University of Florida/Egm4313/s12.team14.report2

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Problem 1[edit]

Given[edit]

Find[edit]

Find the non-homogeneous L2-ODE-CC in standard form and the solution in terms of the initial conditions and the general excitation

Consider no excitation:


Plot the solution.

Solution[edit]

Characteristic equation:


Substituting into :



Non-homogeneous solution:


Homogeneous solution:


Overall solution:


No excitation:



From intitial conditions:



Solving for and :

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and

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So, the final solution is:



Team14R22diagram1.png

Problem 2[edit]

Given[edit]

Find[edit]

Find and plot the solution for

Solution[edit]

Due to no excitation, becomes:

Substituting into :

Factoring, and solving for :


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Since is a double root, the general solution:

From intitial conditions:



Solving for and :

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and

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So, the final solution is:



Team14R22diagram2.png

Problem 3[edit]

Given[edit]

(a)

(b)

Find[edit]

General solution to the differential equation

Solution[edit]

(a)

Characteristic equation:

Using the quadratic equation:

where:

Place values into quadratic equation:


(b)

Characteristic equation:

Using the quadratic equation:

where:

Place values into quadratic equation:


Problem 4[edit]

Given[edit]

(5)


(6)

h

Find[edit]

For both (5) and (6), find a general solution. Then, check answers by substitution.

Solution[edit]

(5)

To obtain the general solution for (Eq. 1), we let

This yields:

Therefore, we have:

Where is a solution to the characteristic equation

We can see that

and
.

Next, we must examine the quadratic formula to determine whether the system is over-, under-, or critically-damped. The discriminant is:

Therefore, the equation is critically-damped.

The general solution is therefore represented by:

Where .

We can determine the value of by again using the quadratic equation, this time in full:

for which we find that .

Therefore, the general solution to the equation is:


Verification by substitution:

Plugging into Eq. 1 and combining terms, it is found:

All the terms cancel, so the statement is true, and the solution is verified.


(6)

To obtain the general solution for (Eq. 2), let

So then, we have

as a solution to the characteristic equation. All terms must be divided by 10, to put the equation in standard form:

Examining the discriminant, we find that

Therefore, the equation is critically-damped.

The general solution is represented by:

and .

To determine the value of , we set:

for which we find that .

Therefore, the general solution to the equation is:


Verification by substitution:

Plugging into Eq. 1 and combining terms, it is found:

All the terms cancel, so the statement is true, and the solution is verified.

Problem 5[edit]

Given[edit]



Given the basis:

(a)



(b)



Find[edit]

For the given Information above, Find an ODE for both (a) and (b)

Solution[edit]

(a)The general solution can be written as:



The characteristic equation can be written in the following way:



Giving the ODE:



(b) The general solution can be written as:



The characteristic equation can be written in the following way:



Giving the ODE:


Problem 6[edit]

Given[edit]

alt text
Mass Spring Dashpot system FBD's

Spring-dashpot equation of motion from sec 1-5



Find[edit]

Spring-dashpot-mass system in series. Find the values for the parameters k,c,m with a double real root of

Solution[edit]

Consider the double real root

Characteristic equation is

Recall

Thus

Solve for c

Therefore


Problem 7[edit]

Given[edit]

Taylor Series at t=0

Find[edit]

Develop the MacLaurin Series for

Solution[edit]

Taylor series is defined as

The MacLaurin series occurs when t=0

Development of MacLaurin series for

Final Answer



Explicit form can be written as


Development of MacLaurin series for

Final Answer



Explicit form can be written as


Development of MacLaurin series for

Final Answer



Explicit form can be written as

Problem 8[edit]

Given[edit]

(8)



(15)

Find[edit]

Solution[edit]

(8)

Assume that the solution is of the form



And



Substituting equations 2, 3, and 4 into equation 1a yields



Since



We must use the quadratic formula to obtain values for lambda





Where



In order to verify solution, we must evaluate it's first and second derivative, then plug them into the equation
Let





Plugging into Eq. 1 and collecting terms gives



Therefore the solution holds for any values of




(15)

Assume that the solution is of the form



And



Substituting equations 2, 3, and 4 into equation 1b yields



Since



We must use the quadratic formula to obtain values for lambda




Where



In order to verify solution, we must evaluate it's first and second derivative, then plug them into the equation
Let





Plugging into Eq. 1 and collecting terms gives



Therefore the solution holds for any values of

Problem 9[edit]

Given[edit]

Find[edit]

Solution[edit]

Problem 10[edit]

Given[edit]

Differential equation, initial conditions, and forcing function as shown:



Find[edit]

Find[edit]

The solution to Eq. 1

Solution[edit]

Substituting Equations 2, 3, and 4 into Equation 1 yields

To solve for the constants:

Solving Equations 6-9 yields

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The homogeneous solution:

The total solution:

Setting yields

Solving for yields

Finding yields

Setting yields

Solving for yields

Therefore, the total solution is


Team Member Tasks[edit]

Name Responsibilities Checked by
Bo Turano Problem R2. --
David Parsons Problem R2.4 --
Dean Pickett Problem R2.8 --
Giacomo Savardi Problem R2. --
Isaac Kimiagarov Problem R2.3 --
Kyle Steiner Problem R2. --
Tony Han Problem R2.6, R2.7 --

All team members contributed to the coding of this page.