative of ($$) wrt p,

substitute ($$), ($$) and ($$) into RHS of ($$),

thus,

This verified the 1st relation in the 2nd exactness condition.

To verify the 2nd relation in the 2nd exactness condition:

take partial derivative of ($$) wrt p,

take partial derivative of ($$) wrt p,

substitute ($$), ($$) and ($$) into LHS of ($$),

take partial derivative of ($$) wrt p,

take partial derivative of ($$) wrt p,

substitute ($$) into RHS of ($$),

thus,

This verified the 2nd relation in the 2nd exactness condition.

Thus, ($$) satisfies the 2nd exactness condition.

=Reference=

= Contribution =