The variables x and y satisfy the differential equation (x + 1)(3x + 1) dy/dx = y, and it is given that y = 1 when x = 1.

Mark scheme

Marking points from the official mark scheme (Feb/Mar 2022). The mark or mark type is shown in brackets.

1. [1] Correctly separate variables and integrate at least one side
2. [1] Obtain term ln y from integral of 1/y
3. [1] State or imply the form A/(x+1) + B/(3x+1) and use a correct method to find a constant
4. [1] Obtain A = –½ and B = ¾
5. [1] Obtain term –½ ln(x+1) or –½ ln(2x+2)
6. [1] Obtain term ¾ ln(3x+1) or ¾ ln(6x+2)
7. [1] Use x = 1 and y = 1 to evaluate a constant, or expression for a constant, (decimal equivalent of ln terms allowed) or as limits, in a solution containing terms a ln y, b ln(x+1) and c ln(3x+1), where abc ≠ 0
8. [1] Obtain an expression for y or y² and substitute x = 3
9. [1] Obtain answer y = ½√(5/4) or ½√(10/8) or simplified equivalent