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

Write each equation as a pair of coupled first order equations, using y = dx/dt.

1. (a) \(\frac{d^2x}{dt^2} + 7\frac{dx}{dt} + 2x = 0\)

   • Let y = dx/dt, so d²x/dt² = dy/dt
   • \(\frac{dy}{dt} = -7y - 2x\)

2. (b) \(\frac{d^2x}{dt^2} = 3x - 2\frac{dx}{dt} + 4\)

   • Replace dx/dt by y and d²x/dt² by dy/dt
   • \(\frac{dy}{dt} = 3x - 2y + 4\)

A motion satisfies \(\frac{d^2x}{dt^2} + 3\frac{dx}{dt} + 5x = 0\), with x(0) = 2 and x′(0) = −3. Use Euler's method with step length 0.2.

1. (a) Write down the recurrences for \(x_{n+1}\) and \(y_{n+1}\).

   • \(t_{n+1} = t_n + 0.2\)
   • \(x_{n+1} = x_n + 0.2y_n,\quad y_{n+1} = y_n + 0.2(-3y_n - 5x_n)\)

2. (b) Work out x and y after two steps, and so estimate x(0.4).

   • Step 1: x_1 = 1.400, y_1 = −3.200
   • Step 2: x_2 = 0.7600, y_2 = −2.680
   • x(0.4) ≈ 0.7600

Given \(\frac{d^2x}{dt^2} + 2\frac{dx}{dt} + 8x = 0\), x(0) = 3 and x′(0) = 0, use Euler's method with step length 0.1 and the calculator to estimate x(2).

• \(\frac{dx}{dt} = y,\quad \frac{dy}{dt} = -2y - 8x\)
• \(x_{n+1} = x_n + 0.1y_n,\quad y_{n+1} = y_n + 0.1(-2y_n - 8x_n)\)
• Starting values x_0 = 3, y_0 = 0
• From the calculator: x_20 = 0.53670
• x(2) ≈ 0.537

A function satisfies \(\frac{d^2x}{dt^2} + 2\frac{dx}{dt} + 5x = 5t\), with x(0) = 0 and x′(0) = 1. Use Euler's method with step length 0.1 to estimate x(1.5).

• \(\frac{dx}{dt} = y,\quad \frac{dy}{dt} = -2y - 5x + 5t\)
• \(t_{n+1} = t_n + 0.1,\quad x_{n+1} = x_n + 0.1y_n,\quad y_{n+1} = y_n + 0.1(-2y_n - 5x_n + 5t_n)\)
• From the calculator: x_15 = 0.97475
• x(1.5) ≈ 0.975

The displacement x mm of the arm of a vibration sensor, t seconds after it is knocked, satisfies \(\frac{d^2x}{dt^2} + 1.5\frac{dx}{dt} + 7x = 0\), with x(0) = 0 and x′(0) = 5.

1. (a) Write the equation as a pair of coupled first order equations.

   • \(y = \frac{dx}{dt}\), so \(\frac{dx}{dt} = y,\quad \frac{dy}{dt} = -1.5y - 7x\)

2. (b) Use Euler's method with step length 0.1 and technology to graph x for 0 ≤ t ≤ 4. Estimate the maximum displacement to 2 significant figures.

   • \(x_{n+1} = x_n + 0.1y_n,\quad y_{n+1} = y_n + 0.1(-1.5y_n - 7x_n)\)
   • Graph x against t: the highest point is at t = 0.50
   • maximum ≈ 1.6 mm

The sideways displacement x cm of a model footbridge deck, t seconds after wind gusts begin, satisfies \(\frac{d^2x}{dt^2} + 0.5\frac{dx}{dt} + 4x = 3\sin t\), with x(0) = 0 and x′(0) = 0.

1. (a) Write the equation as a pair of coupled first order equations.

   • \(\frac{dx}{dt} = y,\quad \frac{dy}{dt} = -0.5y - 4x + 3\sin t\)

2. (b) Use Euler's method with step length 0.1 to estimate x(3).

   • \(t_{n+1} = t_n + 0.1,\quad x_{n+1} = x_n + 0.1y_n,\quad y_{n+1} = y_n + 0.1(-0.5y_n - 4x_n + 3\sin t_n)\)
   • From the calculator (radians): x_30 = 0.48776
   • x(3) ≈ 0.488

A motion satisfies \(\frac{d^2x}{dt^2} + 4\frac{dx}{dt} + 3x = 0\), with x(0) = 1 and x′(0) = 0.

1. (a) Use Euler's method to estimate x(4) with step length (i) 1 and (ii) 0.5.

   • \(\frac{dx}{dt} = y,\quad \frac{dy}{dt} = -4y - 3x\)
   • Step length 1: 4 steps, x_4 = −8.00
   • Step length 0.5: 8 steps, x_8 = 0.00391

2. (b) Which estimate is more accurate? Justify your answer.

   • Check with step length 0.1: x(4) ≈ 0.0222
   • 0.00391 is much closer to 0.0222 than −8.00 is
   • So step length 0.5 is more accurate: a smaller step follows the curve more closely

A child on a swing is pulled back and released. The angle \(\theta\) radians that the swing makes with the vertical after \(t\) seconds satisfies \(\dfrac{d^2\theta}{dt^2}+0.3\dfrac{d\theta}{dt}+9\sin\theta=0\), with \(\theta(0)=0.6\) and \(\theta'(0)=0\).

1. (a) Write the equation as a pair of coupled first order equations.

   • \(\omega=\dfrac{d\theta}{dt}\), so \(\dfrac{d\theta}{dt}=\omega,\quad \dfrac{d\omega}{dt}=−0.3\omega−9\sin\theta\)

2. (b) Use Euler's method with step length 0.1 to estimate \(\theta(1)\), to 3 significant figures.

   • \(\theta_{n+1}=\theta_n+0.1\omega_n,\quad \omega_{n+1}=\omega_n+0.1(−0.3\omega_n-9\sin\theta_n)\)
   • Start at \(\theta_0=0.6\), \(\omega_0=0\), in radians
   • 10 steps: \(\theta_{10}=−0.75176\)
   • \(\theta(1)\approx−0.752\)

3. (c) Replace \(\sin\theta\) by \(\theta\) and repeat the estimate. Find the difference between the two estimates, to 2 significant figures.

   • Use \(\omega_{n+1}=\omega_n+0.1(−0.3\omega_n-9\theta_n)\)
   • 10 steps: \(\theta_{10}=−0.78115\)
   • Difference \(=0.029\)

The height x cm of a door-closer piston above its rest position, t seconds after release, satisfies \(\frac{d^2x}{dt^2} + 3\frac{dx}{dt} + 2x = 0\), with x(0) = 2 and x′(0) = 1. The motion is heavily damped: x rises a little at first, then falls towards 0 and never becomes negative.

1. (a) Use Euler's method to estimate x(6) with step length (i) 1, (ii) 0.5 and (iii) 0.25.

   • \(\frac{dx}{dt} = y,\quad \frac{dy}{dt} = -3y - 2x\)
   • h = 1: 6 steps, x = −3.00
   • h = 0.5: 12 steps, x = 0.00122
   • h = 0.25: 24 steps, x = 0.00502

2. (b) One estimate cannot be right. Say which, give a reason, and say which estimate you trust most.

   • The step length 1 table swings between positive and negative values: x = 2.00, 3.00, −3.00, 3.00, −3.00, …
   • That cannot be the motion described, so step length 1 is too long
   • Step length 0.25 is the smallest step, so it is the most accurate