A-Level Mathematics May/June 2024 Q11(b): Using partial fractions, solve the differential equation and obtain an expression for t…

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  The differential equation to be solved is $\frac{dx}{dt} = \frac{1}{1495}x(300-x)$.

  Step 1: Separate the variables.

  Separate the terms involving $x$ and $t$ to opposite sides of the equation.

  $\int \frac{1}{x(300-x)} \, dx = \int \frac{1}{1495} \, dt$

  Step 2: Integrate the right-hand side.

  $\int \frac{1}{1495} \, dt = \frac{1}{1495}t + C$

  Step 3: Use partial fractions for the left-hand side.

  Let the integrand be expressed in partial fractions:

  $\frac{1}{x(300-x)} \equiv \frac{A}{x} + \frac{B}{300-x}$

  Multiply by $x(300-x)$ to get the identity:

  $1 \equiv A(300-x) + Bx$

  To find the constants $A$ and $B$:

  Let $x=0$: $1 = A(300) \implies A = \frac{1}{300}$

  Let $x=300$: $1 = B(300) \implies B = \frac{1}{300}$

  So, the integral becomes:

  $\int \left( \frac{1/300}{x} + \frac{1/300}{300-x} \right) \, dx$

  Step 4: Integrate the left-hand side.

  $\frac{1}{300} \int \left( \frac{1}{x} + \frac{1}{300-x} \right) \, dx = \frac{1}{300} (\ln|x| - \ln|300-x|)$

  Step 5: Combine the integrated parts and find the constant of integration.

  The general solution is:

  $\frac{1}{300}(\ln x - \ln(300-x)) = \frac{1}{1495}t + C$

  Given the initial condition $t=0$ when $x=1$:

  $\frac{1}{300}(\ln 1 - \ln(300-1)) = \frac{1}{1495}(0) + C$

  $\frac{1}{300}(0 - \ln 299) = C$

  $C = -\frac{1}{300}\ln 299$

  Step 6: Substitute C and rearrange to find t.

  Substitute the value of $C$ back into the general solution:

  $\frac{1}{300}(\ln x - \ln(300-x)) = \frac{1}{1495}t - \frac{1}{300}\ln 299$

  Now, rearrange to make $t$ the subject. First, isolate the term with $t$:

  $\frac{1}{1495}t = \frac{1}{300}(\ln x - \ln(300-x)) + \frac{1}{300}\ln 299$

  Factor out $\frac{1}{300}$ and use logarithm laws:

  $\frac{1}{1495}t = \frac{1}{300} [(\ln x - \ln(300-x)) + \ln 299]$

  $\frac{1}{1495}t = \frac{1}{300} [\ln\left(\frac{x}{300-x}\right) + \ln 299]$

  $\frac{1}{1495}t = \frac{1}{300} \ln\left(\frac{299x}{300-x}\right)$

  Finally, solve for $t$:

  $t = \frac{1495}{300} \ln\left(\frac{299x}{300-x}\right)$

  Simplify the fraction $\frac{1495}{300} = \frac{299 \times 5}{60 \times 5} = \frac{299}{60}$.

  $t = \frac{299}{60} \ln\left(\frac{299x}{300-x}\right)$

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  How the marks are awarded

  • B1 — Correctly separating the variables to get ∫ (1 / (x(300-x))) dx = ∫ k dt (where k is a constant, here 1/1495).
  • B1 — Correctly integrating the right-hand side, ∫ (1/1495) dt, to obtain (1/1495)t.
  • B1 — Stating or implying the correct form for the partial fractions, A/x + B/(300-x).
  • M1 — Using a correct method to find the constants A and B, such as substituting values for x (e.g., x=0 and x=300) into the identity 1 ≡ A(300-x) + Bx.
  • A1 — Correctly integrating the partial fractions to get the terms (1/300)ln(x) - (1/300)ln(300-x). The minus sign from integrating 1/(300-x) is crucial.
  • M1 — Using the initial conditions t=0, x=1 in an equation containing ln(x), ln(300-x), and t terms to find the constant of integration, C.
  • A1 — Obtaining a correct equation relating x and t in any form, for example, (1/300)(ln x - ln(300-x)) = (1/1495)t - (1/300)ln(299).
  • M1 — Correctly using logarithm laws twice: first to combine ln(x) - ln(300-x) and then to combine the result with ln(299).
  • A1 — Obtaining the final correct expression for t as a single logarithm, t = (299/60) ln(299x/(300-x)), including the simplified constant.

  Common mistakes

  • A sign error when integrating 1/(300-x), resulting in +ln(300-x) instead of the correct -ln(300-x). This is a frequent chain rule mistake.
  • Incorrectly handling the constant of integration, either by forgetting it entirely or by making an error when combining it with the logarithmic terms, e.g., writing ln(A) + C = ln(A+C).
  • Errors in algebraic rearrangement when making t the subject, particularly when dealing with the multiple constants and logarithmic terms.
  • Making a slip in the partial fraction calculation, for example by finding incorrect values for A or B, which then carries through the rest of the problem.

  Examiner tip: Solving complex differential equations often involves a chain of standard techniques; master each one individually—variable separation, partial fractions, and logarithm laws—so you can apply them confidently in sequence.

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