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

Separable Differential Equations — Full Worked Solutions

  1. Solve dy/dx = 3y. Fluency

    Step 1 — Identify: This is separable with f(x) = 3, g(y) = y.

    Step 2 — Separate: dy/y = 3 dx

    Step 3 — Integrate both sides: ∫ dy/y = ∫ 3 dx

    ln|y| = 3x + C

    Step 4 — Solve for y: |y| = e3x + C = eCe3x

    Writing A = ±eC (arbitrary nonzero constant):

    y = Ae3x

    This is the general solution. A = y(0) gives the initial value.

  2. Solve dy/dx = x/y with initial condition y(0) = 2. Fluency

    Step 1 — Separate: Multiply both sides by y and by dx:

    y dy = x dx

    Step 2 — Integrate: ∫ y dy = ∫ x dx

    y²/2 = x²/2 + C

    Multiply by 2: y² = x² + K where K = 2C.

    Step 3 — Apply IC y(0) = 2: 4 = 0 + K, so K = 4.

    y² = x² + 4.

    Step 4 — Solve for y: y = √(x² + 4) (positive root since y(0) = 2 > 0).

    Particular solution: y = √(x² + 4)

  3. Solve dy/dx = xy². Fluency

    Step 1 — Separate: Divide both sides by y² and multiply by dx:

    y−2 dy = x dx

    Step 2 — Integrate: ∫ y−2 dy = ∫ x dx

    −y−1 = x²/2 + C

    That is: −1/y = x²/2 + C

    Step 3 — Solve for y:

    1/y = −x²/2 − C

    y = −1/(x²/2 + C) = −2/(x² + K) where K = 2C.

    Note: y = 0 is a singular solution (lost when we divided by y²).

  4. A population satisfies dP/dt = 0.05P with P(0) = 1000. Find P(t). Fluency

    Step 1 — Recognise: dP/dt = kP with k = 0.05. This is exponential growth.

    Step 2 — Separate: dP/P = 0.05 dt

    Step 3 — Integrate: ln P = 0.05t + C (P > 0 so |P| = P)

    Step 4 — General solution: P = Ae0.05t

    Step 5 — Apply IC: P(0) = A = 1000.

    P(t) = 1000e0.05t

    Doubling time: t = ln 2 / 0.05 ≈ 13.86 units.

  5. Solve dy/dx = (x + 1)/(y − 1). Understanding

    Step 1 — Separate: (y − 1) dy = (x + 1) dx

    Step 2 — Integrate both sides:

    ∫ (y − 1) dy = ∫ (x + 1) dx

    (y − 1)²/2 = (x + 1)²/2 + C

    Step 3 — General solution (implicit):

    (y − 1)² = (x + 1)² + K where K = 2C.

    This represents a family of hyperbolas shifted to centre (−1, 1).

  6. Solve dy/dx = ex − y. Understanding

    Step 1 — Rewrite using index laws: ex − y = ex × e−y.

    So dy/dx = ex × e−y (separable: f(x) = ex, g(y) = e−y).

    Step 2 — Separate: Multiply both sides by ey:

    ey dy = ex dx

    Step 3 — Integrate: ey = ex + C

    Step 4 — Solve for y: y = ln(ex + C)

    y = ln(ex + C)   (valid for ex + C > 0)

  7. Newton’s cooling: dT/dt = −k(T − 20). Coffee starts at 90°C and cools to 60°C after 5 min. Find T(t) and T(10). Understanding

    Step 1 — Substitute u = T − 20: du/dt = dT/dt = −ku. This is dP/dt = kP with k replaced by −k.

    Step 2 — Solve: u = Ae−kt ⇒ T = 20 + Ae−kt.

    Step 3 — Apply T(0) = 90: 90 = 20 + A ⇒ A = 70. So T = 20 + 70e−kt.

    Step 4 — Apply T(5) = 60:

    60 = 20 + 70e−5k ⇒ e−5k = 40/70 = 4/7

    ⇒ k = −(1/5) ln(4/7) = (1/5) ln(7/4) ≈ 0.1120 min−1

    T(t) = 20 + 70e−kt with k = (1/5)ln(7/4).

    Step 5 — Find T(10):

    e−10k = (e−5k)² = (4/7)² = 16/49

    T(10) = 20 + 70 × (16/49) = 20 + 1120/49 ≈ 20 + 22.86 ≈ 42.9°C

  8. Solve dy/dx = y sin(x) with y(0) = 2. Understanding

    Step 1 — Separate: dy/y = sin(x) dx

    Step 2 — Integrate: ∫ dy/y = ∫ sin(x) dx

    ln|y| = −cos(x) + C

    Step 3 — General solution: y = Ae−cos(x)

    Step 4 — Apply y(0) = 2:

    2 = Ae−cos(0) = Ae−1 = A/e ⇒ A = 2e

    y = 2e × e−cos(x) = 2e1 − cos(x)

    Check: y(0) = 2e1−1 = 2e0 = 2 ✓

  9. Logistic growth: dP/dt = 0.002P(1000 − P) with P(0) = 100. Find P(t). Problem Solving

    Step 1 — Separate: dP / [P(1000 − P)] = 0.002 dt

    Step 2 — Partial fractions: 1/[P(1000−P)] = (1/1000)(1/P + 1/(1000−P))

    Verify: (1/1000) × [(1000−P + P)/(P(1000−P))] = 1/[P(1000−P)] ✓

    Step 3 — Integrate:

    (1/1000)[ln|P| − ln|1000 − P|] = 0.002t + C

    ln|P/(1000−P)| = 2t + K

    Step 4: P/(1000−P) = Ae2t

    Step 5 — Apply P(0) = 100: 100/900 = A ⇒ A = 1/9

    P/(1000−P) = e2t/9

    9P = e2t(1000−P) ⇒ P(9 + e2t) = 1000e2t

    P(t) = 1000e2t / (9 + e2t) = 1000 / (1 + 9e−2t)

    As t → ∞, P → 1000 (the carrying capacity).

  10. Salt-tank problem: 50 L tank, 5 kg salt, pure water in at 2 L/min, drains at 2 L/min. Find when x = 1 kg. Problem Solving

    Step 1 — Set up DE:

    Rate in = 2 L/min × 0 kg/L = 0

    Rate out = 2 L/min × (x/50) kg/L = x/25 kg/min

    dx/dt = 0 − x/25 = −x/25

    Step 2 — Separate: dx/x = −dt/25

    Step 3 — Integrate: ln x = −t/25 + C (x > 0)

    Step 4 — Solve: x = Ae−t/25

    Step 5 — Apply IC x(0) = 5: A = 5.

    x(t) = 5e−t/25

    Step 6 — Find when x = 1:

    1 = 5e−t/25 ⇒ e−t/25 = 1/5 ⇒ −t/25 = −ln 5

    t = 25 ln 5 ≈ 40.2 minutes