Mixed practice: Euler's method for second order equations Problem set B One question at a time

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

1. (a) \(\frac{d^2x}{dt^2} + 5\frac{dx}{dt} + 9x = 0\)

   • Let y = dx/dt, so d²x/dt² = dy/dt
   • \(\frac{dy}{dt} = -5y - 9x\)

2. (b) \(\frac{d^2x}{dt^2} = 4x + 5\frac{dx}{dt} - 3\)

   • Replace dx/dt by y and d²x/dt² by dy/dt
   • \(\frac{dy}{dt} = 4x + 5y - 3\)

A motion satisfies \(\frac{d^2x}{dt^2} + 4\frac{dx}{dt} + 7x = 0\), with x(0) = 1 and x′(0) = 2. Use Euler's method with step length 0.25.

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

   • \(t_{n+1} = t_n + 0.25\)
   • \(x_{n+1} = x_n + 0.25y_n,\quad y_{n+1} = y_n + 0.25(-4y_n - 7x_n)\)

2. (b) Work out x and y after two steps, and so estimate x(0.5).

   • Step 1: x_1 = 1.500, y_1 = −1.750
   • Step 2: x_2 = 1.0625, y_2 = −2.625
   • x(0.5) ≈ 1.0625

Given \(\frac{d^2x}{dt^2} + 3\frac{dx}{dt} + 10x = 0\), x(0) = 1 and x′(0) = 4, use Euler's method with step length 0.1 and the calculator to estimate x(1.5).

• \(\frac{dx}{dt} = y,\quad \frac{dy}{dt} = -3y - 10x\)
• \(x_{n+1} = x_n + 0.1y_n,\quad y_{n+1} = y_n + 0.1(-3y_n - 10x_n)\)
• Starting values x_0 = 1, y_0 = 4
• From the calculator: x_15 = −0.36375
• x(1.5) ≈ −0.364

A function satisfies \(\frac{d^2x}{dt^2} + 4\frac{dx}{dt} + 6x = 3e^{-t}\), with x(0) = 1 and x′(0) = 0. Use Euler's method with step length 0.1 to estimate x(1.5).

• \(\frac{dx}{dt} = y,\quad \frac{dy}{dt} = -4y - 6x + 3e^{-t}\)
• \(t_{n+1} = t_n + 0.1,\quad x_{n+1} = x_n + 0.1y_n,\quad y_{n+1} = y_n + 0.1(-4y_n - 6x_n + 3e^{-t_n})\)
• From the calculator: x_15 = 0.23002
• x(1.5) ≈ 0.230

The height x cm of a diving board tip above its rest position, t seconds after a jump, satisfies \(\frac{d^2x}{dt^2} + 2\frac{dx}{dt} + 12x = 0\), with x(0) = 0 and x′(0) = 7.

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

   • \(y = \frac{dx}{dt}\), so \(\frac{dx}{dt} = y,\quad \frac{dy}{dt} = -2y - 12x\)

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

   • \(x_{n+1} = x_n + 0.1y_n,\quad y_{n+1} = y_n + 0.1(-2y_n - 12x_n)\)
   • Graph x against t: the highest point is at t = 0.40
   • maximum ≈ 1.8 cm

The displacement x mm of a speaker cone, t milliseconds after a signal begins, satisfies \(\frac{d^2x}{dt^2} + 0.8\frac{dx}{dt} + 5x = 2\cos 2t\), with x(0) = 0 and x′(0) = 0.

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

   • \(\frac{dx}{dt} = y,\quad \frac{dy}{dt} = -0.8y - 5x + 2\cos 2t\)

2. (b) Use Euler's method with step length 0.1 to estimate x(2.5).

   • \(t_{n+1} = t_n + 0.1,\quad x_{n+1} = x_n + 0.1y_n,\quad y_{n+1} = y_n + 0.1(-0.8y_n - 5x_n + 2\cos 2t_n)\)
   • From the calculator (radians): x_25 = −0.89400
   • x(2.5) ≈ −0.894

A motion satisfies \(\frac{d^2x}{dt^2} + 2.5\frac{dx}{dt} + x = 0\), with x(0) = 3 and x′(0) = 0.

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

   • \(\frac{dx}{dt} = y,\quad \frac{dy}{dt} = -2.5y - x\)
   • Step length 1: 4 steps, x_4 = −0.750
   • Step length 0.5: 8 steps, x_8 = 0.400

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

   • Check with step length 0.1: x(4) ≈ 0.514
   • 0.400 is much closer to 0.514 than −0.750 is
   • So step length 0.5 is more accurate: a smaller step follows the curve more closely

A lamp hangs from a long chain in a hall and is nudged. The angle \(\theta\) radians that the chain makes with the vertical after \(t\) seconds satisfies \(\dfrac{d^2\theta}{dt^2}+0.5\dfrac{d\theta}{dt}+16\sin\theta=0\), with \(\theta(0)=0.4\) and \(\theta'(0)=0\).

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

   • \(\omega=\dfrac{d\theta}{dt}\), so \(\dfrac{d\theta}{dt}=\omega,\quad \dfrac{d\omega}{dt}=−0.5\omega−16\sin\theta\)

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

   • \(\theta_{n+1}=\theta_n+0.1\omega_n,\quad \omega_{n+1}=\omega_n+0.1(−0.5\omega_n-16\sin\theta_n)\)
   • Start at \(\theta_0=0.4\), \(\omega_0=0\), in radians
   • 10 steps: \(\theta_{10}=−0.53912\)
   • \(\theta(1)\approx−0.539\)

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

   • Use \(\omega_{n+1}=\omega_n+0.1(−0.5\omega_n-16\theta_n)\)
   • 10 steps: \(\theta_{10}=−0.52421\)
   • Difference \(=0.015\)

The depth x cm of liquid in a slowly draining tube above its final level, t seconds after the tap opens, satisfies \(\frac{d^2x}{dt^2} + 4\frac{dx}{dt} + 3x = 0\), with x(0) = 2 and x′(0) = 0. The motion is heavily damped: x falls steadily 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.

   • \(\frac{dx}{dt} = y,\quad \frac{dy}{dt} = -4y - 3x\)
   • h = 1: 6 steps, x = −64.0
   • h = 0.5: 12 steps, x = 0.000488
   • h = 0.25: 24 steps, x = 0.00301

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, 2.00, −4.00, 8.00, −16.0, …
   • 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