Mixed practice: second order equations, exact solutions and models Problem set A Worked solutions

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

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

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

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

   • \(\frac{dx}{dt} = y,\quad \frac{dy}{dt} = -3y + 4x\)
   • Matrix \(\begin{pmatrix} 0 & 1 \\ 4 & -3 \end{pmatrix}\)

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

   • \(\lambda^2 + 3\lambda - 4 = 0\)
   • 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 = 3; y(0): A − 4B = −2
   • A = 2, B = 1
   • \(x = 2e^{t} + e^{-4t}\)

4. (d) Find x(1).

   • x(1) = 2(2.718) + 1(0.01832) = 5.45

A function satisfies \(\frac{d^2x}{dt^2} + 1.5\frac{dx}{dt} + 0.5x = 0\), with x(0) = 2 and x′(0) = 0. Its exact solution is \(x = 4e^{-0.5t} - 2e^{-t}\).

1. (a) Use Euler's method to estimate x(3) with step length 1 and with step length 0.5.

   • \(\frac{dx}{dt} = y,\quad \frac{dy}{dt} = -1.5y - 0.5x\)
   • h = 1: 3 steps, x_3 = 0.50000
   • h = 0.5: 6 steps, x_6 = 0.68066

2. (b) Find the exact value of x(3).

   • x(3) = 0.79295

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

   • h = 1: |0.50000 − 0.79295| ÷ 0.79295 × 100 = 36.9%
   • h = 0.5: |0.68066 − 0.79295| ÷ 0.79295 × 100 = 14.2%
   • The smaller step length has the smaller error

The displacement x cm of the tip of a tuning fork, t seconds after it is struck, ignoring air resistance, satisfies \(\frac{d^2x}{dt^2} + 9x = 0\), with x(0) = 0 and x′(0) = 3.

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

   • \(\frac{dx}{dt} = y,\quad \frac{dy}{dt} = -9x\)
   • \(x_{n+1} = x_n + 0.1y_n,\quad y_{n+1} = y_n + 0.1(-9x_n)\)
   • 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 = 1.27 at t = 0.6
   • Second peak at t = 2.7
   • amplitude ≈ 1.3; period ≈ 2.7 − 0.6 = 2.1

The displacement x cm of a loudspeaker cone satisfies \(\frac{d^2x}{dt^2} + 0.4\frac{dx}{dt} + 5x = 2\sin(\omega t)\). 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.

   • \(\frac{dx}{dt} = y,\quad \frac{dy}{dt} = -0.4y - 5x + 2\sin t\)
   • Run the table to t = 8 from x_0 = 0, y_0 = 0 (radians)
   • largest x = 0.827 at t = 7.8

2. (b) Find the maximum displacement when ω = 2.2.

   • \(\frac{dx}{dt} = y,\quad \frac{dy}{dt} = -0.4y - 5x + 2\sin(2.2t)\)
   • largest x = 3.80 at t = 7.2

3. (c) Which driving frequency gives the larger motion? The natural frequency is √5 ≈ 2.24.

   • 3.8 is larger than 0.83
   • ω = 2.2 is much closer to the natural frequency √5 ≈ 2.24, so the driving is more in step with the motion

The number n thousand of bacteria in a dish, t hours after an antibiotic is added, satisfies \(\frac{d^2n}{dt^2} + 1.2\frac{dn}{dt} + 9n = 6\), with n(0) = 2 and n′(0) = −12.

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

   • \(\frac{dn}{dt} = v,\quad \frac{dv}{dt} = -1.2v - 9n + 6\)
   • Run the table to t = 7 from n_0 = 2, v_0 = −12
   • Lowest value −3.0 at t = 0.6; the next turning points are at t = 1.6, 2.7, 3.7
   • The values oscillate about c/b = 0.67

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

   • The minimum −3.0 is negative
   • The number of bacteria cannot be negative, so the model fails here

The price of shares in an engineering company is modelled by \(\frac{d^2p}{dt^2} + 0.8\frac{dp}{dt} + 3p = 12 - 2e^{-0.5t}\), where p is the price in dollars and t is the time in years. The price is now 6 dollars and is falling at 1.8 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

   • \(\frac{dp}{dt} = v,\quad \frac{dv}{dt} = -0.8v - 3p + 12 - 2e^{-0.5t}\)
   • Graph p against t from the table (to t = 20)
   • Lowest price 1.86 at t = 1.6: buy then

2. (b) how long to hold

   • Next peak 5.02 at t = 3.4
   • Hold for 3.4 − 1.6 = 1.8 years

3. (c) the profit per share

   • 5.02 − 1.86 = 3.2 dollars

4. (d) the long-term price

   • The graph levels off; at t = 20 the price is 3.99, so about 4.0 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 \(\frac{d^2x}{dt^2} + \left(\frac{dx}{dt}\right)^2 + x = 2 + 5t^2\), 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).

   • \(\frac{dx}{dt} = y,\quad \frac{dy}{dt} = 2 + 5t^2 - y^2 - x\)
   • From the calculator: x_10 = 0.90813, x_20 = 3.8298

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

   • \(x' = 2t,\quad x'' = 2\)
   • \(2 + (2t)^2 + t^2 = 2 + 5t^2\)
   • x(0) = 0 and x′(0) = 0

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

   • Exact: x(1) = 1, x(2) = 4
   • x(1): |0.90813 − 1| ÷ 1 × 100 = 9.19%
   • x(2): |3.8298 − 4| ÷ 4 × 100 = 4.25%