Practice Maths

← Rates of Change and Differential Equations › Applications of Differential Equations › Solutions

Applications of Differential Equations — Full Worked Solutions

  1. Bacterial culture: N(0) = 500, N(2) = 2000. Find N(5). Fluency

    Set up DE: dN/dt = kN, a separable equation with solution N(t) = N0ekt.

    Apply N(0) = 500: N0 = 500, so N(t) = 500ekt.

    Find k using N(2) = 2000:

    2000 = 500e2k ⇒ e2k = 4 ⇒ 2k = ln4 ⇒ k = ln2.

    Find N(5):

    N(5) = 500e5 ln2 = 500 × 25 = 500 × 32 = 16,000 bacteria

  2. Half-life 10 years: fraction remaining after 25 years. Fluency

    DE: dN/dt = kN. Half-life T1/2 = 10 years ⇒ k = −ln2/10.

    Solution: N(t) = N0 e(−ln2/10)t = N0 × 2−t/10.

    At t = 25:

    N/N0 = 2−25/10 = 2−5/2 = 1/(4√2) = √2/8 ≈ 0.1768.

    Approximately 17.7% remains after 25 years.

    Intuition: After 10 years, half remains; after 20 years, a quarter; after 25 years, between a quarter and an eighth.

  3. Newton’s cooling: body at 25°C, room at 20°C, cools to 22°C in 1 hour. Find k. Fluency

    DE: dT/dt = −k(T − 20).

    Solution: T(t) = 20 + (T0 − 20)e−kt = 20 + 5e−kt.

    Apply T(1) = 22:

    22 = 20 + 5e−k ⇒ 5e−k = 2 ⇒ e−k = 2/5.

    k = −ln(2/5) = ln(5/2).

    k = ln(2.5) ≈ 0.916 per hour

  4. Logistic DE with K = 500, r = 0.2, P(0) = 50. State DE and solution. Fluency

    Differential equation: dP/dt = 0.2P(1 − P/500).

    Equilibrium solutions: P = 0 (unstable) and P = 500 (stable).

    Constant A: A = (K − P0)/P0 = (500 − 50)/50 = 450/50 = 9.

    Solution: P(t) = 500 / (1 + 9e−0.2t)

    As t → ∞, e−0.2t → 0 so P → 500 = K, confirming the carrying capacity is the stable equilibrium.

  5. Tea at 95°C, room at 22°C; 65°C after 10 min. Find T after 30 min. Understanding

    Solution form: T(t) = 22 + 73e−kt.

    Find k using T(10) = 65:

    65 = 22 + 73e−10k ⇒ e−10k = 43/73.

    Find T(30):

    T(30) = 22 + 73e−30k = 22 + 73(e−10k)3 = 22 + 73(43/73)3.

    = 22 + 433/732 = 22 + 79507/5329 ≈ 22 + 14.92.

    T(30) ≈ 36.9°C

    Note the convenient cancellation when using T(10) as a ratio — no need to find k explicitly.

  6. Mixing: 200 L tank, brine 0.05 kg/L at 4 L/min in and out. Find Q(t) and limiting amount. Understanding

    Set up DE:

    Rate in = 0.05 kg/L × 4 L/min = 0.2 kg/min.

    Rate out = [Q(t)/200] kg/L × 4 L/min = Q/50 kg/min.

    dQ/dt = 0.2 − Q/50.

    Standard form: dQ/dt + (1/50)Q = 0.2.

    Integrating factor: μ = e∫(1/50) dt = et/50.

    Integrate:

    d/dt[et/50Q] = 0.2et/50.

    et/50Q = 0.2 × 50 et/50 + C = 10et/50 + C.

    Q = 10 + Ce−t/50.

    Apply Q(0) = 0: 0 = 10 + C ⇒ C = −10.

    Q(t) = 10(1 − e−t/50) kg

    As t → ∞: Q → 10 kg (= 0.05 kg/L × 200 L, the equilibrium amount at brine concentration).

  7. Logistic: K = 800, r = 0.3, P(0) = 200. Find t when P = 600. Understanding

    Solution: A = (800 − 200)/200 = 3.   P(t) = 800/(1 + 3e−0.3t).

    Set P = 600:

    800/(1 + 3e−0.3t) = 600.

    1 + 3e−0.3t = 4/3.

    3e−0.3t = 1/3 ⇒ e−0.3t = 1/9.

    −0.3t = −ln9 ⇒ t = ln9/0.3 = 2ln3/0.3 = 20ln3/3.

    t = (20 ln 3)/3 ≈ 7.32 years

    Note: P = 600 = 3K/4 > K/2 = 400, so this is past the inflection point of the logistic curve.

  8. RL circuit: L = 1, R = 2, V = 4e−t, I(0) = 1. Find I(t). Understanding

    DE: dI/dt + 2I = 4e−t.   P(t) = 2.

    Integrating factor: μ = e2t.

    Multiply: d/dt[e2tI] = 4e−te2t = 4et.

    Integrate: e2tI = 4et + C ⇒ I = 4e−t + Ce−2t.

    Apply I(0) = 1: 1 = 4 + C ⇒ C = −3.

    I(t) = 4e−t − 3e−2t amperes

    Both terms decay to zero as t → ∞, which makes physical sense since the applied voltage also decays.

  9. Forensic cooling: room 18°C, body 32°C at 9:00 am, 28°C at 10:00 am. Estimate time of death. Problem Solving

    Let t = 0 at time of death. Normal body temperature = 37°C.

    Solution: T(t) = 18 + 19e−kt.

    Let 9:00 am correspond to t = t1 and 10:00 am to t = t1 + 1.

    Equations:

    19e−kt1 = 14   … (i)

    19e−k(t1+1) = 10   … (ii)

    Divide (ii) by (i): e−k = 10/14 = 5/7.

    k = ln(7/5) ≈ 0.3365 per hour.

    Find t1 from (i):

    e−0.3365 t1 = 14/19 ⇒ t1 = −ln(14/19)/0.3365 = ln(19/14)/ln(7/5).

    ln(19/14) ≈ 0.3054,   ln(7/5) ≈ 0.3365.

    t1 ≈ 0.3054/0.3365 ≈ 0.908 hours ≈ 54.5 minutes before 9:00 am.

    Estimated time of death: approximately 8:05–8:06 am

  10. Carbon-14 dating: 62% remaining. Half-life 5730 years. Find age. Problem Solving

    DE: dN/dt = −(ln2/5730)N.   Solution: N(t) = N0 × 2−t/5730.

    Set N/N0 = 0.62:

    2−t/5730 = 0.62.

    Take logarithm:

    −(t/5730) ln2 = ln(0.62).

    t = −5730 × ln(0.62) / ln2.

    Calculate:

    ln(0.62) ≈ −0.4780,   ln2 ≈ 0.6931.

    t = 5730 × 0.4780 / 0.6931 ≈ 5730 × 0.6897 ≈ 3952.

    Age of sample ≈ 3952 years

    This is less than one half-life (5730 years), which is consistent with 62% > 50% remaining.