Question 3(c) Merit

Opens with a numerical estimate from measured depths, then a log integral left in terms of a limit, a separable differential equation, an area that must be turned into a cubic, and a balance point found from a ratio of two integrals.

Consider the differential equation dy/dx = (√(4y + 1))/x². Given that y = 2 when x = 2/3, find the value(s) of y when x = 4/5.

The 2025 paper, page 11 ↗ Marking schedule, page 5 ↗

Work it out

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

1. 1 Separate the variables: everything with y goes with dy, everything with x with dx.

   int (4y + 1)^−1/2 dy = int x⁻² dx

2. 2 Integrate both sides. On the left, divide by the 4 inside the bracket as well.

   (1/2)(4y + 1)^1/2 = −1/x + c

3. 3 Use y = 2 when x = 2/3.

   (1/2)√9 = −3/2 + c, so 3/2 = −3/2 + c and c = 3.

4. 4 Now put x = 4/5 in and solve for y.

   (1/2)√(4y + 1) = −5/4 + 3 = 7/4
   √(4y + 1) = 7/2
   4y + 1 = 49/4, so y = 45/16

5. 5 The question says “value(s)”. Is there a second one?

   No. A square root is never negative, and 7/2 is positive, so squaring adds nothing: one value.

What the marker wanted

• Achieved Both integrations correct.
• Merit The value of y, 2.8125, with the integration shown.

In our words, from NZQA’s 2025 marking schedule.

How did it go?

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