AS 91579 · 2024 paper · expansion, constants, DE, mass

Question 2(c) Merit

Expand-then-integrate, a lower limit recovered from a definite integral, a separable equation in y², an area between a curve and a horizontal line, and a physical integral left entirely in symbols.

Consider the differential equation dy/dx = 12y²e^3x. Given that y = 0.5 when x = 0, find the value of y when x = 1/3.

The 2024 paper, page 7 ↗ Marking schedule, page 3 ↗

Work it out

Try it on paper first. Stuck? Take one hint at a time.

1 Separate the variables: y² goes to the left with dy.
int (1/y²) dy = int 12e^3x dx
2 Integrate both sides. Write 1/y² as y⁻² first.
−1/y = 4e^3x + c
3 Use y = 0.5 when x = 0.
−2 = 4 + c, so c = −6, and −1/y = 4e^3x − 6.
4 Now x = 1/3. What is e^3x then?
e¹ = e, so −1/y = 4e − 6 and y = −1/(4e − 6) ≈ −0.2052.

Answer y = −1/(4e − 6) ≈ −0.2052

What the marker wanted

Achieved Reaching −1/y = 4e^3x + c.
Merit The value of y, −0.2052, from correct integration. Any sensible rounding is accepted.

Strictly, the solution through (0, 0.5), y = 1/(6 − 4e^3x), shoots off to infinity at x = (1/3) ln 1.5 ≈ 0.135, before reaching x = 1/3. The formula still gives −0.2052 there, and that is the value the schedule marks.

In our words, from NZQA’s 2024 marking schedule.

Common mistake

Slipping with the indices when moving 12y² across. NZQA’s 2024 report, listing index errors among the Not Achieved answers, gives rewriting 1/12y² wrongly as its example. Keeping the 12 on the right, as here, leaves only y⁻² to integrate.

How did it go?

Got it moves it further out. Review brings it back tomorrow.