mth 256 – sec 6.4 sample problems -...
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MTH 256 – Sec 6.4 Sample Problems
Ex. Find the solution of the initial value problem
y ′′ + y = u2π(t), y(0) = 1, y ′(0) = 0,
where u2π(t) is the unit step function (Heaviside function) with jumpdiscontinuity at t = 2π, using the method of Laplace Transforms.
Sol.Tranform both sides of the ODE which results in
s2Y (s)− sy(0)− y ′(0) + Y (s) =e−2πs
s.
Substitute in the initial conditions to get
s2Y (s)− s + Y (s) =e−2πs
s.
Prof. Gibson (Math 256) Section 6.4 Fall 2011 1 / 9
MTH 256 – Sec 6.4 Sample Problems
Ex. Find the solution of the initial value problem
y ′′ + y = u2π(t), y(0) = 1, y ′(0) = 0,
where u2π(t) is the unit step function (Heaviside function) with jumpdiscontinuity at t = 2π, using the method of Laplace Transforms.
Sol.Tranform both sides of the ODE which results in
s2Y (s)− sy(0)− y ′(0) + Y (s) =e−2πs
s.
Substitute in the initial conditions to get
s2Y (s)− s + Y (s) =e−2πs
s.
Prof. Gibson (Math 256) Section 6.4 Fall 2011 1 / 9
MTH 256 – Sec 6.4 Sample Problems
Ex. Find the solution of the initial value problem
y ′′ + y = u2π(t), y(0) = 1, y ′(0) = 0,
where u2π(t) is the unit step function (Heaviside function) with jumpdiscontinuity at t = 2π, using the method of Laplace Transforms.
Sol.Tranform both sides of the ODE which results in
s2Y (s)− sy(0)− y ′(0) + Y (s) =e−2πs
s.
Substitute in the initial conditions to get
s2Y (s)− s + Y (s) =e−2πs
s.
Prof. Gibson (Math 256) Section 6.4 Fall 2011 1 / 9
Substitute in the initial conditions to get
s2Y (s)− s + Y (s) =e−2πs
s.
Solve for Y (s)
Y (s) =s
s2 + 1+ e−2πs 1
s(s2 + 1).
Re-write it in a “useable” form to prepare for an inverse Laplacetransform. In particular,
1
s(s2 + 1)=
A
s+
Bs
s2 + 1=
As2 + A + Bs2
s(s2 + 1)
which is true for A = 1 and B = −1. This allows us to rewrite
Y (s) =s
s2 + 1+ e−2πs
(1
s− s
s2 + 1
). (1)
Prof. Gibson (Math 256) Section 6.4 Fall 2011 2 / 9
Substitute in the initial conditions to get
s2Y (s)− s + Y (s) =e−2πs
s.
Solve for Y (s)
Y (s) =s
s2 + 1+ e−2πs 1
s(s2 + 1).
Re-write it in a “useable” form to prepare for an inverse Laplacetransform. In particular,
1
s(s2 + 1)=
A
s+
Bs
s2 + 1=
As2 + A + Bs2
s(s2 + 1)
which is true for A = 1 and B = −1. This allows us to rewrite
Y (s) =s
s2 + 1+ e−2πs
(1
s− s
s2 + 1
). (1)
Prof. Gibson (Math 256) Section 6.4 Fall 2011 2 / 9
Substitute in the initial conditions to get
s2Y (s)− s + Y (s) =e−2πs
s.
Solve for Y (s)
Y (s) =s
s2 + 1+ e−2πs 1
s(s2 + 1).
Re-write it in a “useable” form to prepare for an inverse Laplacetransform. In particular,
1
s(s2 + 1)=
A
s+
Bs
s2 + 1=
As2 + A + Bs2
s(s2 + 1)
which is true for A = 1 and B = −1. This allows us to rewrite
Y (s) =s
s2 + 1+ e−2πs
(1
s− s
s2 + 1
). (1)
Prof. Gibson (Math 256) Section 6.4 Fall 2011 2 / 9
Invert the Laplace transform. To use the formula
L−1{e−csF (s)} = uc(t)f (t − c),
where in our case c = 2π and
F (s) =1
s− s
s2 + 1,
means that we need the inverse Laplace transform of F (s),
f (t) = 1− cos(t).
Therefore
L−1
{e−2πs
(1
s− s
s2 + 1
)}= u2π(t)(1− cos(t − 2π)).
Finally, adding in the inverse Laplace transform of the first term inEquation (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 3 / 9
Invert the Laplace transform. To use the formula
L−1{e−csF (s)} = uc(t)f (t − c),
where in our case c = 2π and
F (s) =1
s− s
s2 + 1,
means that we need the inverse Laplace transform of F (s),
f (t) = 1− cos(t).
Therefore
L−1
{e−2πs
(1
s− s
s2 + 1
)}= u2π(t)(1− cos(t − 2π)).
Finally, adding in the inverse Laplace transform of the first term inEquation (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 3 / 9
y(t) = cos(t) + u2π(t)(1− cos(t − 2π)).
Note that this function could be written
y(t) =
{cos(t), 0 ≤ t ≤ 2π
1, t > 2π
0 1 2 3 4 5 6 7 8 9 10
−1
−0.5
0
0.5
1
t
cos(t)+(t>2 π) (1−cos(t))
Prof. Gibson (Math 256) Section 6.4 Fall 2011 4 / 9
3. Find the solution of the initial value problem
y ′′ + 4y = sin(t)− u2π(t) sin(t − 2π), y(0) = 0, y ′(0) = 0
using the method of Laplace Transforms.
Sol.Tranform both sides of the ODE which results in
s2Y (s)− sy(0)− y ′(0) + 4Y (s) =1
s2 + 1− e−2πs 1
s2 + 1
Substitute in the initial conditions to get
s2Y (s) + 4Y (s) =1
s2 + 1− e−2πs 1
s2 + 1.
Prof. Gibson (Math 256) Section 6.4 Fall 2011 5 / 9
3. Find the solution of the initial value problem
y ′′ + 4y = sin(t)− u2π(t) sin(t − 2π), y(0) = 0, y ′(0) = 0
using the method of Laplace Transforms.
Sol.Tranform both sides of the ODE which results in
s2Y (s)− sy(0)− y ′(0) + 4Y (s) =1
s2 + 1− e−2πs 1
s2 + 1
Substitute in the initial conditions to get
s2Y (s) + 4Y (s) =1
s2 + 1− e−2πs 1
s2 + 1.
Prof. Gibson (Math 256) Section 6.4 Fall 2011 5 / 9
3. Find the solution of the initial value problem
y ′′ + 4y = sin(t)− u2π(t) sin(t − 2π), y(0) = 0, y ′(0) = 0
using the method of Laplace Transforms.
Sol.Tranform both sides of the ODE which results in
s2Y (s)− sy(0)− y ′(0) + 4Y (s) =1
s2 + 1− e−2πs 1
s2 + 1
Substitute in the initial conditions to get
s2Y (s) + 4Y (s) =1
s2 + 1− e−2πs 1
s2 + 1.
Prof. Gibson (Math 256) Section 6.4 Fall 2011 5 / 9
Substitute in the initial conditions to get
s2Y (s) + 4Y (s) =1
s2 + 1− e−2πs 1
s2 + 1.
Solve for Y (s)
Y (s) =1
(s2 + 1)(s2 + 4)− e−2πs 1
(s2 + 1)(s2 + 4).
Re-write it in a “useable” form to prepare for an inverse Laplacetransform. In particular,
1
(s2 + 1)(s2 + 4)=
A
s2 + 1+
B
s2 + 4=
As2 + 4A + Bs2 + B
(s2 + 1)(s2 + 4)
which is true for A = 1/3 and B = −1/3. This allows us to rewrite
Y (s) =
(1
3
1
s2 + 1− 1
3
1
s2 + 4
)− e−2πs
(1
3
1
s2 + 1− 1
3
1
s2 + 4
)(2)
Prof. Gibson (Math 256) Section 6.4 Fall 2011 6 / 9
Substitute in the initial conditions to get
s2Y (s) + 4Y (s) =1
s2 + 1− e−2πs 1
s2 + 1.
Solve for Y (s)
Y (s) =1
(s2 + 1)(s2 + 4)− e−2πs 1
(s2 + 1)(s2 + 4).
Re-write it in a “useable” form to prepare for an inverse Laplacetransform. In particular,
1
(s2 + 1)(s2 + 4)=
A
s2 + 1+
B
s2 + 4=
As2 + 4A + Bs2 + B
(s2 + 1)(s2 + 4)
which is true for A = 1/3 and B = −1/3. This allows us to rewrite
Y (s) =
(1
3
1
s2 + 1− 1
3
1
s2 + 4
)− e−2πs
(1
3
1
s2 + 1− 1
3
1
s2 + 4
)(2)
Prof. Gibson (Math 256) Section 6.4 Fall 2011 6 / 9
Substitute in the initial conditions to get
s2Y (s) + 4Y (s) =1
s2 + 1− e−2πs 1
s2 + 1.
Solve for Y (s)
Y (s) =1
(s2 + 1)(s2 + 4)− e−2πs 1
(s2 + 1)(s2 + 4).
Re-write it in a “useable” form to prepare for an inverse Laplacetransform. In particular,
1
(s2 + 1)(s2 + 4)=
A
s2 + 1+
B
s2 + 4=
As2 + 4A + Bs2 + B
(s2 + 1)(s2 + 4)
which is true for A = 1/3 and B = −1/3. This allows us to rewrite
Y (s) =
(1
3
1
s2 + 1− 1
3
1
s2 + 4
)− e−2πs
(1
3
1
s2 + 1− 1
3
1
s2 + 4
)(2)
Prof. Gibson (Math 256) Section 6.4 Fall 2011 6 / 9
Invert the Laplace transform.Consider the first term in parenthesis, re-written as follows,
F (s) =1
3
1
s2 + 1− 1
6
2
s2 + 4.
The inverse Laplace transform is
f (t) = L−1{F (s)} =1
3sin(t)− 1
6sin(2t).
For the second term in Equation (2), we wish to use the formula
L−1{e−csF (s)} = uc(t)f (t − c),
where in this case c = 2π and F (s) is as above.Therefore
L−1
{e−2πs
(1
3
1
s2 + 1− 1
3
1
s2 + 4
)}=
u2π(t)
(1
3sin(t − 2π)− 1
6sin(2(t − 2π))
).
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,
F (s) =1
3
1
s2 + 1− 1
6
2
s2 + 4.
The inverse Laplace transform is
f (t) = L−1{F (s)} =1
3sin(t)− 1
6sin(2t).
For the second term in Equation (2), we wish to use the formula
L−1{e−csF (s)} = uc(t)f (t − c),
where in this case c = 2π and F (s) is as above.Therefore
L−1
{e−2πs
(1
3
1
s2 + 1− 1
3
1
s2 + 4
)}=
u2π(t)
(1
3sin(t − 2π)− 1
6sin(2(t − 2π))
).
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,
F (s) =1
3
1
s2 + 1− 1
6
2
s2 + 4.
The inverse Laplace transform is
f (t) = L−1{F (s)} =1
3sin(t)− 1
6sin(2t).
For the second term in Equation (2), we wish to use the formula
L−1{e−csF (s)} = uc(t)f (t − c),
where in this case c = 2π and F (s) is as above.
Therefore
L−1
{e−2πs
(1
3
1
s2 + 1− 1
3
1
s2 + 4
)}=
u2π(t)
(1
3sin(t − 2π)− 1
6sin(2(t − 2π))
).
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,
F (s) =1
3
1
s2 + 1− 1
6
2
s2 + 4.
The inverse Laplace transform is
f (t) = L−1{F (s)} =1
3sin(t)− 1
6sin(2t).
For the second term in Equation (2), we wish to use the formula
L−1{e−csF (s)} = uc(t)f (t − c),
where in this case c = 2π and F (s) is as above.Therefore
L−1
{e−2πs
(1
3
1
s2 + 1− 1
3
1
s2 + 4
)}=
u2π(t)
(1
3sin(t − 2π)− 1
6sin(2(t − 2π))
).
Prof. Gibson (Math 256) Section 6.4 Fall 2011 7 / 9
Finally, subtracting these two inverse Laplace transforms gives thesolution to the IVP to be
y(t) =1
3sin(t)−1
6sin(2t)−u2π(t)
(1
3sin(t − 2π)− 1
6sin(2(t − 2π))
).
This answer is sufficient, however, if we wish to simplify terms tomatch more closely with the answer in the back of the book, note thatsin(t − 2π) = sin(t), thus we really have
y(t) =
(1
3sin(t)− 1
6sin(2t)
)(1− u2π(t)) .
Prof. Gibson (Math 256) Section 6.4 Fall 2011 8 / 9
Finally, subtracting these two inverse Laplace transforms gives thesolution to the IVP to be
y(t) =1
3sin(t)−1
6sin(2t)−u2π(t)
(1
3sin(t − 2π)− 1
6sin(2(t − 2π))
).
This answer is sufficient, however, if we wish to simplify terms tomatch more closely with the answer in the back of the book, note thatsin(t − 2π) = sin(t), thus we really have
y(t) =
(1
3sin(t)− 1
6sin(2t)
)(1− u2π(t)) .
Prof. Gibson (Math 256) Section 6.4 Fall 2011 8 / 9
y(t) =
(1
3sin(t)− 1
6sin(2t)
)(1− u2π(t)) .
0 1 2 3 4 5 6 7 8 9 10
−0.8
−0.6
−0.4
−0.2
0
0.2
0.4
0.6
0.8
t
(sin(t)/3−sin(2 t)/6) (1−(t>2 π))
g(t)y(t)
Prof. Gibson (Math 256) Section 6.4 Fall 2011 9 / 9