Mixed practice: second order equations, exact solutions and models Problem set B One question at a time

Show that \(x = 2e^{2t} + e^{-3t}\) satisfies \(\frac{d^2x}{dt^2} + \frac{dx}{dt} - 6x = 0\) and the starting values x(0) = 3 and x′(0) = 1.

• \(x' = 4e^{2t} - 3e^{-3t},\quad x'' = 8e^{2t} + 9e^{-3t}\)
• Coefficient of \(e^{2t}\): 8 + (1)(4) + (−6)(2) = 0
• Coefficient of \(e^{-3t}\): 9 + (1)(−3) + (−6)(1) = 0
• x(0) = 2 + 1 = 3 and x′(0) = 4 − 3 = 1

A function satisfies \(\frac{d^2x}{dt^2} + 5\frac{dx}{dt} + 4x = 0\), with x(0) = 5 and x′(0) = −8.

1. (a) Write the equation as a system and give its matrix.

   • \(\frac{dx}{dt} = y,\quad \frac{dy}{dt} = -5y - 4x\)
   • Matrix \(\begin{pmatrix} 0 & 1 \\ -4 & -5 \end{pmatrix}\)

2. (b) Find the eigenvalues and an eigenvector for each.

   • \(\lambda^2 + 5\lambda + 4 = 0\)
   • λ = −1 and −4
   • An eigenvector for λ is (1, λ): (1, −1) and (1, −4)

3. (c) Find the exact solution for x in terms of t.

   • \(x = Ae^{-t} + Be^{-4t}\)
   • x(0): A + B = 5; y(0): −A − 4B = −8
   • A = 4, B = 1
   • \(x = 4e^{-t} + e^{-4t}\)

4. (d) Find x(1).

   • x(1) = 4(0.3679) + 1(0.01832) = 1.49

A function satisfies \(\frac{d^2x}{dt^2} + 2\frac{dx}{dt} + 0.75x = 0\), with x(0) = 3 and x′(0) = −1. Its exact solution is \(x = 3.5e^{-0.5t} - 0.5e^{-1.5t}\).

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

   • \(\frac{dx}{dt} = y,\quad \frac{dy}{dt} = -2y - 0.75x\)
   • h = 1: 4 steps, x_4 = 0.18750
   • h = 0.5: 8 steps, x_8 = 0.35039

2. (b) Find the exact value of x(4).

   • x(4) = 0.47243

3. (c) Find the percentage error of each estimate.

   • h = 1: |0.18750 − 0.47243| ÷ 0.47243 × 100 = 60.3%
   • h = 0.5: |0.35039 − 0.47243| ÷ 0.47243 × 100 = 25.8%
   • The smaller step length has the smaller error

The displacement x cm of the end of a clamped ruler, t seconds after it is flicked, ignoring friction at the clamp, satisfies \(\frac{d^2x}{dt^2} + 6.25x = 0\), with x(0) = 0 and x′(0) = 2.5.

1. (a) Use Euler's method with step length 0.1 to sketch the motion for 0 ≤ t ≤ 4.

   • \(\frac{dx}{dt} = y,\quad \frac{dy}{dt} = -6.25x\)
   • \(x_{n+1} = x_n + 0.1y_n,\quad y_{n+1} = y_n + 0.1(-6.25x_n)\)
   • Graph x against t from the table

2. (b) Estimate the amplitude, using the first peak, and the period, using the first two peaks. Give each to 2 significant figures.

   • First peak: x = 1.22 at t = 0.7
   • Second peak at t = 3.3
   • amplitude ≈ 1.2; period ≈ 3.3 − 0.7 = 2.6

The vertical displacement x mm of a washing machine casing satisfies \(\frac{d^2x}{dt^2} + 0.6\frac{dx}{dt} + 8x = 3\cos(\omega t)\). It starts at rest at x = 0. Use Euler's method with step length 0.1 and technology to find the maximum displacement in the first 8 seconds.

1. (a) Find the maximum displacement when ω = 1.5.

   • \(\frac{dx}{dt} = y,\quad \frac{dy}{dt} = -0.6y - 8x + 3\cos(1.5t)\)
   • Run the table to t = 8 from x_0 = 0, y_0 = 0 (radians)
   • largest x = 1.48 at t = 7.9

2. (b) Find the maximum displacement when ω = 2.8.

   • \(\frac{dx}{dt} = y,\quad \frac{dy}{dt} = -0.6y - 8x + 3\cos(2.8t)\)
   • largest x = 5.82 at t = 7.4

3. (c) Which driving frequency gives the larger motion? The natural frequency is √8 ≈ 2.83.

   • 5.8 is larger than 1.5
   • ω = 2.8 is much closer to the natural frequency √8 ≈ 2.83, so the driving is more in step with the motion

The concentration n mg per litre of a drug in a patient's blood, t hours after a dose, satisfies \(\frac{d^2n}{dt^2} + 0.6\frac{dn}{dt} + 5n = 4\), with n(0) = 1.5 and n′(0) = −8.

1. (a) Use Euler's method with step length 0.1 to describe the behaviour of n for 0 ≤ t ≤ 7.

   • \(\frac{dn}{dt} = v,\quad \frac{dv}{dt} = -0.6v - 5n + 4\)
   • Run the table to t = 7 from n_0 = 1.5, v_0 = −8
   • Lowest value −2.6 at t = 0.8; the next turning points are at t = 2.2, 3.6, 5.0
   • The values oscillate about c/b = 0.80

2. (b) Explain why the model is unsatisfactory.

   • The minimum −2.6 is negative
   • A concentration cannot be negative, so the model fails here

The price of copper on a metals market is modelled by \(\frac{d^2p}{dt^2} + \frac{dp}{dt} + 4p = 14 - 3e^{-0.4t}\), where p is the price in dollars and t is the time in years. The price is now 5 dollars and is falling at 2 dollars per year. Use Euler's method with step length 0.1 to find, to 1 decimal place, according to the model:

1. (a) when to buy

   • \(\frac{dp}{dt} = v,\quad \frac{dv}{dt} = -v - 4p + 14 - 3e^{-0.4t}\)
   • Graph p against t from the table (to t = 20)
   • Lowest price 1.50 at t = 1.3: buy then

2. (b) how long to hold

   • Next peak 4.20 at t = 2.9
   • Hold for 2.9 − 1.3 = 1.6 years

3. (c) the profit per tonne

   • 4.20 − 1.50 = 2.7 dollars

4. (d) the long-term price

   • The graph levels off; at t = 20 the price is 3.50, so about 3.5 dollars

5. (e) Suggest why the model is unlikely to be accurate.

   • The model gives a regular rise and fall to a fixed price, with no effect of news or chance
   • Real prices do not follow a rule like this, so the predicted profit is unlikely to be achieved

Consider \(\frac{d^2x}{dt^2} + \left(\frac{dx}{dt}\right)^2 + x = 4 + 18t^2\), with x(0) = 0 and x′(0) = 0.

1. (a) Use Euler's method with step length 0.1 to estimate x(1) and x(2).

   • \(\frac{dx}{dt} = y,\quad \frac{dy}{dt} = 4 + 18t^2 - y^2 - x\)
   • From the calculator: x_10 = 1.8119, x_20 = 7.6351

2. (b) Show that \(x = 2t^2\) satisfies the equation and the starting values.

   • \(x' = 4t,\quad x'' = 4\)
   • \(4 + (4t)^2 + 2t^2 = 4 + 18t^2\)
   • x(0) = 0 and x′(0) = 0

3. (c) Hence find the percentage error in each estimate in part (a).

   • Exact: x(1) = 2, x(2) = 8
   • x(1): |1.8119 − 2| ÷ 2 × 100 = 9.40%
   • x(2): |7.6351 − 8| ÷ 8 × 100 = 4.56%