Home / A-Level Maths 9709 / Differential equations / Solving by separating the variables
Cambridge International AS & A Level · Topic 3.8
A-Level Maths 9709 / Topic 3.8 / AS & A Level

Sort the variables, then integrate.

Paper 3 only. When the right-hand side of a differential equation splits into a function of \(x\) times a function of \(y\), you can sort the variables onto opposite sides and integrate each one. An initial condition then picks out the one curve you need.

Topic 3.8All differential equations lessons →
The key idea

If \(\dfrac{dy}{dx} = f(x)\,g(y)\), then \(\displaystyle\int \frac{1}{g(y)}\,dy = \int f(x)\,dx\). Add ONE constant after integrating, use the initial condition to find it, and only then rearrange to make \(y\) the subject.

t N 024681012 5101520 N = 20 (0, 4) (2.77, 10) N = 20 ÷ (1 + 4e −t/2 )
Fig. 1: the solution of dN/dt = N(20 − N)/40 with N = 4 when t = 0; it passes N = 10 at t = 4 ln 2 = 2.77 and levels off towards N = 20
Section 01

Separate, integrate, add one constant.

Treat \(\dfrac{dy}{dx}\) as if it were a fraction: divide by the \(y\)-part and 'multiply' by \(dx\), so that each side has only one variable. Then put an integral sign in front of each side.

Worked example A first separable equation
Solve \(\dfrac{dy}{dx} = \dfrac{2x}{y}\), given that \(y = 3\) when \(x = 1\). Find \(y\) when \(x = 3\).
  1. Separate: \(y\,dy = 2x\,dx\), so \(\displaystyle\int y\,dy = \int 2x\,dx\).
  2. Integrate: \(\tfrac{1}{2}y^{2} = x^{2} + C\).
  3. Use \(x = 1\), \(y = 3\): \(\tfrac{9}{2} = 1 + C\), so \(C = \tfrac{7}{2}\).
  4. Rearrange: \(y^{2} = 2x^{2} + 7\). Since \(y = 3 > 0\) at the start, take the positive root: \(y = \sqrt{2x^{2} + 7}\).
  5. At \(x = 3\): \(y = \sqrt{18 + 7} = 5\).
Answer: \(y^{2} = 2x^{2} + 7\); \(y = 5\) when \(x = 3\).
Worked example Exponential growth
A population satisfies \(\dfrac{dP}{dt} = 0.04P\), with \(P = 500\) when \(t = 0\) (\(t\) in years). Find \(P\) in terms of \(t\) and the time for the population to double.
  1. Separate: \(\displaystyle\int \frac{1}{P}\,dP = \int 0.04\,dt\), so \(\ln P = 0.04t + C\) (\(P > 0\)).
  2. \(t = 0\), \(P = 500\): \(C = \ln 500\).
  3. \(\ln P - \ln 500 = 0.04t\), so \(\ln\dfrac{P}{500} = 0.04t\) and \(P = 500e^{0.04t}\).
  4. Double: \(1000 = 500e^{0.04t}\), so \(e^{0.04t} = 2\) and \(t = \dfrac{\ln 2}{0.04}\).
Answer: \(P = 500e^{0.04t}\); doubling time \(25\ln 2 = 17.3\) years (3 s.f.).
Worked example Exponentials: split the power first
Solve \(\dfrac{dy}{dx} = e^{2x - y}\), given that \(y = 0\) when \(x = 0\), giving \(y\) in terms of \(x\).
  1. Split using index laws: \(e^{2x - y} = e^{2x} \times e^{-y}\).
  2. Separate: \(\dfrac{1}{e^{-y}}\,dy = e^{2x}\,dx\), i.e. \(\displaystyle\int e^{y}\,dy = \int e^{2x}\,dx\).
  3. Integrate: \(e^{y} = \tfrac{1}{2}e^{2x} + C\).
  4. \(x = 0\), \(y = 0\): \(1 = \tfrac{1}{2} + C\), so \(C = \tfrac{1}{2}\).
  5. \(e^{y} = \tfrac{1}{2}(e^{2x} + 1)\); take logarithms.
Answer: \(y = \ln\left(\dfrac{e^{2x} + 1}{2}\right)\)
Your turn
Solve \(\dfrac{dy}{dx} = y\cos x\), given that \(y = 2\) when \(x = 0\).
Examiner trap

\(\ln y = 2x + C\) gives \(y = e^{2x + C} = Ae^{2x}\), NOT \(y = e^{2x} + C\). The constant becomes a multiplier when you exponentiate, so find \(C\) BEFORE removing the logarithm, or write \(A = e^{C}\).

Section 02

Particular solutions in context.

In a modelling question the initial condition is usually hidden in words ('initially', 'at the start', 'when \(t = 0\)'), and a second piece of data may fix the constant of proportionality. Answer the actual question at the end: a time, a value, or a long-term behaviour.

Worked example Newton's law of cooling
Tea cools according to \(\dfrac{d\theta}{dt} = -0.05(\theta - 20)\), where \(\theta\) is in °C and \(t\) in minutes. Initially \(\theta = 80\). Find \(\theta\) in terms of \(t\), and the time taken to cool to \(40^\circ\)C.
  1. Separate: \(\displaystyle\int \frac{1}{\theta - 20}\,d\theta = \int -0.05\,dt\).
  2. \(\ln(\theta - 20) = -0.05t + C\) (here \(\theta > 20\)).
  3. \(t = 0\), \(\theta = 80\): \(C = \ln 60\), so \(\ln\dfrac{\theta - 20}{60} = -0.05t\).
  4. \(\theta = 20 + 60e^{-0.05t}\).
  5. \(\theta = 40\): \(60e^{-0.05t} = 20\), so \(e^{-0.05t} = \tfrac{1}{3}\), \(-0.05t = \ln\tfrac{1}{3}\), \(t = 20\ln 3\).
Answer: \(\theta = 20 + 60e^{-0.05t}\); it reaches \(40^\circ\)C after \(20\ln 3 = 22.0\) minutes.
Worked example Finding the constant of proportionality
A liquid cools according to \(\dfrac{d\theta}{dt} = -k(\theta - 20)\). Initially \(\theta = 100\), and after 10 minutes \(\theta = 60\). Find \(k\).
  1. As above: \(\theta - 20 = 80e^{-kt}\) (using \(\theta = 100\) at \(t = 0\)).
  2. \(t = 10\): \(40 = 80e^{-10k}\), so \(e^{-10k} = \tfrac{1}{2}\).
  3. \(-10k = \ln\tfrac{1}{2} = -\ln 2\).
Answer: \(k = \dfrac{\ln 2}{10} = 0.0693\) (3 s.f.)
Examiner trap

Do not drop the 20: \(\theta = 80e^{-0.05t}\) solves a different equation, and it cools to zero instead of to room temperature. Keep \(\theta - 20\) together throughout.

Stage 1 · Learn

Check your understanding

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Which correctly separates \(\dfrac{dy}{dx} = x^{2}y\)?

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What is the general solution of \(\dfrac{dy}{dx} = 3y\)?

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After integrating, \(\ln y = 2x + C\), and \(y = 5\) when \(x = 0\). Which is the particular solution?

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Which differential equation can NOT be solved by separating the variables?

Section 03

Using partial fractions and other integrals.

The integral on the \(y\)-side may need any technique from section 3.5: partial fractions, \(\dfrac{1}{x^{2} + a^{2}}\), trigonometric identities. Logistic growth, where growth slows as the population fills its space, is the standard case (Fig. 1).

Worked example Logistic growth
\(\dfrac{dN}{dt} = \dfrac{N(20 - N)}{40}\), with \(N = 4\) when \(t = 0\). Find \(N\) in terms of \(t\), the time when \(N = 10\), and the long-term value of \(N\).
  1. Separate: \(\displaystyle\int \frac{40}{N(20 - N)}\,dN = \int 1\,dt\).
  2. Partial fractions: \(\dfrac{40}{N(20 - N)} = \dfrac{2}{N} + \dfrac{2}{20 - N}\) (check: \(2(20 - N) + 2N = 40\)).
  3. Integrate: \(2\ln N - 2\ln(20 - N) = t + C\), i.e. \(2\ln\dfrac{N}{20 - N} = t + C\).
  4. \(t = 0\), \(N = 4\): \(C = 2\ln\tfrac{4}{16} = -2\ln 4\). So \(2\ln\dfrac{4N}{20 - N} = t\).
  5. \(\dfrac{4N}{20 - N} = e^{t/2}\), so \(4N = 20e^{t/2} - Ne^{t/2}\) and \(N = \dfrac{20e^{t/2}}{4 + e^{t/2}} = \dfrac{20}{1 + 4e^{-t/2}}\).
  6. \(N = 10\): \(\dfrac{40}{10} = e^{t/2}\), so \(t = 2\ln 4 = 4\ln 2\).
  7. As \(t \to \infty\), \(e^{-t/2} \to 0\), so \(N \to 20\).
Answer: \(N = \dfrac{20}{1 + 4e^{-t/2}}\); \(N = 10\) at \(t = 4\ln 2 = 2.77\); \(N\) tends to 20.
Your turn
Solve \(\dfrac{dy}{dx} = x(y^{2} + 1)\), given that \(y = 1\) when \(x = 0\).
Examiner trap

\(\displaystyle\int \frac{1}{20 - N}\,dN = -\ln(20 - N)\), not \(+\ln(20 - N)\): the inside has derivative \(-1\). Losing this sign turns the logistic curve into nonsense.

Stage 2 · Exam

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Solve \(\dfrac{dy}{dx} = 3x^{2}y^{2}\), given that \(y = 1\) when \(x = 0\).

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A liquid cools according to \(\dfrac{d\theta}{dt} = -k(\theta - 20)\). Initially \(\theta = 100\), and \(\theta = 60\) when \(t = 10\). Find \(k\).

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\(\dfrac{dx}{dt} = kx(5 - x)\). After separating, which is a correct result of integrating?

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Given \(\dfrac{dy}{dx} = e^{x + 2y}\) and \(y = 0\) when \(x = 0\), find \(y\) when \(x = \ln 1.25\).

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Variables separated, equations solved.

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Six original Cambridge-style questions
General and particular solutions, an inverse tangent, cooling to a target temperature, logistic growth with partial fractions, and a draining tank.