1. For the system dt dy =1, the curve Y(1) = (cost, sint) is a solution. This solution is periodic. Its initial position is Y(0) = (1,0), and it returns to this position when t = 2r. So Y(2r) = (1,0) and Y( + 2) = Y(t) for all t. (a) Check that Y(1) = (cost, sin r) is a solution. (b) Use Euler's method with step size 0.5 to approximate this solution, and check how close the approximate solution is to the real solution when t = 4, t = 6, and t = 10.
1. For the system dt dy =1, the curve Y(1) = (cost, sint) is a solution. This solution is periodic. Its initial position is Y(0) = (1,0), and it returns to this position when t = 2r. So Y(2r) = (1,0) and Y( + 2) = Y(t) for all t. (a) Check that Y(1) = (cost, sin r) is a solution. (b) Use Euler's method with step size 0.5 to approximate this solution, and check how close the approximate solution is to the real solution when t = 4, t = 6, and t = 10.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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Section 2.5
Problem 1
Only (a.) and (b.) please
![**EXERCISES FOR SECTION 2.5**
1. For the system
\[
\frac{dx}{dt} = -y \\
\frac{dy}{dt} = x,
\]
the curve \( Y(t) = (\cos t, \sin t) \) is a solution. This solution is periodic. Its initial position is \( Y(0) = (1, 0) \), and it returns to this position when \( t = 2\pi \). So \( Y(2\pi) = (1, 0) \) and \( Y(t + 2\pi) = Y(t) \) for all \( t \).
(a) Check that \( Y(t) = (\cos t, \sin t) \) is a solution.
(b) Use Euler’s method with step size 0.5 to approximate this solution, and check how close the approximate solution is to the real solution when \( t = 4, t = 6, \) and \( t = 10 \).
(c) Use Euler’s method with step size 0.1 to approximate this solution, and check how close the approximate solution is to the real solution when \( t = 4, t = 6, \) and \( t = 10 \).
(d) The points on the solution curve \( Y(t) \) are all 1 unit distance from the origin. Is this true for the approximate solutions? Are they too far from the origin or too close to it? What will happen for other step sizes (that is, will approximate solutions formed with other step sizes be too far or too close to the origin)?
[Use a computer or calculator to perform Euler’s method.]
2. For the system
\[
\frac{dx}{dt} = 2x \\
\frac{dy}{dt} = y,
\]
we claim that the curve \( Y(t) = (e^{2t}, 3e^t) \) is a solution. Its initial position is \( Y(0) = (1, 3) \).
(a) Check that \( Y(t) = (e^{2t}, 3e^t) \) is a solution.
(b) Use Euler’s method with step size \(\Delta t = 0.5\) to approximate this solution, and check how close the approximate solution is to the real solution when](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8091f38e-d53b-4700-9beb-62676d968b37%2Fd47a83e6-174d-40ac-a7b9-70d9863c300a%2Fyntbo2h_processed.png&w=3840&q=75)
Transcribed Image Text:**EXERCISES FOR SECTION 2.5**
1. For the system
\[
\frac{dx}{dt} = -y \\
\frac{dy}{dt} = x,
\]
the curve \( Y(t) = (\cos t, \sin t) \) is a solution. This solution is periodic. Its initial position is \( Y(0) = (1, 0) \), and it returns to this position when \( t = 2\pi \). So \( Y(2\pi) = (1, 0) \) and \( Y(t + 2\pi) = Y(t) \) for all \( t \).
(a) Check that \( Y(t) = (\cos t, \sin t) \) is a solution.
(b) Use Euler’s method with step size 0.5 to approximate this solution, and check how close the approximate solution is to the real solution when \( t = 4, t = 6, \) and \( t = 10 \).
(c) Use Euler’s method with step size 0.1 to approximate this solution, and check how close the approximate solution is to the real solution when \( t = 4, t = 6, \) and \( t = 10 \).
(d) The points on the solution curve \( Y(t) \) are all 1 unit distance from the origin. Is this true for the approximate solutions? Are they too far from the origin or too close to it? What will happen for other step sizes (that is, will approximate solutions formed with other step sizes be too far or too close to the origin)?
[Use a computer or calculator to perform Euler’s method.]
2. For the system
\[
\frac{dx}{dt} = 2x \\
\frac{dy}{dt} = y,
\]
we claim that the curve \( Y(t) = (e^{2t}, 3e^t) \) is a solution. Its initial position is \( Y(0) = (1, 3) \).
(a) Check that \( Y(t) = (e^{2t}, 3e^t) \) is a solution.
(b) Use Euler’s method with step size \(\Delta t = 0.5\) to approximate this solution, and check how close the approximate solution is to the real solution when
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