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Show all of your work. Full credit may not be given for an answer alone. This is a non-calculator exam. You may use the backs of the pages for scratch work. Do not unstaple or remove pages.

By taking this exam you are agreeing to abide by Rutgers University's Academic Integrity Policy.

 Problem Possible Points Number Points Earned 1 15 2 15 3 15 4 20 5 15 6 20 Total 100

1. (15 points)
1. (10 points) Compute , using only the definition of Laplace transform.

Proof. [Solution]

2. (5 points) Compute , using either the definition or by differentiating part (a).

Proof. [Solution]One can do integration by parts to find , but the easiest way is with differentiation.

2. (15 points) Compute .

Hint 1   .

Proof. [Solution]Using the hint, we have

So

According to the table, . As for , write it as , so we can just integrate . Thus

3. (15 points) Use the Laplace transform to solve the ordinary differential equation

Proof. [Solution]The Laplace transform of the equation is

We factor and thus we have a partial fraction decomposition

This solves to give , . Thus

4. (20 points) Let and suppose is a solution to the ordinary differential equation

1. (15 points) Show for some constant .

Proof. [Solution]Remember that transforms to , so the transformed equation is

The integrating factor for this ODE in is

Thus

2. (5 points) Suppose . What is in terms of ?

Proof. [Solution]Notice that

Thus .

5. (15 points) Let on . Calculate the Fourier coefficients of and write down the Fourier series.

Proof. [Solution]This function is even, so all of its sine Fourier coefficients are zero. So we need only integrate between 0 and , and count it twice as much. Moreover, on , . So

Furthermore, if ,

Therefore the Fourier series is

6. (20 points) Let and define on . Calculate the Fourier coefficients of and write down the Fourier series.

Proof. [Solution] is intended to be an arbitrary number between 0 and . I should have excluded the case , but many people ignored altogether and integrated between and . As I said in class, in fact during the review, the Fourier series for on is . You will get and all other .

Assume then. Notice is an even function so all are zero, and we can integrate over half the interval.

Moreover, if ,

Let , . Then

Now let , . We have

Thus

Therefore

Matthew Leingang 2002-10-10