(a) The variables x and y satisfy the differential equation (dy/dx) = (4 + 9y²)/e^(2x+1). It is given that y = 0 when x = 1.

Mark scheme

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

1. B1 Separate variables correctly [∫(1/(4+9y²)) dy = ∫e^-(2x+1)dx. Condone missing integral signs or dx and dy missing.]
2. B1 Obtain term -½e^-(2x+1) [OE e.g. -½e^(-2x)/e.]
3. M1 Obtain term of the form a tan⁻¹(by) [or a tan⁻¹(y/b)]
4. A1 Obtain term (1/6)tan⁻¹(3y/2) [OE e.g. (1/9)x(3/2)tan⁻¹(3y/2).]
5. M1 Use x = 1, y = 0 to evaluate a constant or as limits in a solution containing or derived from terms of the form a tan⁻¹(by) and ce^±(2x+1) [If they rearrange before evaluating the constant, the constant must be of the correct form.]
6. A1 Obtain correct answer in any form [e.g. (1/6)tan⁻¹(3y/2) = -½e^-(2x+1) - (1/6)tan⁻¹(0) = -½e^-3.]
7. A1 Obtain final answer y = (2/3)tan(3e^-3 - 3e^-(2x+1)) [OE Allow with 3e^-3 = 0.149....]
8. B1 FT State that y approaches (2/3)tan(3e^-3) [Or exact equivalent. The FT is on correct work on a solution containing e^-(2x+1). Condone y = .... Accept correct answer stated with minimal wording. 0.10032... is not exact so B0.]