Question 2: The function y(t) satisfies the ordinary differential equation dy dt = f(t, y) for t> 0. (4) The variable t takes discrete values, tη with a constant step-size h = tn+1 − tn, for all n € N. The approximation of y(tn) is denoted yn. (a) Derive an implicit scheme by integrating (4) over the interval [tn, tn+1], using the trapezium quadrature rule. (b) Consider the case f(y) = -λy where > 0 is a constant. (i) Find the exact solution of equation (4), subject to the initial condition y(0) = yo, and identify the behaviour of the solution as t→ +∞. (ii) Write down the difference equation corresponding to the implicit scheme derived in part (a) and show that Yn+1 = 1 - Xh/2 1+ Xh/2 Yn. (iii) Solve the difference equation (i.e. write down yn in terms of yo) and show that this implicit scheme is unconditionally stable. = (c) Verify that y(t) tlnt2t is a solution of equation (4) for f(t, y) initial condition y(1) = 2. = 1+y/t, with Write down the implicit scheme obtained in part (a) in an explicit form (i.e. express Yn+1 in terms of yn). Then, starting at t = 1 with h = 1 and yo = 2, approximate y at t = 2 and calculate the error of the approximation. (d) The global error of this numerical scheme can be analysed for f(y) = -λy. The global error is the difference between the exact solution y(t), obtained in part ((b)i), and the approximate solution, obtained in ((b)iii), when n = t/h. Show, by expanding the global error as a Taylor series in powers of h, that the method is of second order.
Question 2: The function y(t) satisfies the ordinary differential equation dy dt = f(t, y) for t> 0. (4) The variable t takes discrete values, tη with a constant step-size h = tn+1 − tn, for all n € N. The approximation of y(tn) is denoted yn. (a) Derive an implicit scheme by integrating (4) over the interval [tn, tn+1], using the trapezium quadrature rule. (b) Consider the case f(y) = -λy where > 0 is a constant. (i) Find the exact solution of equation (4), subject to the initial condition y(0) = yo, and identify the behaviour of the solution as t→ +∞. (ii) Write down the difference equation corresponding to the implicit scheme derived in part (a) and show that Yn+1 = 1 - Xh/2 1+ Xh/2 Yn. (iii) Solve the difference equation (i.e. write down yn in terms of yo) and show that this implicit scheme is unconditionally stable. = (c) Verify that y(t) tlnt2t is a solution of equation (4) for f(t, y) initial condition y(1) = 2. = 1+y/t, with Write down the implicit scheme obtained in part (a) in an explicit form (i.e. express Yn+1 in terms of yn). Then, starting at t = 1 with h = 1 and yo = 2, approximate y at t = 2 and calculate the error of the approximation. (d) The global error of this numerical scheme can be analysed for f(y) = -λy. The global error is the difference between the exact solution y(t), obtained in part ((b)i), and the approximate solution, obtained in ((b)iii), when n = t/h. Show, by expanding the global error as a Taylor series in powers of h, that the method is of second order.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
![Question 2:
The function y(t) satisfies the ordinary differential equation
dy
dt
= f(t, y) for t> 0.
(4)
The variable t takes discrete values, tη with a constant step-size h = tn+1 − tn, for all n € N.
The approximation of y(tn) is denoted yn.
(a) Derive an implicit scheme by integrating (4) over the interval [tn, tn+1], using the
trapezium quadrature rule.
(b) Consider the case f(y) = -λy where > 0 is a constant.
(i) Find the exact solution of equation (4), subject to the initial condition y(0) = yo,
and identify the behaviour of the solution as t→ +∞.
(ii) Write down the difference equation corresponding to the implicit scheme derived
in part (a) and show that
Yn+1 =
1 - Xh/2
1+ Xh/2
Yn.
(iii) Solve the difference equation (i.e. write down yn in terms of yo) and show that
this implicit scheme is unconditionally stable.
=
(c) Verify that y(t) tlnt2t is a solution of equation (4) for f(t, y)
initial condition y(1) = 2.
=
1+y/t, with
Write down the implicit scheme obtained in part (a) in an explicit form (i.e. express
Yn+1 in terms of yn). Then, starting at t = 1 with h = 1 and yo = 2, approximate y at
t = 2 and calculate the error of the approximation.
(d) The global error of this numerical scheme can be analysed for f(y) = -λy. The global
error is the difference between the exact solution y(t), obtained in part ((b)i), and the
approximate solution, obtained in ((b)iii), when n = t/h. Show, by expanding the
global error as a Taylor series in powers of h, that the method is of second order.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F69cce4ac-4bf6-4e6b-8636-bf160e045b58%2Fa22e60ed-5c28-4f62-9033-1b07e8c7c74a%2Fk12ktd_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Question 2:
The function y(t) satisfies the ordinary differential equation
dy
dt
= f(t, y) for t> 0.
(4)
The variable t takes discrete values, tη with a constant step-size h = tn+1 − tn, for all n € N.
The approximation of y(tn) is denoted yn.
(a) Derive an implicit scheme by integrating (4) over the interval [tn, tn+1], using the
trapezium quadrature rule.
(b) Consider the case f(y) = -λy where > 0 is a constant.
(i) Find the exact solution of equation (4), subject to the initial condition y(0) = yo,
and identify the behaviour of the solution as t→ +∞.
(ii) Write down the difference equation corresponding to the implicit scheme derived
in part (a) and show that
Yn+1 =
1 - Xh/2
1+ Xh/2
Yn.
(iii) Solve the difference equation (i.e. write down yn in terms of yo) and show that
this implicit scheme is unconditionally stable.
=
(c) Verify that y(t) tlnt2t is a solution of equation (4) for f(t, y)
initial condition y(1) = 2.
=
1+y/t, with
Write down the implicit scheme obtained in part (a) in an explicit form (i.e. express
Yn+1 in terms of yn). Then, starting at t = 1 with h = 1 and yo = 2, approximate y at
t = 2 and calculate the error of the approximation.
(d) The global error of this numerical scheme can be analysed for f(y) = -λy. The global
error is the difference between the exact solution y(t), obtained in part ((b)i), and the
approximate solution, obtained in ((b)iii), when n = t/h. Show, by expanding the
global error as a Taylor series in powers of h, that the method is of second order.
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