Module MA2E02 (Frolov), Multivariable Calculus
Tutorial Sheet 9
Due: at the end of the tutorial session Tuesday/Thursday, 5/7 April 2016
Name and student number:
1.
Consider the following initial value problem
y 00 + 16 y = 16 u(t − π) − 4δ(t − 2π) ,
y(0) = 1 ,
y 0 (0) = 4 .
(a) Plot the function
1 1
16 u(t − π) − 4 u(t − 2π + ) − u(t − 2π − )
2
2
(b) Solve the initial value problem by the Laplace transform.
(c) Plot the solution.
Show the details of your work.
Solution :
(a) The plot of the function is shown below.
15
10
5
2
4
6
8
(b) We denote Y (s) = L(y), and then using the formulae
L(y 00 ) = s2 Y (s) − sy(0) − y 0 (0) , L(u(t − a)) =
e−as
, L(δ(t − a)) = e−as ,
s
we get the algebraic equation
(s2 + 16)Y (s) = 16
e−πs
− 4e−2πs + s + 4 .
s
1
Solving the equation for Y , we get
Y (s) =
4
16e−πs
4e−2πs
s
+
+
−
.
s2 + 16 s2 + 16 s (s2 + 16) s2 + 16
Then we represent
16
1
s
=
−
.
s(s2 + 16)
s s2 + 16
Finally we use the formulae of the inverse Laplace transform
s
ω
1
−1
−1
−1
= 1, L
= cos ωt , L
= sin ωt ,
L
s
s2 + ω 2
s2 + ω 2
L−1 e−as F (s) = f (t − a)u(t − a) ,
(0.1)
to get
y(t) = cos(4t) + sin 4t − u(t − π)(cos(4t) − 1) − u(t − 2π) sin(4t)

 cos 4t + sin 4t if 0 < t < π
1 + sin 4t
if π < t < 2π .
=

1
if t > 2π
(0.2)
(c) The plot of the solution is shown below.
2.0
1.5
1.0
0.5
2
4
6
8
-0.5
-1.0
-1.5
2. Applying convolution,
(a) find the solution.
(b) sketch the input function and the solution.
Show the details of your work.
6 if 0 < t < 1
00
0
y + 4y + 3y =
,
0 if t > 1
y(0) = 0 ,
y 0 (0) = 2 .
Solution : We denote Y (s) = L(y) and represent the function on the right-hand side of the
equation as
r(t) = 6 u(t) − u(t − 1) .
2
By using the formulas
L(y 0 ) = sY (s)−y(0) , L(y 00 ) = s2 Y (s)−sy(0)−y 0 (0) , L(u(t−a)) =
e−as
6
e−s
⇒ L(r) = −6
,
s
s
s
we get the algebraic equation
6
e−s
−6
s
s
(s2 + 4s + 3)Y (s) − 2 =
⇒
Y (s) = 2Q(s) + R(s)Q(s) ,
where
R(s) =
6
e−s
−6
,
s
s
Q(s) =
s2
1
1
1
1
=
=
−
.
+ 4s + 3
(s + 1)(s + 3)
2(s + 1) 2(s + 3)
By using the convolution theorem we get the integral representation
Z t
y(t) = 2q(t) +
q(t − τ )r(τ ) dτ ,
0
where
1
1
q(t) = L−1 (Q(s)) = e−t − e−2t ,
2
2
r(t) = L−1 (R(s)) = 6 u(t) − u(t − 1) .
Computing the integral, we get
Z t
Z t
Z t
q(t−τ )r(τ ) dτ = 6
q(t−τ ) dτ = 3
(e−t+τ −e−3t+3τ ) dτ = 2−3e−t +e−3t
0
Z
0
t
1
Z
Z
q(t−τ ) dτ = 3
q(t−τ )r(τ ) dτ = 6
0
if 0 < t < 1 ,
0
1
(e−t+τ −e−3t+3τ ) dτ = −e3−3t +3e1−t +e−3t −3e−t if t > 1 ,
0
0
Adding 2q(t), we get
 −t
if t < 0
 e − e−3t
−t
2 − 2e
if 0 < t < 1 ,
y(t) =

−t
3−3t
(3e − 2)e − e
if t > 1
The plot of the input function and the solution is shown below.
6
4
2
0.5
-0.5
1.0
1.5
-2
3
2.0
2.5
3.0
3. Using Laplace transforms, solve the integral equation. (Show the details of your work.)
Z t
y(τ ) sin(t − τ ) dτ .
y(t) = cos t +
0
Solution : We denote Y (s) = L(y) and represent the equation in the following form
y = cos t + y ∗ sin(t) .
By the convolution theorem,
Y (s) =
s2
s
1
+ Y (s) 2
.
+1
s +1
Solving for Y (s), we obtain
Y (s) =
1
,
s
and therefore
y(t) = 1 .
4