The variables y and θ satisfy the differential equation `(1+y)(1+cos2θ) dy/dθ = e^(3y)`. It is given that y = 0 when θ = (1/2)π. Solve the differential equation and find the exact value of tanθ when y = 1.

Mark scheme

Marking points from the official mark scheme (Feb/Mar 2024). The mark or mark type is shown in brackets.

1. [B1] Separate variables correctly
2. [M1] Integrate to obtain p(1+y)e^(-3y) + [qe^(-3y)dy
3. [A1] Obtain (-1/3)(1+y)e^(-3y) + (1/9)e^(-3y)
4. [A1] Obtain (-1/3)(1+y)e^(-3y) - (1/9)e^(-3y) (+A)
5. [B1] Use correct double angle formula to obtain 1/(2cos²θ) dθ
6. [B1] Obtain ktanθ[+B]
7. [M1*] Use y=0, θ = (1/2)π to evaluate a constant of integration in an expression of the form aye^(-3y), Be^(-3y) and γtanθ only.
8. [DM1] Use y=1
9. [A1] Obtain tanθ = (17/9) - (14/9)e^(-3)