Problem 1
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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:
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Plot the solution.
Solution
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Characteristic equation:
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Substituting into :
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Non-homogeneous solution:
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Homogeneous solution:
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Overall solution:
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No excitation:
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From intitial conditions:
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Solving for and :
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and
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So, the final solution is:
Problem 2
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Find and plot the solution for
Solution
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Due to no excitation, becomes:
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Substituting into :
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Factoring, and solving for :
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Since is a double root, the general solution:
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From intitial conditions:
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Solving for and :
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and
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So, the final solution is:
Problem 3
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(a)
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(b)
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General solution to the differential equation
Solution
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(a)
Characteristic equation:
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Using the quadratic equation:
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where:
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Place values into quadratic equation:
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(b)
Characteristic equation:
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Using the quadratic equation:
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where:
Place values into quadratic equation:
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Problem 4
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(5)
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(6)
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h
For both (5) and (6), find a general solution. Then, check answers by substitution.
Solution
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(5)
To obtain the general solution for (Eq. 1), we let
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This yields:
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Therefore, we have:
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Where is a solution to the characteristic equation
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We can see that
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Next, we must examine the quadratic formula to determine whether the system is over-, under-, or critically-damped.
The discriminant is:
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Therefore, the equation is critically-damped.
The general solution is therefore represented by:
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Where .
We can determine the value of by again using the quadratic equation, this time in full:
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for which we find that .
Therefore, the general solution to the equation is:
Verification by substitution:
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Plugging into Eq. 1 and combining terms, it is found:
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All the terms cancel, so the statement is true, and the solution is verified.
(6)
To obtain the general solution for (Eq. 2), let
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So then, we have
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as a solution to the characteristic equation. All terms must be divided by 10, to put the equation in standard form:
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Examining the discriminant, we find that
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Therefore, the equation is critically-damped.
The general solution is represented by:
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and .
To determine the value of , we set:
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for which we find that .
Therefore, the general solution to the equation is:
Verification by substitution:
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Plugging into Eq. 1 and combining terms, it is found:
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All the terms cancel, so the statement is true, and the solution is verified.
Problem 5
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Given the basis:
(a)
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(b)
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For the given Information above, Find an ODE for both (a) and (b)
Solution
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(a)The general solution can be written as:
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The characteristic equation can be written in the following way:
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Giving the ODE:
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(b) The general solution can be written as:
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The characteristic equation can be written in the following way:
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Giving the ODE:
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Problem 6
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Spring-dashpot equation of motion from sec 1-5
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Spring-dashpot-mass system in series. Find the values for the parameters k,c,m with a double real root of
Solution
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Consider the double real root
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Characteristic equation is
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Recall
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Thus
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Solve for c
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Therefore
Problem 7
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Taylor Series at t=0
Develop the MacLaurin Series for
Solution
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Taylor series is defined as
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The MacLaurin series occurs when t=0
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Development of MacLaurin series for
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Final Answer
Explicit form can be written as
Development of MacLaurin series for
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Final Answer
Explicit form can be written as
Development of MacLaurin series for
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Final Answer
Explicit form can be written as
Problem 8
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(8)
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(15)
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Solution
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(8)
Assume that the solution is of the form
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And
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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
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And
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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
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Problem 10
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Differential equation, initial conditions, and forcing function as shown:
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The solution to Eq. 1
Solution
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Substituting Equations 2, 3, and 4 into Equation 1 yields
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To solve for the constants:
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Solving Equations 6-9 yields
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The homogeneous solution:
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The total solution:
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Setting yields
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Solving for yields
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Finding yields
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Setting yields
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Solving for yields
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Therefore, the total solution is
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Team Member Tasks
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All team members contributed to the coding of this page.