University of Florida/Egm4313/s12.team7/Report2

Since (Template:EquationNote) follows the same form as the quadratic equation, we can solve for $λ_{1}$ and $λ_{2}$ as follows: Template:NumBlk Template:NumBlk

Referring back to algebra, we know that the solution to these two equations can be one of three cases:

Case 1

Two real roots if...
Template:NumBlk These two roots give us two solutions:

y_{1} = e^{λ_{1} x}

and

y_{2} = e^{λ_{2} x}

The corresponding general solution then takes the form of the following: Template:NumBlk

Case 2

A real double root if...
Template:NumBlk This yields only one solution:

y_{1} = e^{λ x}

In order to form a basis, a set of two linearly independent solutions, we must find another solution. To do so, we will use the method of reduction of order, where:

y_{2} = u y_{1}

y'_{2} = u^{'} y_{1} + u y'_{1}

and

y^{'}'_{2} = u^{″} y_{1} + 2 u^{'} y'_{1} + u y^{'}'_{1}

yeilding

(u^{″} y_{1} + 2 u^{'} y'_{1} + u y^{'}'_{1}) + a (u^{'} y_{1} + u y'_{1}) + b u y_{1} = 0

u^{″} y_{1} + u^{'} (2 y'_{1} + a y_{1}) + u (y^{'}'_{1} + a^{'} y_{1} + b y_{1}) = 0

Since $y_{1}$ is a solution of (1-1), the last set of parentheses is zero.
Similarly, the first parentheses is zero as well, because

2 y'_{1} = - a e^{a x / 2} = - a y_{1}

From integration, we get the solution:

u = c_{1} x + c_{2}

If we set $c_{1} = 1$ and $c_{2} = 0$

y_{2} = x y_{1}

Thus, the general solution is: Template:NumBlk

Case 3

Complex conjugate roots if...
Template:NumBlk In this case, the roots of (Template:EquationNote) are complex:

y_{1} = e^{- a x / 2} c o s (ω x)

and

y_{2} = e^{- a x / 2} s i n (ω x)

Thus, the corresponding general solution is of the form: Template:NumBlk

Contributions

Typed by - Mark James 2/6/2012 12:08 PM UTC
Reviewed By - Jennifer Melroy 2/6/2012 2:15 UTC
Edited By - N/A

References

↑ Template:Cite book Pg 52-58
↑ Team 7: Report 1

R.2.1

Problem Statement

For this problem, the task was to derive the linear second-order non-homogeneous ODE and its solution given the two roots, initial conditions, and general excitation. However, for this problem it was assumed that r(x) = 0. Then, three non-standard and non-homogeneous ODE's were generated from the two given roots.

Roots:
Template:NumBlk

Initial conditions:
Template:NumBlk

Background Theory

To solve this problem, it is necessary to know the general form for a linear second-order ODE:
Template:NumBlk and that the characteristic equation for this type of ODE can be written as
Template:NumBlk

Solution

Substitution both roots into the characteristic equation,
Template:NumBlk and distributing fully yields
Template:NumBlk The characteristic equation can be used to form the second-order ODE:
Template:NumBlk
Therefore, the solution for the homogeneous function is in the form of
Template:NumBlk and its derivative is
Template:NumBlk Algebraic substitution yields: Template:NumBlk Template:NumBlk Template:NumBlk Template:NumBlk Template:NumBlk Template:NumBlk

Plugging in initial conditions,
Template:NumBlk and Template:NumBlk

Algebraic substitution and solving for each constant term yields: Template:NumBlk Template:NumBlk Template:NumBlk Template:NumBlk Template:NumBlk Template:NumBlk

Therefore, since r(x)=0,
Template:NumBlk and the graph of y(x) is
File:Capture111.png

Generated ODEs

Contributions

Typed by - Maxwell Shuman 2/6/2012 15:10 PM UTC
Reviewed By - Jennifer Melroy 2/6/2012 2:15 UTC
Edited By - N/A

R.2.2

Problem Statement

Find and plot the solution for

With the initial conditions of Template:NumBlk Template:NumBlk Template:NumBlk

Background Theory

From the background theory at the beginning of this report, using the quadratic formula we find that

In this case

This would make this problem under the category of Case 2, in which the general solution becomes equation (Template:EquationNote) Template:NumBlk

Solution

To find ʎ we use Template:NumBlk and we find that Template:NumBlk

Because the problem is in the category of Case 2, the general solution is Template:NumBlk

Differentiating this general solution gives us Template:NumBlk

Using equation(Template:EquationNote) and plugging in the inital condition of y(0)= 1 , we find that

Using equation (Template:EquationNote) and plugging in the initial condition of y'(0)=0 and $c_{1} = 1$ , we find that Template:NumBlk Template:NumBlk Template:NumBlk Template:NumBlk

The general solution for this problem will be Template:NumBlk

File:Report2Plot.JPEG

Contributions

Typed by ---Dalwin Marrero 21:59, 6 February 2012 (UTC)
Reviewed By - Jennifer Melroy 2/6/2012 2:15 UTC
Edited By - N/A

R.2.3

P.2.2.3

Problem Statement

For this problem, we were tasked with finding the general solution to the homogeneous linear ODE given below:
Template:NumBlk

Solution

Following the process that yields (Template:EquationNote), we find the equation: Template:NumBlk

To find the form of the general solution, we must solve for the values of $λ$ using the quadratic formula:

File:P2 2 3.png

Since the solutions to $λ$ are complex numbers, our solution takes the form of (Template:EquationNote):
Template:NumBlk

$c_{1}$ and $c_{2}$ are constant coefficients of unknown value. Had we been given initial values of $y (x)$ and $y^{'} (x)$ , we would solve for those values as well.

Confirmation of Solution

To confirm that the solution is true, we will substitute $y$ , $y^{'}$ , and $y^{″}$ into (Template:EquationNote). If the final result is a tautology^[1] then the solution is confirmed. If the result is a contradiction, a statement that is never true, then our solution is disproved.

Substituting (Template:EquationNote), (Template:EquationNote), and (Template:EquationNote) into (Template:EquationNote), we are left with:

Template:NumBlk Template:NumBlk Template:NumBlk Template:NumBlk

Combining similar terms allows us to clean up this solution and check our answer:

Final Solution

Since the outcome of our check is a tautology, the solution is confirmed for all values of A and B and we affirm that the final solution is:

y = c_{1} e^{- 2.8 x} + c_{2} e^{- 3.2 x}

P.2.2.4^[2]

Problem Statement

For this problem, we were tasked with finding the general solution to the homogeneous linear ODE given below:
Template:NumBlk

Solution

Following the process that yields (Template:EquationNote), we find the equation: Template:NumBlk

To find the form of the general solution, we must solve for the values of $λ$ using the quadratic formula:

Since the solutions to $λ$ are complex numbers, our solution takes the form of (Template:EquationNote):
Template:NumBlk

A and B are constant coefficients of unknown value. Had we been given initial values of $y (x)$ and $y^{'} (x)$ , we would solve for those values as well.

Confirmation of Solution

To confirm that the solution is true, we will substitute $y$ , $y^{'}$ , and $y^{″}$ into (Template:EquationNote). If the final result is a tautology^[1] then the solution is confirmed. If the result is a contradiction, a statement that is never true, then our solution is disproved.

Template:NumBlk Template:NumBlk Template:NumBlk Template:NumBlk

Substituting (Template:EquationNote), (Template:EquationNote), and (Template:EquationNote) into (Template:EquationNote), we are left with:

Template:NumBlk Template:NumBlk Template:NumBlk Template:NumBlk

Combining similar terms allows us to clean up this solution and check our answer:

Final Solution

Since the outcome of our check is a tautology, the solution is confirmed for all values of A and B and we affirm that the final solution is:

y = e^{- 2 x} (A c o s (π x) + B s i n (π x))

Contributions

Typed by - Mark James 2/6/2012 12:08 PM UTC
Reviewed By - Jennifer Melroy 2/6/2012 2:15 UTC
Edited By - N/A

References

↑ ^1.0 ^1.1 Introduction to Propositional Logic
↑ Team 7: Report 1

R.2.4

P.2.2.5

Problem Statement

For this problem, we were asked to find the general solutions for problem 5 on page 59 of (need to site book; do not know how to) Template:NumBlk

Solution

Following the process that yields (Template:EquationNote), we find the equation: Template:NumBlk
To find the form of the general solution, we must solve for the values of $λ$ . Instead of going through the quadratic formula, we can analyze the discriminant. Template:NumBlk

Thus we see that the discriminant follows Case 2 and the general solution must follow the form of (Template:EquationNote), and we have the following: Template:NumBlk where $c_{1}$ and $c_{2}$ are unknown constant coefficients. If this were an initial value problem with values of $y (x)$ and $y^{'} (x)$ , we would solve for those values as well.

Confirmation of Solution

To confirm that the solution is true, we will substitute $y$ , $y^{'}$ , and $y^{″}$ and check to see if the final result is a tautology or a contradiction.
Template:NumBlk Template:NumBlk
Substituting, we are left with: Template:NumBlk After combining similar terms, we have the form: Template:NumBlk Template:NumBlk

Final Solution

Since the above result is a tautology for all values of $c_{1}$ and $c_{2}$ ,the general solution is: Template:NumBlk

P2.2.6

Problem Statement

For this problem, we were asked to find the general solutions for problem 6 on page 59 of (need to site book; do not know how to) Template:NumBlk

Solution

For this problem, we have to divide by 10 to put the equation in standard form, making it Template:NumBlk

Following the process that yields (Template:EquationNote), we find the equation: Template:NumBlk
To find the form of the general solution, we must solve for the values of $λ$ . Instead of going through the quadratic formula, we can analyze the discriminant. Template:NumBlk

Thus we see that the discriminant follows Case 2 and the general solution must follow the form of (Template:EquationNote), and we have the following: Template:NumBlk Template:NumBlk where $c_{1}$ and $c_{2}$ are unknown constant coefficients. If this were an initial value problem with values of $y (x)$ and $y^{'} (x)$ , we would solve for those values as well.

Confirmation of Solution

To confirm that the solution is true, we will substitute $y$ , $y^{'}$ , and $y^{″}$ and check to see if the final result is a tautology or a contradiction.
Template:NumBlk Template:NumBlk
Substituting, we are left with: Template:NumBlk After combining similar terms, we have the form: Template:NumBlk Template:NumBlk

Final Solution

Since the above result is a tautology for all values of $c_{1}$ and $c_{2}$ ,the general solution is: Template:NumBlk

Contributions

Typed by - Mark James 2/6/2012 12:08 PM UTC
Reviewed By - Jennifer Melroy 2/6/2012 2:15 UTC
Edited By - N/A

R.2.5

Background Theory

A basis is the root of a ODE. They take the form of

They can be used to identify which background case can be applied.

It can also be noted that Template:NumBlk

P.2.2.16

Problem Statement

For this problem find the ordinary differential equation in the from of (Template:EquationNote)

using the following basis the fit the form of (Template:EquationNote)
Template:NumBlk Template:NumBlk

Solution

For this solution, it is assumed that Template:NumBlk Template:NumBlk

Since(Template:EquationNote) and (Template:EquationNote) match the given roots in Case 1, the general solution in the form of (Template:EquationNote) can be found by substituting (Template:EquationNote) and (Template:EquationNote). This results in Template:NumBlk

From (Template:EquationNote), $λ_{1}$ and $λ_{2}$ can be found Template:NumBlk Template:NumBlk

Using the relationship in (Template:EquationNote), and proprieties of quadratic equations it can be assumed Template:NumBlk

Multiplying (Template:EquationNote) out give the form Template:NumBlk

Since (Template:EquationNote) is in the form of (Template:EquationNote), the relationship in (Template:EquationNote) can be used to get the form of Template:NumBlk

P.2.2.17

Problem Statement

For this problem find the ordinary differential equation in the from of (Template:EquationNote)

using the following basis the fit the form of (Template:EquationNote)
Template:NumBlk Template:NumBlk

Solution

For this solution, it is assumed that Template:NumBlk Template:NumBlk

Since(Template:EquationNote) and (Template:EquationNote) match the given roots in Case 2, the general solution in the form of (Template:EquationNote) can be found by substituting (Template:EquationNote) and (Template:EquationNote). This results in Template:NumBlk

From (Template:EquationNote), $λ$ can be found Template:NumBlk

Using the relationship in (Template:EquationNote), and proprieties of quadratic equations it can be assumed Template:NumBlk

Multiplying (Template:EquationNote) out give the form Template:NumBlk

Since (Template:EquationNote) is in the form of (Template:EquationNote), the relationship in (Template:EquationNote) can be used to get the form of Template:NumBlk

Contributions

Solved and Typed by - Jennifer Melroy 2/4/2012 9:20 PM UTC
Reviewed By -
Edited By - N/A

References

R.2.6

Problem Statement

Given this system:
File:Spring-dashpot-ball2.PNG
It was found that the equation for this system is Template:NumBlk The parameters of this equation can be found by the characteristic equation for a double real root if given a value for $λ$ .
Find the values of m , k , and c with $λ = - 3$ .

Solution

With $λ = - 3$ then the characteristic equation becomes Template:NumBlk

If we relate these equations we can say that the coefficients in (Template:EquationNote) and (Template:EquationNote) are the same. With that said then,

Template:NumBlk Template:NumBlk Template:NumBlk Template:NumBlk

We are fortunate that with this method m and k are solved without any need for calculation.

This leaves c to be solved using (Template:EquationNote). Template:NumBlk Template:NumBlk Template:NumBlk

So m = 1, k = 9 and c = 1.5

Contributions

Typed by - Yamil Herrera 2/6/2012 12:08 PM UTC
Reviewed By - Jennifer Melroy 2/6/2012 2:15 UTC
Edited By - N/A

R.2.7

Background

Taylor series is a method used to find the value of a equation using a set of infinite sums of the its derivatives at a certain point a. Taylor Series: Template:NumBlk The MacLaurin Series is the most basic form of the Taylor series by making a=0. Maclaurin Series: Template:NumBlk

Problem Statement

Develop a MacLaurin series for the following equations: $e^{t}$ , cos(t), and sin(t).

Solution

The first step for finding the solution is to solve for f(a) and its derivatives.
For $y = e^{t}$ (where y = f(a)) the first three derivatives come out to: Template:NumBlk Template:NumBlk Template:NumBlk Template:NumBlk Once the derivatives are found, make t=0, and plug back into (Template:EquationRef) which results in the the equation: Template:NumBlk Doing the same thing for y= cos(t)the derivatives come out to: Template:NumBlk Template:NumBlk Template:NumBlk Template:NumBlk Template:NumBlk Again, plugging in zero and placing the ys into (Template:EquationRef) you get: Template:NumBlk Repeating the same method for y= sin(t) as the previous two equations the derivatives come out as: Template:NumBlk Template:NumBlk Template:NumBlk Template:NumBlk Template:NumBlk Template:NumBlk Using (Template:EquationRef) you get the equation: Template:NumBlk

R.2.8

P.2.2.8

Problem Statement

For this problem, find the general solution for the following equation: Template:NumBlk

Solution

From the equation (Template:EquationNote) our a=1 and b=3.25

With this information we can decide what kind of root we have.

Template:NumBlk So with a=1 and b=3.25

$(1)^{2} - 4 (3.25) = - 12$

Since (Template:EquationNote) is less than zero, we can use Case 3 from our Background Theory.
Our general solution will be: Template:NumBlk

To find $ω$ we need to use the characteristic equation (Template:EquationNote) and solve for the root $λ$ Template:NumBlk where Template:NumBlk Using (Template:EquationNote) we can solve for $ω$ Template:NumBlk Now that we have $ω$ we can plug it in to get our general solution from (Template:EquationNote) Template:NumBlk

Confirmation of Solution

To confirm that the solution is true, we will find $y^{'}$ and $y^{″}$ and plug it into (Template:EquationNote). If the result agrees and equals zero then it it considered true. Template:NumBlk Template:NumBlk Template:NumBlk Template:NumBlk

Substituting (Template:EquationNote), (Template:EquationNote), and (Template:EquationNote) into (Template:EquationNote), we are left with:

Combining similar terms allows us to clean up this solution and check our answer:

P.2.2.15

Problem Statement

For this problem, find the general solution for the following equation:

Solution

From the equation (Template:EquationNote) our a=0.54 and b=0.0729+ $π$

With this information we can decide what kind of root we have.

Template:NumBlk So with a=0.54 and b=0.0729+ $π$ 3.25

$(0.54)^{2} - 4 (0.0729 + π) = - 12.57$

Since (Template:EquationNote) is less than zero, we can use Case 3 from our Background Theory.
Our general solution will be: Template:NumBlk

To find $ω$ we need to use the characteristic equation (Template:EquationNote) and solve for the root $λ$ Template:NumBlk where Template:NumBlk Using (Template:EquationNote) we can solve for $ω$ Template:NumBlk Now that we have $ω$ we can plug it in to get our general solution from (Template:EquationNote) Template:NumBlk

Confirmation of Solution

To confirm that the solution is true, we will find $y^{'}$ and $y^{″}$ and plug it into (Template:EquationNote). If the result agrees and equals zero then it it considered true. Template:NumBlk Template:NumBlk Template:NumBlk Template:NumBlk

Substituting (Template:EquationNote), (Template:EquationNote), and (Template:EquationNote) into (Template:EquationNote), we are left with:

Combining similar terms allows us to clean up this solution and check our answer:

Contributions

Typed by - Yamil Herrera 2/6/2012 12:08 PM UTC
Reviewed By - Jennifer Melroy 2/6/2012 2:15 UTC
Edited By - N/A

R.2.9

Problem Statement

For this problem we were to find and plot a solution to the following problem Template:NumBlk We also had to superpose three figures: one for this problem, and problems .

Background Theory

In the problem we were given the initial conditions
$y (0) = 1$ ,
$y^{'} (0) = 0$ ,
and that there is no excitation, having
$r (x) = 0$ .

There are a few steps that we have to go through before we can solve this problem. We first have to solve for $λ_{1}$ and $λ_{2}$ . To find out what $λ_{1}$ and $λ_{2}$ equal we can apply pythagorean theorem, Template:NumBlk We will know if this problem has complex roots or not by applying $a^{2} - 4 b$ . If the answer we get is greater than zero we will use case one. If it is equal to zero we will use vase two and if it is less than zero we will use case 3. Template:NumBlk , so we will use case 3.
The standard form for case 3 is (Template:EquationNote) We will also have to find out what $a$ and $w$ are equal to and we use this equation to find out what they are Template:NumBlk

Solution

Part 1

This first step in solving this problem is to factor in the fact that there is no excitation. Template:NumBlk Once we have does this we need to solve for $λ_{1}$ and $λ_{2}$ . When apply (Template:EquationNote) we get Template:NumBlk. After reducing it down we end up with Template:NumBlk Now that we know what $λ_{1}$ and $λ_{2}$ are equal to, we apply (Template:EquationNote) to find out what $a$ and $w$ are.
We find that $a = 2$ and $w = 3$ Knowing that we have complex roots, we can now plug the known values of $a$ and $w$ , into (Template:EquationNote).
Once we do this we have Template:NumBlk The next step we have to do is to take the derivative of (Template:EquationNote). We end up getting Template:NumBlk We know have to find what $C_{1}$ and $C_{2}$ are equal to. From what we already know about our initial conditions we can apply those and what we found $a$ and $w$ to be into both (Template:EquationNote) Template:NumBlk Template:NumBlk and (Template:EquationNote).