The variables x and y satisfy the differential equation (x²+1) dy/dx = kxe^(-2y), where k is a constant. It is given that y = 0 when x = 0 and that y = -1/2 when x = 1.

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2025). The mark or mark type is shown in brackets.

1. [B1] Separate variables correctly
2. [B1] Obtain term Ae^(-2y)
3. [B1] Obtain term B ln(x²+1)
4. [B1] Obtain -1/2 e^(-2y) = 1/2 k ln(x²+1) (+C)
5. [M1] Use y = 0 when x = 0 to evaluate c or as limits in a solution containing terms of the form e^(-2y) and ln(x²+1)
6. [M1] Use y = -1/2 when x = 1 to evaluate k or as limits in a solution containing terms of the form e^(-2y) and ln(x²+1)
7. [A1] Obtain c = 1/2 and k = (1-e)/ln 2
8. [A1] Obtain final answer y = -1/2 ln(2-1)