The variables x and y are related by the differential equation y² (d²y)/(dx²) + 2y (dy/dx)² - 5y³ = 8e⁻ˣ .

Mark scheme

Marking points from the official mark scheme (May/June 2011). The mark or mark type is shown in brackets.

1. [B1B1] Differentiates twice. \( v = y^3 \Rightarrow v' = 3y^2y' \Rightarrow v'' = 3y^2y'' + 6y(y')^2 \)
2. [M1] Substitutes. 1/3 v'' - 2/3 y(y')² + 2/3 y(y')² - 5v = 8e⁻ˣ OR v'' - 2y(y')² + 2y(y')² - 5v = 8e⁻ˣ
3. [A1] Obtains v-x equation. ⇒ (d²v)/(dx²) + 2dv/dx - 15v = 24e⁻ˣ (AG)
4. [M1] Solves AQE \( m^2 + 2m - 15 = 0 \Rightarrow m = -5, 3 \)
5. [A1] Finds CF. \( CF: Ae^{-5x} + Be^{3x} \)
6. [M1] Differentiates form for PI. \( PI: v = ke^{-x} \Rightarrow v' = -ke^{-x} \Rightarrow v'' = ke^{-x} \)
7. [M1] Substitutes. \( ke^{-x} - 2ke^{-x} - 15ke^{-x} = 24e^{-x} \)
8. [A1] Obtains PI ⇒ -16ke⁻ˣ = 24e⁻ˣ ⇒ k = -3/2
9. [A1] Obtains GS GS: v = Ae⁻⁵ˣ + Be³ˣ - 3/2 e⁻ˣ
10. [A1 checkmark ] Obtains y in terms of x (from a complete, if incorrect, GS). y = ( Ae⁻⁵ˣ + Be³ˣ - 3/2 e⁻ˣ )^(1/3)