Mixed practice: Euler's method for second order equations Problem set A Answers

1. (a) \(\frac{dx}{dt} = y,\quad \frac{dy}{dt} = -7y - 2x\)
2. (b) \(\frac{dx}{dt} = y,\quad \frac{dy}{dt} = 3x - 2y + 4\)

1. (a) \(x_{n+1} = x_n + 0.2y_n,\quad y_{n+1} = y_n + 0.2(-3y_n - 5x_n)\)
2. (b) x(0.4) ≈ 0.7600 (and y = −2.680)

x(2) ≈ 0.537

x(1.5) ≈ 0.975

1. (a) \(\frac{dx}{dt} = y,\quad \frac{dy}{dt} = -1.5y - 7x\)
2. (b) maximum ≈ 1.6 mm (at about t = 0.50)

1. (a) \(\frac{dx}{dt} = y,\quad \frac{dy}{dt} = -0.5y - 4x + 3\sin t\)
2. (b) x(3) ≈ 0.488

1. (a) (i) −8.00, (ii) 0.00391
2. (b) Step length 0.5. A run with step length 0.1 gives 0.0222, and the step length 0.5 estimate is much closer to it than the step length 1 estimate is

1. (a) \(\dfrac{d\theta}{dt}=\omega,\quad \dfrac{d\omega}{dt}=−0.3\omega−9\sin\theta\)
2. (b) \(\theta(1)\approx−0.752\)
3. (c) \(\theta(1)\approx−0.781\); difference 0.029

1. (a) (i) −3.00, (ii) 0.00122, (iii) 0.00502
2. (b) Step length 1 gives a negative x, which the motion never has. Trust step length 0.25: the smallest step