Math 4564 Practice Test 1 solution
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Math 4564
Practice Test 1 solution
Exercise 1: (25 pts) Use Laplace transform to solve the following IVP
y′ (t) + y(t) = f(t), y(0) = 0,
where
f(t) =
Solution: First, we need to find F(s). There are at least two ways to do that:
1. Notice that f(t) = (1 − U (t − 1)) − U (t − 1) = 1 − 2U (t − 1). Therefore, F(s) = .
2. The definition of LT states that
F(s) = 0 ∞ f(t)e−st dt = 0 1 e −st dt − 1 ∞ e −st dt = .
Now, we apply Laplace transform to the IVP to get
(s + 1)Y (s) = =⇒ Y (s) = − 2 e −s
We apply a simple PFD to get
L−1 = L−1 − = 1 − e −t .
The ILT of the second term can be obtained via the second translation theorem
L−1 e−s =1 − e− (t− 1) U (t − 1)
We combine both results to get
y(t) = 1 − e−t − 2 1 − e− (t− 1) U (t − 1) |
Exercise 2: (25 pts) Find the Laplace transform of this periodic function
Solution: The period of f is 4. On [0, 4], we have
f(t) =
Therefore
0 ≤ t ≤ 2
2 ≤ t ≤ 4.
F(s) = Z0 2 (1 − t)e−st dt + Z2 4 (t − 3)e−st dt .
Integration by parts gives
F(s) = − + − − +
Exercise 3: (25 pts) Use Laplace transform to solve the following differential equation y′′ (t) + 2y′ (t) + y(t) = δ(t − 3), y(0) = y′ (0) = 0.
Solution: Apply LT to both sides of the equation to get
(s + 1)2 Y (s) = e −3s .
Which leads to
Y (s) = e−3s
Using the the first translation theorem (or the table) L−1 = te−t . Using the second
translation theorem we get
y(t) = L−1 e−3s = (t − 3)e− (t−3) U (t − 3). |
Exercise 4: (25 pts) Find
1. L−1 .
2. L r sin(r) dr . 0
3. L{sin(t)cos(t)}.
Solution:
1.
L−1 = L−1 = L−1 {1} − L−1 = δ(t) − sin(t).
2. Use theorem 7.4.3 and then theorem 7.4.1 to get
L Z0 t r sin(r) dr = L{tsin(t)} = L{sin(t)} = =
Note: There is another (lengthier) way to find this LT by evaluating the integral
Z0 t r sin(r) dr = − r cos(r) 0(t) + Z0 t sin(t) dr = −tcos(t) + sin(t)
and then applying 7.4.1 to compute the LT.
3. We have
L{sin(t)cos(t)} = 1 L{sin(2t)} = 1 · 2 = 1
Note: You CANNOT apply the convolution theorem here.
2022-06-02