The variables x and y satisfy the differential equation dy/dx = (y² + 4)/(x(y + 4)) for x > 0. It is given that x = 4 when y = 2√3. Solve the differential equation to obtain the value of x when y = 2.

Mark scheme

Marking points from the official mark scheme (May/June 2023). The mark or mark type is shown in brackets.

1. [B1] Separate the variables correctly; (y+4)/(y²+4) dy = 1/x dx.
2. [B1] Obtain lnx
3. [M1*] Split the fraction and integrate to obtain pln(y² + 4) or q tan⁻¹(y/2) correctly; Only following subdivision into y/(y²+4) + 4/(y²+4). If no subdivision seen then both terms pln(y²+4) and q tan⁻¹(y/2) must be present.
4. [A1] Obtain (1/2)ln(y² + 4)
5. [A1] Obtain 2tan⁻¹(y/2)
6. [DM1] Use (4, 2√3) in an expression containing at least 2 of a lnx, b ln (y² + 4) and c tan⁻¹(y/2) to obtain constant of integration; Allow one sign or arithmetic error e.g. 2π/3. May use (4, 2√3) and (x, 2) as limits to find x for the final 3 marks.
7. [A1] Correct solution (any form); e.g. (1/2)ln(y²+4) + 2tan⁻¹(y/2) = lnx + 2π/3 or (1/2)ln(x²+4)+2tan⁻¹(y/2) = lnx+2tan⁻¹(√3)+1n16-1n4.
8. [A1] Obtain √8e^(π/6) or 1.68 or more accurate or 2√2e^(π/6) or (√8)/e^(π/6) or e^(0.516); ISW Must remove ln so x = e^(ln(2√2)-π/6) A0. Alternative method for first *M1 A1 A1: [M1*] ∫(p/(y²+4) + q/(y²+4)) dy; Allow sign error. [A1] ∫(y/(y²+4))dy = (1/2)ln(y²+4) [A1] ∫(4/(y²+4))dy = 2tan⁻¹(y/2).