Zadanie 9, Paper 1, maj 2025, TZ1, Matematyka AA IB HL

With y = vx, $x\frac{dv}{dx}+v=\frac{2-v}{1+v}$, so $\frac{1+v}{2-2v-v^2}\,dv=\frac{dx}{x}$. Integrating gives $-\frac12\ln|2-2v-v^2|=\ln x+C$. Therefore x²∣2 − 2v − v²∣ = K. At (3, 0), v = 0 and K = 18. The expression inside the absolute value is positive at this point and cannot change sign on the connected solution, hence 2x² − 2xy − y² = 18. The branch through the initial point is $y=-x+\sqrt{3x^2-18}$, on $x>\sqrt6$; the other branch does not pass through (3, 0).

A1: correct transformed differential equation. M1: separate variables. A1: integral ln x on the right. M1: recognize the derivative of 2 − 2v − v² or use substitution. A1: logarithmic integral with factor $-\frac12$. M1: use the initial condition to determine a constant. A1: x²∣2 − 2v − v²∣ = 18 or an equivalent correct constant equation. A1: remove the absolute value with the correct sign and recover the stated implicit equation. No extra mark is attached to the explicit branch, which clarifies the domain.