IBDP Maths AHL 5.18 differential equations AA HL Paper 2- Exam Style Questions- New Syllabus
Question
Syllabus Topic Codes (IB Mathematics AA HL):
• SL 5.5: Introduction to integration as antidifferentiation; link between anti-derivatives, definite integrals and area
• SL 5.10: Indefinite integral of \( e^x \) and composites with the linear function \( ax + b \)
• AHL 5.18: First order differential equations; variables separable
▶️ Answer/Explanation
METHOD 1 (Using definite integral for net change)
The start of 2026 corresponds to \( t = 4 \) (since \( t=0 \) at start of 2022).
Net change in population from \( t=0 \) to \( t=4 \): \( \int_{0}^{4} -104000e^{-0.0145t} \, dt \)
Integrate: \( \left[ \frac{-104000}{-0.0145} e^{-0.0145t} \right]_{0}^{4} = \left[ \frac{104000}{0.0145} e^{-0.0145t} \right]_{0}^{4} \)
Let \( k = \frac{104000}{0.0145} \approx 7172413.79 \).
Then net change \( = k e^{-0.0145(4)} – k e^{0} = k(e^{-0.058} – 1) \approx -404165.81 \)
Initial population \( = 6.78 \times 10^6 = 6780000 \).
Population at start of 2026: \( 6780000 – 404165.81 = 6375834.19 \)
Rounded to three significant figures: \( \boxed{6.38 \times 10^{6}} \) (or \( 6380000 \))
METHOD 2 (Solving the differential equation)
Separate variables: \( dP = -104000 e^{-0.0145t} \, dt \).
Integrate: \( P = \int -104000 e^{-0.0145t} \, dt = \frac{-104000}{-0.0145} e^{-0.0145t} + C \)
\( P = 7172413.79 e^{-0.0145t} + C \)
Using \( t=0, P=6780000 \):
\( 6780000 = 7172413.79 + C \)
\( C = -392413.79 \)
So \( P(t) = 7172413.79 e^{-0.0145t} – 392413.79 \).
At \( t=4 \):
\( P(4) = 7172413.79 e^{-0.058} – 392413.79 \approx 6375834.19 \)
Rounded to three significant figures: \( \boxed{6.38 \times 10^{6}} \)