A-Level Mathematics May/June 2025 Q8: The variables x and θ satisfy the differential equation sin 2θ dx/dθ = (4x+3) cos 2θ, a…

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  The given differential equation is:

  $sin 2\theta \frac{dx}{d\theta} = (4x+3) cos 2\theta$

  Step 1: Separate the variables.

  We rearrange the equation to group terms involving $x$ on one side and terms involving $\theta$ on the other.

  $\frac{1}{4x+3} dx = \frac{cos 2\theta}{sin 2\theta} d\theta$

  Step 2: Integrate both sides.

  $\int \frac{1}{4x+3} dx = \int \frac{cos 2\theta}{sin 2\theta} d\theta$

  The left-hand side (LHS) integrates to:

  $\int \frac{1}{4x+3} dx = \frac{1}{4} \ln(4x+3)$

  The right-hand side (RHS) is of the form $\int \frac{f'(u)}{f(u)} du$. Let $u = \sin 2\theta$, then $\frac{du}{d\theta} = 2\cos 2\theta$. The integral becomes $\int \frac{1}{2u} du = \frac{1}{2}\ln(u)$.

  So, $\int \frac{cos 2\theta}{sin 2\theta} d\theta = \frac{1}{2} \ln(\sin 2\theta)$.

  Combining the integrated sides gives the general solution:

  $\frac{1}{4} \ln(4x+3) = \frac{1}{2} \ln(\sin 2\theta) + C$

  Step 3: Use the initial condition to find the constant C.

  We are given that $x = 0$ when $\theta = \frac{1}{12}\pi$. First, calculate $2\theta = 2 \times \frac{1}{12}\pi = \frac{\pi}{6}$.

  Substitute the values into the general solution:

  $\frac{1}{4} \ln(4(0)+3) = \frac{1}{2} \ln(\sin(\frac{\pi}{6})) + C$

  $\frac{1}{4} \ln(3) = \frac{1}{2} \ln(\frac{1}{2}) + C$

  Now, solve for C:

  $C = \frac{1}{4} \ln(3) - \frac{1}{2} \ln(\frac{1}{2})$

  Step 4: Substitute C back and solve for x.

  $\frac{1}{4} \ln(4x+3) = \frac{1}{2} \ln(\sin 2\theta) + \frac{1}{4} \ln(3) - \frac{1}{2} \ln(\frac{1}{2})$

  To simplify, multiply the entire equation by 4:

  $\ln(4x+3) = 2 \ln(\sin 2\theta) + \ln(3) - 2 \ln(\frac{1}{2})$

  Using logarithm laws ($n \ln a = \ln a^n$ and $\ln a + \ln b = \ln ab$, $\ln a - \ln b = \ln(a/b)$):

  $\ln(4x+3) = \ln((\sin 2\theta)^2) + \ln(3) - \ln((\frac{1}{2})^2)$

  $\ln(4x+3) = \ln(\sin^2 2\theta) + \ln(3) - \ln(\frac{1}{4})$

  $\ln(4x+3) = \ln\left(\frac{3 \sin^2 2\theta}{1/4}\right)$

  $\ln(4x+3) = \ln(12 \sin^2 2\theta)$

  Exponentiate both sides to remove the logarithms:

  $4x+3 = 12 \sin^2 2\theta$

  Finally, make $x$ the subject:

  $4x = 12 \sin^2 2\theta - 3$

  $x = \frac{12 \sin^2 2\theta - 3}{4}$

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  How the marks are awarded

  • B1 — Correctly separating the variables to get ∫(1/(4x+3))dx = ∫(cos2θ/sin2θ)dθ.
  • B1 — Correctly integrating the x-term to obtain 1/4 ln(4x+3).
  • M1 — Integrating the θ-term to obtain a result of the form A ln(sin2θ), demonstrating understanding of the integration method for cot(2θ).
  • A1 — Obtaining the correct θ-term integral, which is 1/2 ln(sin2θ).
  • M1 — Substituting the initial conditions x = 0 and θ = π/12 into an equation containing ln(4x+3), ln(sin2θ), and a constant of integration to find the constant.
  • A1 — Obtaining a correct expression for the constant or a correct equation after substitution, such as (1/4)ln(3) = (1/2)ln(1/2) + C.
  • A1 — Correctly manipulating the logarithmic equation and rearranging to obtain the final answer for x in the required form, x = (12sin²2θ-3)/4.

  Common mistakes

  • Forgetting the factor of 1/4 when integrating 1/(4x+3) or the factor of 1/2 when integrating cos(2θ)/sin(2θ), due to overlooking the chain rule.
  • Making errors with logarithm laws during the final rearrangement, such as incorrectly combining terms, e.g., ln(A) + ln(B) = ln(A+B), or mishandling coefficients, e.g., 2ln(A) = (ln A)², instead of ln(A²).
  • Incorrectly evaluating sin(2θ) for θ = π/12. Common errors include forgetting to double θ to get π/6, or using degrees instead of radians, leading to an incorrect value for the constant C.
  • Placing the constant of integration incorrectly, for example, writing ln(4x+3+C) instead of ln(4x+3) + C, which makes it impossible to solve correctly.

  Examiner tip: Mastering separation of variables requires correctly isolating x and θ terms, and being meticulous with the chain rule during integration and with logarithm laws during algebraic rearrangement.

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