MTH 256 – Sec 6.4 Sample Problems
Ex. Find the solution of the initial value problem
y 00 + y = u2π (t),
y (0) = 1,
y 0 (0) = 0,
where u2π (t) is the unit step function (Heaviside function) with jump
discontinuity at t = 2π, using the method of Laplace Transforms.
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Section 6.4
Fall 2011
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MTH 256 – Sec 6.4 Sample Problems
Ex. Find the solution of the initial value problem
y 00 + y = u2π (t),
y (0) = 1,
y 0 (0) = 0,
where u2π (t) is the unit step function (Heaviside function) with jump
discontinuity at t = 2π, using the method of Laplace Transforms.
Sol.
Tranform both sides of the ODE which results in
s 2 Y (s) − sy (0) − y 0 (0) + Y (s) =
Prof. Gibson (Math 256)
Section 6.4
e−2πs
.
s
Fall 2011
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MTH 256 – Sec 6.4 Sample Problems
Ex. Find the solution of the initial value problem
y 00 + y = u2π (t),
y (0) = 1,
y 0 (0) = 0,
where u2π (t) is the unit step function (Heaviside function) with jump
discontinuity at t = 2π, using the method of Laplace Transforms.
Sol.
Tranform both sides of the ODE which results in
s 2 Y (s) − sy (0) − y 0 (0) + Y (s) =
e−2πs
.
s
Substitute in the initial conditions to get
s 2 Y (s) − s + Y (s) =
Prof. Gibson (Math 256)
Section 6.4
e−2πs
.
s
Fall 2011
1/9
Substitute in the initial conditions to get
s 2 Y (s) − s + Y (s) =
e−2πs
.
s
Solve for Y (s)
Y (s) =
Prof. Gibson (Math 256)
s2
s
1
+ e−2πs
.
2
+1
s(s + 1)
Section 6.4
Fall 2011
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Substitute in the initial conditions to get
s 2 Y (s) − s + Y (s) =
e−2πs
.
s
Solve for Y (s)
Y (s) =
s2
s
1
+ e−2πs
.
2
+1
s(s + 1)
Re-write it in a “useable” form to prepare for an inverse Laplace
transform. In particular,
A
Bs
As 2 + A + Bs 2
1
=
+
=
s(s 2 + 1)
s
s2 + 1
s(s 2 + 1)
Prof. Gibson (Math 256)
Section 6.4
Fall 2011
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Substitute in the initial conditions to get
s 2 Y (s) − s + Y (s) =
e−2πs
.
s
Solve for Y (s)
Y (s) =
s2
s
1
+ e−2πs
.
2
+1
s(s + 1)
Re-write it in a “useable” form to prepare for an inverse Laplace
transform. In particular,
A
Bs
As 2 + A + Bs 2
1
=
+
=
s(s 2 + 1)
s
s2 + 1
s(s 2 + 1)
which is true for A = 1 and B = −1. This allows us to rewrite
s
1
s
−2πs
Y (s) = 2
+e
− 2
.
s +1
s
s +1
Prof. Gibson (Math 256)
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(1)
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Invert the Laplace transform. To use the formula
L−1 {e−cs F (s)} = uc (t)f (t − c),
where in our case c = 2π and
F (s) =
1
s
− 2
,
s
s +1
means that we need the inverse Laplace transform of F (s),
f (t) = 1 − cos(t).
Prof. Gibson (Math 256)
Section 6.4
Fall 2011
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Invert the Laplace transform. To use the formula
L−1 {e−cs F (s)} = uc (t)f (t − c),
where in our case c = 2π and
F (s) =
1
s
− 2
,
s
s +1
means that we need the inverse Laplace transform of F (s),
f (t) = 1 − cos(t).
Therefore
−1
L
e
−2πs
1
s
− 2
s
s +1
= u2π (t)(1 − cos(t − 2π)).
Finally, adding in the inverse Laplace transform of the first term in
Equation (1) gives the solution to the IVP to be
y (t) = cos(t) + u2π (t)(1 − cos(t − 2π)).
Prof. Gibson (Math 256)
Section 6.4
Fall 2011
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y (t) = cos(t) + u2π (t)(1 − cos(t − 2π)).
Note that this function could be written
(
cos(t), 0 ≤ t ≤ 2π
y (t) =
1,
t > 2π
cos(t)+(t>2 π) (1−cos(t))
1
0.5
0
−0.5
−1
0
Prof. Gibson (Math 256)
1
2
3
4
5
t
Section 6.4
6
7
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3. Find the solution of the initial value problem
y 00 + 4y = sin(t) − u2π (t) sin(t − 2π),
y (0) = 0,
y 0 (0) = 0
using the method of Laplace Transforms.
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Section 6.4
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3. Find the solution of the initial value problem
y 00 + 4y = sin(t) − u2π (t) sin(t − 2π),
y (0) = 0,
y 0 (0) = 0
using the method of Laplace Transforms.
Sol.
Tranform both sides of the ODE which results in
s 2 Y (s) − sy (0) − y 0 (0) + 4Y (s) =
Prof. Gibson (Math 256)
Section 6.4
1
1
− e−2πs 2
s2 + 1
s +1
Fall 2011
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3. Find the solution of the initial value problem
y 00 + 4y = sin(t) − u2π (t) sin(t − 2π),
y (0) = 0,
y 0 (0) = 0
using the method of Laplace Transforms.
Sol.
Tranform both sides of the ODE which results in
s 2 Y (s) − sy (0) − y 0 (0) + 4Y (s) =
1
1
− e−2πs 2
s2 + 1
s +1
Substitute in the initial conditions to get
s 2 Y (s) + 4Y (s) =
Prof. Gibson (Math 256)
Section 6.4
1
1
− e−2πs 2
.
s2 + 1
s +1
Fall 2011
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Substitute in the initial conditions to get
s 2 Y (s) + 4Y (s) =
1
1
− e−2πs 2
.
s2 + 1
s +1
Solve for Y (s)
Y (s) =
Prof. Gibson (Math 256)
1
1
− e−2πs 2
.
(s 2 + 1)(s 2 + 4)
(s + 1)(s 2 + 4)
Section 6.4
Fall 2011
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Substitute in the initial conditions to get
s 2 Y (s) + 4Y (s) =
1
1
− e−2πs 2
.
s2 + 1
s +1
Solve for Y (s)
Y (s) =
1
1
− e−2πs 2
.
(s 2 + 1)(s 2 + 4)
(s + 1)(s 2 + 4)
Re-write it in a “useable” form to prepare for an inverse Laplace
transform. In particular,
(s 2
1
A
B
As 2 + 4A + Bs 2 + B
= 2
+ 2
=
2
+ 1)(s + 4)
s +1 s +4
(s 2 + 1)(s 2 + 4)
Prof. Gibson (Math 256)
Section 6.4
Fall 2011
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Substitute in the initial conditions to get
s 2 Y (s) + 4Y (s) =
1
1
− e−2πs 2
.
s2 + 1
s +1
Solve for Y (s)
Y (s) =
1
1
− e−2πs 2
.
(s 2 + 1)(s 2 + 4)
(s + 1)(s 2 + 4)
Re-write it in a “useable” form to prepare for an inverse Laplace
transform. In particular,
(s 2
1
A
B
As 2 + 4A + Bs 2 + B
= 2
+ 2
=
2
+ 1)(s + 4)
s +1 s +4
(s 2 + 1)(s 2 + 4)
which is true for A = 1/3 and B = −1/3. This allows us to rewrite
1 1
1 1
1 1
1 1
−2πs
Y (s) =
−
−e
−
(2)
3 s2 + 1 3 s2 + 4
3 s2 + 1 3 s2 + 4
Prof. Gibson (Math 256)
Section 6.4
Fall 2011
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Invert the Laplace transform.
Consider the first term in parenthesis, re-written as follows,
1
1
2
1
−
.
F (s) =
3 s2 + 1
6 s2 + 4
Prof. Gibson (Math 256)
Section 6.4
Fall 2011
7/9
Invert the Laplace transform.
Consider the first term in parenthesis, re-written as follows,
1
1
2
1
−
.
F (s) =
3 s2 + 1
6 s2 + 4
The inverse Laplace transform is
f (t) = L−1 {F (s)} =
Prof. Gibson (Math 256)
Section 6.4
1
1
sin(t) − sin(2t).
3
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Fall 2011
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Invert the Laplace transform.
Consider the first term in parenthesis, re-written as follows,
1
1
2
1
−
.
F (s) =
3 s2 + 1
6 s2 + 4
The inverse Laplace transform is
f (t) = L−1 {F (s)} =
1
1
sin(t) − sin(2t).
3
6
For the second term in Equation (2), we wish to use the formula
L−1 {e−cs F (s)} = uc (t)f (t − c),
where in this case c = 2π and F (s) is as above.
Prof. Gibson (Math 256)
Section 6.4
Fall 2011
7/9
Invert the Laplace transform.
Consider the first term in parenthesis, re-written as follows,
1
1
2
1
−
.
F (s) =
3 s2 + 1
6 s2 + 4
The inverse Laplace transform is
f (t) = L−1 {F (s)} =
1
1
sin(t) − sin(2t).
3
6
For the second term in Equation (2), we wish to use the formula
L−1 {e−cs F (s)} = uc (t)f (t − c),
where in this case c = 2π and F (s) is as above.
Therefore
−1
L
1
1
1
1
−2πs
e
−
=
3 s2 + 1 3 s2 + 4
1
1
u2π (t)
sin(t − 2π) − sin(2(t − 2π)) .
3
6
Prof. Gibson (Math 256)
Section 6.4
Fall 2011
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Finally, subtracting these two inverse Laplace transforms gives the
solution to the IVP to be
1
1
1
1
y (t) = sin(t)− sin(2t)−u2π (t)
sin(t − 2π) − sin(2(t − 2π)) .
3
6
3
6
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Section 6.4
Fall 2011
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Finally, subtracting these two inverse Laplace transforms gives the
solution to the IVP to be
1
1
1
1
y (t) = sin(t)− sin(2t)−u2π (t)
sin(t − 2π) − sin(2(t − 2π)) .
3
6
3
6
This answer is sufficient, however, if we wish to simplify terms to
match more closely with the answer in the back of the book, note that
sin(t − 2π) = sin(t), thus we really have
y (t) =
Prof. Gibson (Math 256)
1
1
sin(t) − sin(2t) (1 − u2π (t)) .
3
6
Section 6.4
Fall 2011
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y (t) =
1
1
sin(t) − sin(2t) (1 − u2π (t)) .
3
6
(sin(t)/3−sin(2 t)/6) (1−(t>2 π))
g(t)
y(t)
0.8
0.6
0.4
0.2
0
−0.2
−0.4
−0.6
−0.8
0
Prof. Gibson (Math 256)
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Section 6.4
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