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numerical analysis

Numerical Analysis 9th edition Richard L. Burden, J. Douglas Faires - Solutions

Repeat Exercise 2 using the Trapezoidal Algorithm with TOL = 10−5.In Exercise 2a. y' = −5y + 6et, 0≤ t ≤ 1, y(0) = 2, with h = 0.1; actual solution y(t) = e−5t + et .b. y' = −10y+10t+1, 0 ≤ t ≤ 1, y(0) = e, with h = 0.1; actual solution y(t) = e−10t+1+t.c. y' = −15(y − t−3)
Solve the following stiff initial-value problem using the Runge-Kutta fourth-order method with (a) h = 0.1 and (b) h = 0.025. u'1 = 32u1 + 66u2 + 2/3 t + 2/3 , 0≤ t ≤ 0.5, u1(0) = 1/3; u'2 = −66u1 − 133u2 - 1/3t - 1/3 , 0≤ t ≤ 0.5, u2(0) = 1/3.
Use Euler's method to approximate the solutions for each of the following initial-value problems. a. y' = te3t − 2y, 0≤ t ≤ 1, y(0) = 0, with h = 0.5 b. y' = 1 + (t − y)2, 2≤ t ≤ 3, y(2) = 1, with h = 0.5 c. y' = 1 + y/t, 1≤ t ≤ 2, y(1) = 2, with h = 0.25 d. y' = cos 2t + sin 3t,
Given the initial-value problemy' = 1/t2 – y/t − y2, 1≤ t ≤ 2, y(1) = −1,With exact solution y(t) = −1/t:a. Use Euler’s method with h = 0.05 to approximate the solution, and compare it with the actual values of y.b. Use the answers generated in part (a) and linear interpolation to
Given the initial-value problemy' = −y + t + 1, 0 ≤ t ≤ 5, y(0) = 1,With exact solution y(t) = e−t + t:a. Approximate y(5) using Euler’s method with h = 0.2, h = 0.1, and h = 0.05.b. Determine the optimal value of h to use in computing y(5), assuming δ = 10−6 and that Eq. (5.14) is
Use the results of Exercise 5 and linear interpolation to approximate the following values of y(t). Compare the approximations obtained to the actual values obtained using the functions given in Exercise 7. a. y(1.25) and y(1.93) b. y(2.1) and y(2.75) c. y(1.3) and y(1.93) d. y(0.54) and y(0.94)
Use the results of Exercise 6 and linear interpolation to approximate the following values of y(t). Compare the approximations obtained to the actual values obtained using the functions given in Exercise 8. a. y(0.25) and y(0.93) b. y(1.25) and y(1.93) c. y(2.10) and y(2.75) d. y(0.54) and y(0.94)
Let E(h) = hM/2 + δ/h. a. For the initial-value problem y' = −y + 1, 0 ≤ t ≤ 1, y(0) = 0, Compute the value of h to minimize E(h). Assume δ = 5 × 10−(n+1) if you will be using n-digit arithmetic in part (c). b. For the optimal h computed in part (a), use Eq. (5.13) to compute the minimal
In a circuit with impressed voltage E having resistance R, inductance L, and capacitance C in parallel, the current i satisfies the differential equationSuppose C = 0.3 farads, R = 1.4 ohms, L = 1.7 henries, and the voltage is given byE(t) = eˆ’0.06Ï€t sin(2t ˆ’ Ï€).If i(0) = 0, find
Use Euler's method to approximate the solutions for each of the following initial-value problems. a. y' = et−y, 0≤ t ≤ 1, y(0) = 1, with h = 0.5 b. y' = (1 + t)/(1 + y) , 1≤ t ≤ 2, y(1) = 2, with h = 0.5 c. y' = −y + ty1/2, 2≤ t ≤ 3, y(2) = 2, with h = 0.25 d. y' = t−2(sin 2t −
The actual solutions to the initial-value problems in Exercise 1 are given here. Compare the actual error at each step to the error bound. a. y(t) = 1/5 te3t - 1/25e3t + 1/25e−2t b. y(t) = t + 1/(1 - t) c. y(t) = t ln t + 2t d. y(t) = 1/2sin 2t - 1/3 cos 3t + 4/3
The actual solutions to the initial-value problems in Exercise 2 are given here. Compute the actual error and compare this to the error bound if Theorem 5.9 can be applied. a. y(t) = ln(et + e − 1) b. y(t) = √(t2 + 2t + 6) − 1 c. y(t) =(t − 2 +√2ee−t/2)2 d. y(t) = (4 + cos 2 − cos
Use Euler's method to approximate the solutions for each of the following initial-value problems. a. y' = y/t − (y/t)2, 1≤ t ≤ 2, y(1) = 1, with h = 0.1 b. y' = 1 + y/t + (y/t)2, 1≤ t ≤ 3, y(1) = 0, with h = 0.2 c. y' = −(y + 1)(y + 3), 0≤ t ≤ 2, y(0) = −2, with h = 0.2 d. y' =
Use Euler's method to approximate the solutions for each of the following initial-value problems. a. y' = (2 − 2ty)/(t2 + 1) , 0≤ t ≤ 1, y(0) = 1, with h = 0.1 b. y' = y2/(1 + t), 1≤ t ≤ 2, y(1) = −(ln 2)−1, with h = 0.1 c. y' = (y2 + y)/t, 1≤ t ≤ 3, y(1) = −2, with h = 0.2 d.
The actual solutions to the initial-value problems in Exercise 5 are given here. Compute the actual error in the approximations of Exercise 5. a. y(t) = t/(1 + ln t) b. y(t) = t tan(ln t) c. y(t) = −3 + 2/(1 + e−2t) d. y(t) = t2 + 1/3 e−5t
The actual solutions to the initial-value problems in Exercise 6 are given here. Compute the actual error in the approximations of Exercise 6. a. y(t) = (2t + 1)/(t2 + 1) b. y(t) = −1/(ln(t + 1)) c. y(t) = 2t(1 − 2t) d. y(t) = √(4 − 3e−t2)
Given the initial-value problemy' = 2/t y + t2et, 1≤ t ≤ 2, y(1) = 0,With exact solution y(t) = t2(et − e) :a. Use Euler’s method with h = 0.1 to approximate the solution, and compare it with the actual values of y.b. Use the answers generated in part (a) and linear interpolation to
Use Taylor's method of order two to approximate the solutions for each of the following initial-value problems. a. y' = te3t − 2y, 0≤ t ≤ 1, y(0) = 0, with h = 0.5 b. y' = 1 + (t − y)2, 2≤ t ≤ 3, y(2) = 1, with h = 0.5 c. y' = 1 + y/t, 1≤ t ≤ 2, y(1) = 2, with h = 0.25 d. y' = cos
Given the initial-value problemy' = 1/t2 – y/t − y2, 1≤ t ≤ 2, y(1) = −1,With exact solution y(t) = −1/t:a. Use Taylor’s method of order two with h = 0.05 to approximate the solution, and compare it with the actual values of y.b. Use the answers generated in part (a) and linear
A projectile of mass m = 0.11 kg shot vertically upward with initial velocity v(0) = 8 m/s is slowed due to the force of gravity, Fg = −mg, and due to air resistance, Fr = −kv|v|, where g = 9.8 m/s2 and k = 0.002 kg/m. The differential equation for the velocity v is given by mv' = −mg −
Use the Taylor method of order two with h = 0.1 to approximate the solution to y' = 1 + t sin(ty), 0≤ t ≤ 2, y(0) = 0.
Use Taylor's method of order two to approximate the solutions for each of the following initial-value problems. a. y' = et−y, 0≤ t ≤ 1, y(0) = 1, with h = 0.5 b. y' = (1 + t)/(1 + y), 1≤ t ≤ 2, y(1) = 2, with h = 0.5 c. y' = −y + ty1/2, 2≤ t ≤ 3, y(2) = 2, with h = 0.25 d. y' =
Repeat Exercise 1 using Taylor's method of order four. In Exercise 1 a. y' = te3t − 2y, 0≤ t ≤ 1, y(0) = 0, with h = 0.5 b. y' = 1 + (t − y)2, 2≤ t ≤ 3, y(2) = 1, with h = 0.5 c. y' = 1 + y/t, 1≤ t ≤ 2, y(1) = 2, with h = 0.25 d. y' = cos 2t + sin 3t, 0≤ t ≤ 1, y(0) = 1, with h
Repeat Exercise 2 using Taylor's method of order four. In Exercise 2 a. y' = et−y, 0≤ t ≤ 1, y(0) = 1, with h = 0.5 b. y' = (1 + t)/(1 + y), 1≤ t ≤ 2, y(1) = 2, with h = 0.5 c. y' = −y + ty1/2, 2≤ t ≤ 3, y(2) = 2, with h = 0.25 d. y' = t−2(sin 2t − 2ty), 1≤ t ≤ 2, y(1) = 2,
Use Taylor's method of order two to approximate the solution for each of the following initial-value problems. a. y' = y/t − (y/t)2, 1≤ t ≤ 1.2, y(1) = 1, with h = 0.1 b. y' = sin t + e−t, 0≤ t ≤ 1, y(0) = 0, with h = 0.5 c. y' = (y2 + y)/t, 1≤ t ≤ 3, y(1) = −2, with h = 0.5 d. y'
Use Taylor's method of order two to approximate the solution for each of the following initial-value problems. a. y' = (2 − 2ty)/(t2 + 1) , 0≤ t ≤ 1, y(0) = 1, with h = 0.1 b. y' = y2/(1 + t) , 1≤ t ≤ 2, y(1) = −(ln 2)−1, with h = 0.1 c. y' = (y2 + y)/t, 1≤ t ≤ 3, y(1) = −2,
Repeat Exercise 5 using Taylor's method of order four. In Exercise 5 a. y' = y/t − (y/t)2, 1≤ t ≤ 1.2, y(1) = 1, with h = 0.1 b. y' = sin t + e−t, 0≤ t ≤ 1, y(0) = 0, with h = 0.5 c. y' = (y2 + y)/t, 1≤ t ≤ 3, y(1) = −2, with h = 0.5 d. y' = −ty + 4ty−1, 0≤ t ≤ 1, y(0) =
Repeat Exercise 6 using Taylor's method of order four. In Exercise 6 a. y' = (2 − 2ty)/(t2 + 1) , 0≤ t ≤ 1, y(0) = 1, with h = 0.1 b. y' = y2/(1 + t) , 1≤ t ≤ 2, y(1) = −(ln 2)−1, with h = 0.1 c. y' = (y2 + y)/t, 1≤ t ≤ 3, y(1) = −2, with h = 0.2 d. y' = −ty + 4t/y, 0≤ t ≤
Given the initial-value problem y' = 2/t y + t2et, 1≤ t ≤ 2, y(1) = 0, Discuss.
Use the Modified Euler method to approximate the solutions to each of the following initial-value problems, and compare the results to the actual values. a. y' = te3t − 2y, 0≤ t ≤ 1, y(0) = 0, with h=0.5; actual solution y(t)=1/5 te3t - 1/25 e3t + 1/25 e−2t . b. y' = 1 + (t − y)2, 2≤ t
Repeat Exercise 2 using Heun’s method.In Exercise 2a. y' = et−y, 0≤ t ≤ 1, y(0) = 1, with h = 0.5; actual solution y(t) = ln(et + e − 1).b. y' = (1 + t)/(1 + y), 1≤ t ≤ 2, y(1) = 2, with h = 0.5; actual solution y(t) =√(t2 + 2t + 6−1).c. y' = −y + ty1/2, 2 ≤ t ≤ 3, y(2) = 2,
Repeat Exercise 3 using Heun’s method.In Exercise 3a. y' = y/t − (y/t)2, 1≤ t ≤ 2, y(1) = 1, with h = 0.1; actual solution y(t) = t/(1 + ln t).b. y' = 1+y/t +(y/t)2, 1≤ t ≤ 3, y(1) = 0, with h = 0.2; actual solution y(t) = t tan(ln t).c. y' = −(y + 1)(y + 3), 0 ≤ t ≤ 2, y(0) =
Repeat Exercise 4 using Heun’s method.In Exercise 4a. y' = (2 − 2ty)/(t2 + 1), 0≤ t ≤ 1, y(0) = 1, with h = 0.1; actual solution y(t) = (2t + 1)/(t2 + 1).b. y' = y2/(1 + t), 1≤ t ≤ 2, y(1) = −(ln 2)−1, with h = 0.1; actual solution y(t) =−1/(ln(t + 1)).c. y' = (y2 + y)/t, 1≤ t
Repeat Exercise 1 using the Runge-Kutta method of order four.In Exercise 1a. y' = te3t − 2y, 0≤ t ≤ 1, y(0) = 0, with h=0.5; actual solution y(t)=1/5 te3t – 1/25 e3t + 1/25 e−2t.b. y' = 1 + (t − y)2, 2≤ t ≤ 3, y(2) = 1, with h = 0.5; actual solution y(t) = t + 1/(1−t).c. y' = 1 +
Repeat Exercise 2 using the Runge-Kutta method of order four.In Exercise 2a. y' = et−y, 0≤ t ≤ 1, y(0) = 1, with h = 0.5; actual solution y(t) = ln(et + e − 1).b. y' = (1 + t)/(1 + y), 1≤ t ≤ 2, y(1) = 2, with h = 0.5; actual solution y(t) =√(t2 + 2t + 6−1).c. y' = −y + ty1/2, 2
Repeat Exercise 3 using the Runge-Kutta method of order four.In Exercise 3a. y' = y/t − (y/t)2, 1≤ t ≤ 2, y(1) = 1, with h = 0.1; actual solution y(t) = t/(1 + ln t).b. y' = 1+y/t +(y/t)2, 1≤ t ≤ 3, y(1) = 0, with h = 0.2; actual solution y(t) = t tan(ln t).c. y' = −(y + 1)(y + 3), 0
Repeat Exercise 4 using the Runge-Kutta method of order four.In Exercise 4a. y' = (2 − 2ty)/(t2 + 1), 0≤ t ≤ 1, y(0) = 1, with h = 0.1; actual solution y(t) = (2t + 1)/(t2 + 1).b. y' = y2/(1 + t), 1≤ t ≤ 2, y(1) = −(ln 2)−1, with h = 0.1; actual solution y(t) =−1/(ln(t + 1)).c. y' =
Use the results of Exercise 3 and linear interpolation to approximate values of y(t), and compare the results to the actual values. a. y(1.25) and y(1.93) b. y(2.1) and y(2.75) c. y(1.3) and y(1.93) d. y(0.54) and y(0.94)
Use the results of Exercise 4 and linear interpolation to approximate values of y(t), and compare the results to the actual values. a. y(0.54) and y(0.94) b. y(1.25) and y(1.93) c. y(1.3) and y(2.93) d. y(0.54) and y(0.94)
Repeat Exercise 17 using the results of Exercise 7. In Exercise 17 a. y(1.25) and y(1.93) b. y(2.1) and y(2.75) c. y(1.3) and y(1.93)\ d. y(0.54) and y(0.94)
Use the Modified Euler method to approximate the solutions to each of the following initial-value problems, and compare the results to the actual values. a. y' = et−y, 0≤ t ≤ 1, y(0) = 1, with h = 0.5; actual solution y(t) = ln(et + e − 1). b. y' = (1 + t)/(1 + y), 1≤ t ≤ 2, y(1) = 2,
Repeat Exercise 18 using the results of Exercise 8. In Exercise 18 a. y(0.54) and y(0.94)\ b. y(1.25) and y(1.93) c. y(1.3) and y(2.93) d. y(0.54) and y(0.94)
Repeat Exercise 17 using the results of Exercise 11. In Exercise 17 a. y(1.25) and y(1.93) b. y(2.1) and y(2.75) c. y(1.3) and y(1.93)\ d. y(0.54) and y(0.94)
Repeat Exercise 18 using the results of Exercise 12. In Exercise 18 a. y(0.54) and y(0.94)\ b. y(1.25) and y(1.93) c. y(1.3) and y(2.93) d. y(0.54) and y(0.94)
Repeat Exercise 17 using the results of Exercise 15. In Exercise 17 a. y(1.25) and y(1.93) b. y(2.1) and y(2.75) c. y(1.3) and y(1.93)\ d. y(0.54) and y(0.94)
Repeat Exercise 18 using the results of Exercise 16. In Exercise 18 a. y(0.54) and y(0.94)\ b. y(1.25) and y(1.93) c. y(1.3) and y(2.93) d. y(0.54) and y(0.94)
Use the results of Exercise 15 and Cubic Hermite interpolation to approximate values of y(t), and compare the approximations to the actual values. a. y(1.25) and y(1.93) b. y(2.1) and y(2.75) c. y(1.3) and y(1.93) d. y(0.54) and y(0.94)
Use the results of Exercise 16 and Cubic Hermite interpolation to approximate values of y(t), and compare the approximations to the actual values. a. y(0.54) and y(0.94) b. y(1.25) and y(1.93) c. y(1.3) and y(2.93) d. y(0.54) and y(0.94)
Show that the Midpoint method and the Modified Euler method give the same approximations to the initial-value problem y' = −y + t + 1, 0 ≤ t ≤ 1, y(0) = 1, For any choice of h. Why is this true?
Use the Modified Euler method to approximate the solutions to each of the following initial-value problems, and compare the results to the actual values. a. y' = y/t − (y/t)2, 1≤ t ≤ 2, y(1) = 1, with h = 0.1; actual solution y(t) = t/(1 + ln t). b. y' = 1+y/t +(y/t)2, 1≤ t ≤ 3, y(1) = 0,
Show that the difference method w0 = α, wi+1 = wi + a1f (ti ,wi) + a2f (ti + α2,w1 + δ2f (ti ,wi)), For each i = 0, 1. . . N − 1, cannot have local truncation error O (h3) for any choice of constants a1, a2, α2, and δ2
Use the Modified Euler method to approximate the solutions to each of the following initial-value problems, and compare the results to the actual values. a. y' = (2 − 2ty)/(t2 + 1), 0≤ t ≤ 1, y(0) = 1, with h = 0.1; actual solution y(t) = (2t + 1)/(t2 + 1). b. y' = y2/(1 + t), 1≤ t ≤ 2,
Repeat Exercise 1 using the Midpoint method.In Exercise 1a. y' = te3t − 2y, 0≤ t ≤ 1, y(0) = 0, with h=0.5; actual solution y(t)=1/5 te3t – 1/25 e3t + 1/25 e−2t .b. y' = 1 + (t − y)2, 2≤ t ≤ 3, y(2) = 1, with h = 0.5; actual solution y(t) = t + 1/(1−t) .c. y' = 1 + y/t, 1≤ t ≤
Repeat Exercise 2 using the Midpoint method.In Exercise 2a. y' = et−y, 0≤ t ≤ 1, y(0) = 1, with h = 0.5; actual solution y(t) = ln(et + e − 1).b. y' = (1 + t)/(1 + y), 1≤ t ≤ 2, y(1) = 2, with h = 0.5; actual solution y(t) =√(t2 + 2t + 6−1).c. y' = −y + ty1/2, 2 ≤ t ≤ 3, y(2)
Repeat Exercise 3 using the Midpoint method.In Exercise 3a. y' = y/t − (y/t)2, 1≤ t ≤ 2, y(1) = 1, with h = 0.1; actual solution y(t) = t/(1 + ln t).b. y' = 1+y/t +(y/t)2, 1≤ t ≤ 3, y(1) = 0, with h = 0.2; actual solution y(t) = t tan(ln t).c. y' = −(y + 1)(y + 3), 0 ≤ t ≤ 2, y(0) =
Repeat Exercise 4 using the Midpoint method.In Exercise 4a. y' = (2 − 2ty)/(t2 + 1), 0≤ t ≤ 1, y(0) = 1, with h = 0.1; actual solution y(t) = (2t + 1)/(t2 + 1).b. y' = y2/(1 + t), 1≤ t ≤ 2, y(1) = −(ln 2)−1, with h = 0.1; actual solution y(t) =−1/(ln(t + 1)).c. y' = (y2 + y)/t, 1≤
Repeat Exercise 1 using Heun’s method.In Exercise 1a. y' = te3t − 2y, 0≤ t ≤ 1, y(0) = 0, with h=0.5; actual solution y(t)=1/5 te3t – 1/25 e3t + 1/25 e−2t .b. y' = 1 + (t − y)2, 2≤ t ≤ 3, y(2) = 1, with h = 0.5; actual solution y(t) = t + 1/(1−t) .c. y' = 1 + y/t, 1≤ t ≤ 2,
Use the Runge-Kutta-Fehlberg method with tolerance TOL = 10−4, hmax = 0.25, and hmin = 0.05 to approximate the solutions to the following initial-value problems. Compare the results to the actual values. a. y' = te3t − 2y, 0≤ t ≤ 1, y(0) = 0; actual solution y(t) = 1/5 te3t - 1/25 e3t +
Use the Runge-Kutta Fehlberg Algorithm with tolerance TOL = 10−4 to approximate the solution to the following initial-value problems. a. y' = (y/t)2 + y/t, 1≤ t ≤ 1.2, y(1) = 1, with hmax = 0.05 and hmin = 0.02. b. y' = sin t + e−t, 0≤ t ≤ 1, y(0) = 0, with hmax = 0.25 and hmin =
Use the Runge-Kutta-Fehlberg method with tolerance TOL = 10−6, hmax = 0.5, and hmin = 0.05 to approximate the solutions to the following initial-value problems. Compare the results to the actual values. a. y' = y/t − (y/t)2, 1≤ t ≤ 4, y(1) = 1; actual solution y(t) = t/(1 + ln t). b. y' = 1
The Runge-Kutta-Verner method (see [Ve]) is based on the formulas wi+1 = wi + 13/160 k1 + 2375/5984 k3 + 5/16 k4 + 12/85 k5 + 3/44 k6 and i+1 = wi + 3/40 k + 875/2244 k3 + 23/72 k4 + 264/1955 k5 + 125/11592 k7 + 43/616 k8, Where k1 = hf (ti ,wi), k2 = hf(ti + h/6 ,wi + 1/6 k1), k3 = hf(ti + 4h/15
In the theory of the spread of contagious disease (see [Ba1] or [Ba2]), a relatively elementary differential equation can be used to predict the number of infective individuals in the population at any time, provided appropriate simplification assumptions are made. In particular, let us assume that
In the previous exercise, all infected individuals remained in the population to spread the disease.
Use all the Adams-Bashforth methods to approximate the solutions to the following initial-value problems. In each case use exact starting values, and compare the results to the actual values. a. y' = te3t − 2y, 0≤ t ≤ 1, y(0) = 0, with h=0.2; actual solution y(t) = 1/5 te3t - 1/25 e3t +1/25
Use the Milne-Simpson Predictor-Corrector method to approximate the solutions to the initial-value problems in Exercise 3. In Exercise 3 a. y' = y/t − (y/t)2, 1≤ t ≤ 2, y(1) = 1, with h = 0.1; actual solution y(t) = t/(1 + ln t). b. y' = 1+y/t+(y/t)2, 1≤ t ≤ 3, y(1) = 0, with h = 0.2;
a. Derive the Adams-Bashforth Two-Step method by using the Lagrange form of the interpolating polynomial. b. Derive the Adams-Bashforth Four-Step method by using Newton's backward-difference form of the interpolating polynomial.
Derive the Adams-Bashforth Three-Step method by the following method. Set y(ti+1) = y(ti) + ahf (ti , y(ti)) + bhf (ti−1, y(ti−1)) + chf (ti−2, y(ti−2)).Expand y(ti+1), f (ti−2, y(ti−2)), and f (ti−1, y(ti−1)) in Taylor series about (ti , y(t=)), and equate the coefficients of h, h2
Derive the Adams-Moulton Two-Step method and its local truncation error by using an appropriate form of an interpolating polynomial.
Derive Simpson's method by applying Simpson's rule to the integral
Derive Milne's method by applying the open Newton-Cotes formula (4.29) to the integral
Verify the entries in Table 5.12 on page 305.Table 5.12
Use each of the Adams-Bashforth methods to approximate the solutions to the following initial-value problems. In each case use starting values obtained from the Runge-Kutta method of order four. Compare the results to the actual values. a. y' = (2 − 2ty)/(t2 + 1), 0≤ t ≤ 1, y(0) = 1, with h =
Use each of the Adams-Bashforth methods to approximate the solutions to the following initial-value problems. In each case use starting values obtained from the Runge-Kutta method of order four. Compare the results to the actual values. a. y' = y/t − (y/t)2, 1≤ t ≤ 2, y(1) = 1, with h = 0.1;
Use all the Adams-Moulton methods to approximate the solutions to the Exercises 1(a), 1(c), and 1(d). In each case use exact starting values, and explicitly solve for wi+1. Compare the results to the actual values. In Exercise 1 a. y' = y/t − (y/t)2, 1≤ t ≤ 2, y(1) = 1, with h = 0.1; actual
Use Algorithm 5.4 to approximate the solutions to the initial-value problems in Exercise 1In Exercise 1a. y' = te3t − 2y, 0≤ t ≤ 1, y(0) = 0, with h=0.2; actual solution y(t) = 1/5 te3t - 1/25 e3t +1/25 e−2t .b. y' = 1 + (t − y)2, 2≤ t ≤ 3, y(2) = 1, with h = 0.2; actual solution y(t)
Use Algorithm 5.4 to approximate the solutions to the initial-value problems in Exercise 2. In Exercise 2 a. y' = (2 − 2ty)/(t2 + 1), 0≤ t ≤ 1, y(0) = 1, with h = 0.1 actual solution y(t) = (2t + 1)/(t2 + 2). b. y' = y2/(1 + t) , 1≤ t ≤ 2, y(1) = −(ln 2)−1, with h = 0.1 actual
Use Algorithm 5.4 to approximate the solutions to the initial-value problems in Exercise 3. In Exercise 3 a. y' = y/t − (y/t)2, 1≤ t ≤ 2, y(1) = 1, with h = 0.1; actual solution y(t) = t/(1 + ln t). b. y' = 1+y/t+(y/t)2, 1≤ t ≤ 3, y(1) = 0, with h = 0.2; actual solution y(t) = t tan(ln
Change Algorithm 5.4 so that the corrector can be iterated for a given number p iterations. Repeat Exercise 7 with p = 2, 3, and 4 iterations. Which choice of p gives the best answer for each initial-value problem?In Exercise 7a. y' = y/t − (y/t)2, 1≤ t ≤ 2, y(1) = 1, with h = 0.1; actual
The initial-value problemy' = ey, 0≤ t ≤ 0.20, y(0) = 1Has solutiony(t) = 1 − ln(1 − et).Applying the three-step Adams-Moulton method to this problem is equivalent to finding the fixed point wi+1 ofg(w) = wi + h/24(9ew + 19ewi − 5ewi−1 + ewi−2 ) .a. With h = 0.01, obtain wi+1 by
Use the Adams Variable Step-Size Predictor-Corrector Algorithm with tolerance TOL = 10−4, hmax = 0.25, and hmin = 0.025 to approximate the solutions to the given initial-value problems. Compare the results to the actual values. a. y' = te3t − 2y, 0≤ t ≤ 1, y(0) = 0; actual solution y(t) =
Use the Adams Variable Step-Size Predictor-Corrector Algorithm with TOL = 10−4 to approximate the solutions to the following initial-value problems: a. y' = (y/t)2 + y/t, 1≤ t ≤ 1.2, y(1) = 1, with hmax = 0.05 and hmin = 0.01. b. y' = sin t + e−t, 0≤ t ≤ 1, y(0) = 0, with hmax = 0.2 and
Use the Adams Variable Step-Size Predictor-Corrector Algorithm with tolerance TOL = 10−6, hmax = 0.5, and hmin = 0.02 to approximate the solutions to the given initial-value problems. Compare the results to the actual values. a. y' = y/t − (y/t)2, 1≤ t ≤ 4, y(1) = 1; actual solution y(t) =
Construct an AdamsVariable Step-Size Predictor-Corrector Algorithm based on the Adams-Bashforth five-step method and the Adams-Moulton four-step method. Repeat Exercise 3 using this newmethod.In Exercise 3a. y' = y/t − (y/t)2, 1≤ t ≤ 4, y(1) = 1; actual solution y(t) = t/(1 + ln t).b. y' = 1
Use the Extrapolation Algorithm with tolerance TOL = 10−4, hmax = 0.25, and hmin = 0.05 to approximate the solutions to the following initial-value problems. Compare the results to the actual values. a. y' = te3t − 2y, 0≤ t ≤ 1, y(0) = 0; actual solution y(t) = 1/5 te3t - 1/25 e3t + 1/25
Use the Extrapolation Algorithm with TOL = 10−4 to approximate the solutions to the following initial-value problems: a. y' = (y/t)2 + y/t, 1≤ t ≤ 1.2, y(1) = 1, with hmax = 0.05 and hmin = 0.02. b. y' = sin t + e−t, 0≤ t ≤ 1, y(0) = 0, with hmax = 0.25 and hmin = 0.02. c. y' = (y2 +
Use the Extrapolation Algorithm with tolerance TOL = 10−6, hmax = 0.5, and hmin = 0.05 to approximate the solutions to the following initial-value problems. Compare the results to the actual values. a. y' = y/t − (y/t)2, 1≤ t ≤ 4, y(1) = 1; actual solution y(t) = t/(1 + ln t). b. y' = 1 +
Use the Runge-Kutta method for systems to approximate the solutions of the following systems of first-order differential equations, and compare the results to the actual solutions. a. u'1 = 3u1 + 2u2 − (2t2 + 1)e2t , u1(0) = 1; u'2 = 4u1 + u2 + (t2 + 2t − 4)e2t , u2(0) = 1; 0 ≤ t ≤ 1; h =
In Exercise 9 we considered the problem of predicting the population in a predator-prey model.Another problem of this type is concerned with two species competing for the same food supply. If the numbers of species alive at time t are denoted by x1(t) and x2(t), it is often assumed that, although
Use the Runge-Kutta method for systems to approximate the solutions of the following systems of first-order differential equations, and compare the results to the actual solutions. a. u'1 = u1 − u2 + 2, u1(0) = −1; u'2 = −u1 + u2 + 4t, u2(0) = 0; 0 ≤ t ≤ 1; h = 0.1; actual solutions u1(t)
Use the Runge-Kutta for Systems Algorithm to approximate the solutions of the following higherorder differential equations, and compare the results to the actual solutions. a. y'' − 2y' + y = tet − t, 0≤ t ≤ 1, y(0) = y'(0) = 0, with h = 0.1; actual solution y(t) = 1/6 t3et − tet + 2et
Use the Runge-Kutta for Systems Algorithm to approximate the solutions of the following higherorder differential equations, and compare the results to the actual solutions. a. y'' − 3y' + 2y = 6e−t, 0≤ t ≤ 1, y(0) = y'(0) = 2, with h = 0.1; actual solution y(t) = 2e2t − et + e−t . b.
Change the Adams Fourth-Order Predictor-Corrector Algorithm to obtain approximate solutions to systems of first-order equations.
Repeat Exercise 2 using the algorithm developed in Exercise 5.In Exercise 2a. u'1 = u1 − u2 + 2, u1(0) = −1; u'2 = −u1 + u2 + 4t, u2(0) = 0; 0 ≤ t ≤ 1; h = 0.1; actual solutions u1(t) = −1/2 e2t + t2 + 2t – ½ and u2(t) = 1/2 e2t + t2 – 1/2.b. u'1 = 1/9 u1 – 2/3 u2
Repeat Exercise 1 using the algorithm developed in Exercise 5.In Exercise 1a. u'1 = 3u1 + 2u2 − (2t2 + 1)e2t , u1(0) = 1; u'2 = 4u1 + u2 + (t2 + 2t − 4)e2t , u2(0) = 1; 0 ≤ t ≤ 1; h = 0.2; actual solutions u1(t) = 1/3 e5t – 1/3 e−t + e2t and u2(t) = 1/3 e5t + 2/3 e−t + t2e2t
Suppose the swinging pendulum described in the lead example of this chapter is 2 ft long and that g = 32.17 ft/s2. With h = 0.1 s, compare the angle θ obtained for the following two initial-value problems at t = 0, 1, and 2 s. a. d2θ/dt2 + g/L sin θ = 0, θ(0) = π/6, θ '(0) = 0, b. d2θ/dt2 +
The study of mathematical models for predicting the population dynamics of competing species has its origin in independent works published in the early part of the 20th century by A. J. Lotka and V. Volterra (see, for example, [Lo1], [Lo2], and [Vo]). Consider the problem of predicting the
For each of the following linear systems, obtain a solution by graphical methods, if possible. Explain the results from a geometrical standpoint. a. x1 + 2x2 = 3, x1 − x2 = 0. b. x1 + 2x2 = 3, 2x1 + 4x2 = 6. c. x1 + 2x2 = 0, 2x1 + 4x2 = 0. d. 2x1 + x2 = −1, 4x1 + 2x2 = −2, x1 − 3x2 = 5.
Given the linear system x1 − x2 + αx3 = −2, −x1 + 2x2 − αx3 = 3, αx1 + x2 + x3 = 2.a. Find value(s) of α for which the system has no solutions.b. Find value(s) of α for which the system has an infinite number of solutions.c. Assuming a unique solution exists for a given α, find the
Show that the operationsa. (λEi) → (Ei) b. (Ei + λEj) → (Ei) c. (Ei) ↔ (Ej)
Gauss-Jordan Method: This method is described as follows. Use the ith equation to eliminate not only xi from the equations Ei+1, Ei+2, . . . , En, as was done in the Gaussian elimination method, but also from E1, E2, . . . , Eiˆ’1. Upon reducing [A, b] to:

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