In each of Problems 9 and 10, use Euler's method to find approximate values of the solution of the given initial value problem at t = 0.5, 1, 1.5, 2, 2.5, and 3: (a) With h = 0.1, (b) With h 0.05, (c) With h = 0.025, (d) With h = 0.01. N 9. y'=5-3√√, y(0) = 2

College Algebra
10th Edition
ISBN:9781337282291
Author:Ron Larson
Publisher:Ron Larson
Chapter8: Sequences, Series,and Probability
Section: Chapter Questions
Problem 16T
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9 (b)

N c. Repeat part a with h = 0.025. Compare the results with
those found in a and b.
N d. Find the solution y = (1) of the given problem and
evaluate (1) at t = 0.1, 0.2, 0.3, and 0.4. Compare these values
with the results of a, b, and c.
y' = 3+1-y, y(0) = 1
y' = 2y = 1, y(0) = 1
1.
2.
3. y' = 0.5-1 + 2y,
y(0) = 1
4. y' = 3 cost-2y, y(0) = 0
In each of Problems 5 through 8, draw a direction field for the given
differential equation and state whether you think that the solutions are
converging or diverging.
5. y'=5-3√//y
6. y' = y(3-ty)
G
7. y'= -ty + 0.1y³
G 8. y' = 1² + y²
In each of Problems 9 and 10, use Euler's method to find approximate
values of the solution of the given initial value problem at t = 0.5,
0.05, (c) With
1, 1.5, 2, 2.5, and 3: (a) With h = 0.1, (b) With h =
h = 0.025, (d) With h = 0.01.
N
y(0) = 2
9. y'=5-3√√y,
N 10. y'= y(3-ty), y(0) = 0.5
11. Consider the initial value problem
31²
3y² - 4'
y'
=
y(1) = 0.
Use Euler's method with h = 0.1, 0.05, 0.025, and 0.01 to explore the
solution of this problem for 0 ≤ t ≤ 1. What is your best estimate
of the value of the solution at t = 0.8? At t = 1? Are your results
consistent with the direction field in Problem 8?
13. Consider the initial value problem -
y' = -ty+0.1y³, y(0) = a,
where a is a given number.
2.8 The Existence and Uniqueness Theorem 83
Ga. Draw a direction field for the differential equation (or
reexamine the one from Problem 7). Observe that there is a
critical value of a in the interval 2 ≤ a ≤ 3 that separates
converging solutions from diverging ones. Call this critical
value ao.
Nb. Use Euler's method with h = 0.01 to estimate ao. Do this
by restricting ao to an interval [a, b], where b - a = 0.01.
14. Consider the initial value problem
y' = y²-1², y(0) = a,
Na. Use Euler's method with h = 0.1 to obtain approximate
0.1 to obtain approximate ouninos Si
values of the solution at t = 1.2, 1.4, 1.6, and 1.8.
N b. Repeat part a with h = 0.05.
c.
that
c. Compare the results of parts a and b. Note that they are
reasonably close for t = 1.2, 1.4, and 1.6 but are quite different
for t = 1.8. Also note (from the differential equation) that
the line tangent to the solution is parallel to the y-axis when
y = ±2/√√3 ±1.155. Explain how this might cause such
a difference in the calculated values.
1
Hint: lim (1+a/n)" = eª.
N 12. Consider the initial value problem
y' = 1² + y², y(0) = 1.
81x
ibnoo lattint on In each of Problems 16 and 17, use the technique discussed in Problem
15 to show that the approximation obtained by the Euler method
converges to the exact solution at any fixed point as h→0.
16. y' = y, y(0) = 1
where a is a given number.
Ga. Draw a direction field for the differential equation. Note
that there is a critical value of a in the interval 0≤ a ≤ 1
that separates converging solutions from diverging ones. Call this
critical value ao.
Nb. Use Euler's method with h = 0.01 to estimate ao. Do this
by restricting ao to an interval [a, b], where b - a = 0.01.
15. Convergence of Euler's Method. It can be shown that
under suitable conditions on f, the numerical approximation
generated by the Euler method for the initial value problem
y' = f(t, y), y(to) = yo converges to the exact solution as the step
size h decreases. This is illustrated by the following example. Consider
the initial value problem
y' = 1-t+y, y(to) = Yo.
a. Show that the exact solution is y = o(t) = (yo-to) e'
Nb. Using the Euler formula, show that
Yk = (1 + h) yk−1+h-htk-1, k = 1,2, ... .
Noting that y₁ = (1 + h) (yo - to) + t₁, show by induction
found
Yn = (1+h)" (yo - to) + tn
for each positive integer n.
d. Consider a fixed point t > to and for a given n choose
h = (t - to)/n. Then th = t for every n. Note also that h→0
as n → ∞. By substituting for h in equation (19) and letting
n→ ∞, show that yn → (t) as n → ∞.
-
17. y' = 2y = 1, y(0) = 1
(19)
THUYẾT TẬT
Hint: y₁ = (1+2h)/2+1/2
Transcribed Image Text:N c. Repeat part a with h = 0.025. Compare the results with those found in a and b. N d. Find the solution y = (1) of the given problem and evaluate (1) at t = 0.1, 0.2, 0.3, and 0.4. Compare these values with the results of a, b, and c. y' = 3+1-y, y(0) = 1 y' = 2y = 1, y(0) = 1 1. 2. 3. y' = 0.5-1 + 2y, y(0) = 1 4. y' = 3 cost-2y, y(0) = 0 In each of Problems 5 through 8, draw a direction field for the given differential equation and state whether you think that the solutions are converging or diverging. 5. y'=5-3√//y 6. y' = y(3-ty) G 7. y'= -ty + 0.1y³ G 8. y' = 1² + y² In each of Problems 9 and 10, use Euler's method to find approximate values of the solution of the given initial value problem at t = 0.5, 0.05, (c) With 1, 1.5, 2, 2.5, and 3: (a) With h = 0.1, (b) With h = h = 0.025, (d) With h = 0.01. N y(0) = 2 9. y'=5-3√√y, N 10. y'= y(3-ty), y(0) = 0.5 11. Consider the initial value problem 31² 3y² - 4' y' = y(1) = 0. Use Euler's method with h = 0.1, 0.05, 0.025, and 0.01 to explore the solution of this problem for 0 ≤ t ≤ 1. What is your best estimate of the value of the solution at t = 0.8? At t = 1? Are your results consistent with the direction field in Problem 8? 13. Consider the initial value problem - y' = -ty+0.1y³, y(0) = a, where a is a given number. 2.8 The Existence and Uniqueness Theorem 83 Ga. Draw a direction field for the differential equation (or reexamine the one from Problem 7). Observe that there is a critical value of a in the interval 2 ≤ a ≤ 3 that separates converging solutions from diverging ones. Call this critical value ao. Nb. Use Euler's method with h = 0.01 to estimate ao. Do this by restricting ao to an interval [a, b], where b - a = 0.01. 14. Consider the initial value problem y' = y²-1², y(0) = a, Na. Use Euler's method with h = 0.1 to obtain approximate 0.1 to obtain approximate ouninos Si values of the solution at t = 1.2, 1.4, 1.6, and 1.8. N b. Repeat part a with h = 0.05. c. that c. Compare the results of parts a and b. Note that they are reasonably close for t = 1.2, 1.4, and 1.6 but are quite different for t = 1.8. Also note (from the differential equation) that the line tangent to the solution is parallel to the y-axis when y = ±2/√√3 ±1.155. Explain how this might cause such a difference in the calculated values. 1 Hint: lim (1+a/n)" = eª. N 12. Consider the initial value problem y' = 1² + y², y(0) = 1. 81x ibnoo lattint on In each of Problems 16 and 17, use the technique discussed in Problem 15 to show that the approximation obtained by the Euler method converges to the exact solution at any fixed point as h→0. 16. y' = y, y(0) = 1 where a is a given number. Ga. Draw a direction field for the differential equation. Note that there is a critical value of a in the interval 0≤ a ≤ 1 that separates converging solutions from diverging ones. Call this critical value ao. Nb. Use Euler's method with h = 0.01 to estimate ao. Do this by restricting ao to an interval [a, b], where b - a = 0.01. 15. Convergence of Euler's Method. It can be shown that under suitable conditions on f, the numerical approximation generated by the Euler method for the initial value problem y' = f(t, y), y(to) = yo converges to the exact solution as the step size h decreases. This is illustrated by the following example. Consider the initial value problem y' = 1-t+y, y(to) = Yo. a. Show that the exact solution is y = o(t) = (yo-to) e' Nb. Using the Euler formula, show that Yk = (1 + h) yk−1+h-htk-1, k = 1,2, ... . Noting that y₁ = (1 + h) (yo - to) + t₁, show by induction found Yn = (1+h)" (yo - to) + tn for each positive integer n. d. Consider a fixed point t > to and for a given n choose h = (t - to)/n. Then th = t for every n. Note also that h→0 as n → ∞. By substituting for h in equation (19) and letting n→ ∞, show that yn → (t) as n → ∞. - 17. y' = 2y = 1, y(0) = 1 (19) THUYẾT TẬT Hint: y₁ = (1+2h)/2+1/2
S
9. a. 2.30800, 2.49006, 2.60023, 2.66773, 2.70939, 2.73521
b. 2.30167, 2.48263, 2.59352, 2.66227, 2.70519, 2.73209
c. 2.29864, 2.47903, 2.59024, 2.65958, 2.70310, 2.73053
d. 2.29686, 2.47691, 2.58830, 2.65798, 2.70185, 2.72959
10. a. 1.70308, 3.06605, 2.44030, 1.77204, 1.37348, 1.11925
b. 1.79548, 3.06051, 2.43292, 1.77807, 1.37795, 1.12191
c. 1.84579, 3.05769, 2.42905, 1.78074, 1.38017, 1.12328
d. 1.87734, 3.05607, 2.42672, 1.78224, 1.38150, 1.12411
11. a. -0.166134, -0.410872, -0.804660, 4.15867
b. -0.174652, -0.434238, -0.889140, -3.09810
12. A reasonable estimate for y at t = 0.8 is between 5.5 and 6. No
reliable estimate is possible at t = 1 from the specified data.
13. b. 2.37 < a < 2.38
Final answers
14. b. 0.67 < αo < 0.68
Section 2.8, page 90
1. dw/ds = (s + 1)² + (w+2)², w(0) = 0
2. dw/ds = 1 -(w+3)³,
w(0) = 0
3. a. ,(1) = ²/4/1
2k tk
k!
n
k=1
c. lim (t) = e²t - 1
818
4. a. (t) =
c. lim
818
n
(-1)*+1+1/(k+1)!2k-1
k=1
(t) = 4e-¹/2+2t-4
t2k-1
1.3.5... (2k-1)
t3k-1
n
5. a. Φ„(t) = Σ
k=1
n
6. a. Φη(t) = - Σ
₁(t)
2.
101
Transcribed Image Text:S 9. a. 2.30800, 2.49006, 2.60023, 2.66773, 2.70939, 2.73521 b. 2.30167, 2.48263, 2.59352, 2.66227, 2.70519, 2.73209 c. 2.29864, 2.47903, 2.59024, 2.65958, 2.70310, 2.73053 d. 2.29686, 2.47691, 2.58830, 2.65798, 2.70185, 2.72959 10. a. 1.70308, 3.06605, 2.44030, 1.77204, 1.37348, 1.11925 b. 1.79548, 3.06051, 2.43292, 1.77807, 1.37795, 1.12191 c. 1.84579, 3.05769, 2.42905, 1.78074, 1.38017, 1.12328 d. 1.87734, 3.05607, 2.42672, 1.78224, 1.38150, 1.12411 11. a. -0.166134, -0.410872, -0.804660, 4.15867 b. -0.174652, -0.434238, -0.889140, -3.09810 12. A reasonable estimate for y at t = 0.8 is between 5.5 and 6. No reliable estimate is possible at t = 1 from the specified data. 13. b. 2.37 < a < 2.38 Final answers 14. b. 0.67 < αo < 0.68 Section 2.8, page 90 1. dw/ds = (s + 1)² + (w+2)², w(0) = 0 2. dw/ds = 1 -(w+3)³, w(0) = 0 3. a. ,(1) = ²/4/1 2k tk k! n k=1 c. lim (t) = e²t - 1 818 4. a. (t) = c. lim 818 n (-1)*+1+1/(k+1)!2k-1 k=1 (t) = 4e-¹/2+2t-4 t2k-1 1.3.5... (2k-1) t3k-1 n 5. a. Φ„(t) = Σ k=1 n 6. a. Φη(t) = - Σ ₁(t) 2. 101
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